Solutions Manual

For

Engineering

Mechanics

Manoj Kumar HarbolaIIT Kanpur

1

Chapter 1

1.1 Rotational speed of the earth is very small (about radians per second). Its

effect on particle motion over small distances is therefore negligible. This will not be

true for intercontinental missiles.

1.2 The net force on the (belt+person) system is zero. This can be seen as follows. To

pull the rope up, the person also pushes the ground and therefore the belt on which he

is standing. This gives zero net force on the belt. For the person, the ground pushes

him up on the feet but the belt pulls him down when he pulls it, giving a zero net

force on him.

1.3 A vector between coordinates (x1, y1, z1) and (x2,y2,z2) is given by

. Thus (i), (ii) and (iv) are equal.

1.4 The vectors are (i) (ii) (iii) (iv)

1.5 (i) The resultant vectors are , and

(ii) The resultant vectors are , and

1.6

1.7 On each reflection, the sign of the vector component perpendicular to the reflecting

mirror changes.

1.8

2

A

BBA

B

A

AB

z

xO

vy

The fly is flying along the vector from (2.5, 2, 0) to (5, 4, 4). This vector is

. The unit vector in this direction is .

The velocity of the fly is therefore

.

1.9 After time t, the position vectors and of particles A and B, respectively, are

Their velocities and are given by differentiating these vectors with respect to

time to get

Velocity of A with respect to B is obtained by subtracting from

1.10 For rotation about the z-axis by an angle

It is given that . Therefore

3

1.11 Component of a vector along an axis is given by its projection on that axis. This is

obtained by taking the dot product of the vector with the unit vector along that axis.

Thus

Also

Substituting the expression for Ax, Ay and Az completes the proof.

1.12 (i) Dot product of two vectors and is

This gives the dot product of the first vector of problem 1.4 each of the other vectors

to be 4, 2 and 25.

(ii) Cross product between two vectors and is

Taking to be the fourth vector and to be the first, second and the third vector

gives the cross products to be

1.13 If the angle between two vectors is , the cosine of this angle is given by

. Thus

Between (i) and (ii)

Between (i) and (iii)

Between (i) and (iv)

Between (ii) and (iii)

Between (ii) and (iv)

4

Between (iii) and (iv)

1.14 Vector can be written as

Similarly

Taking the dot product between the two vectors and using the formula ,

where is the angle between the two vectors, we get the answer.

1.15 Magnitude

Similarly

Equating the two gives which implies that the two vectors are perpendicular

to each other.

1.16

This comes out to be equal to and

1.17 From the expression for it is clear that it is equal to the determinant

Interchange of two rows in a determinant changes the sign of the determinant. This

implies

5

thereby proving the equalities in problem 1.16.

1.18

Therefore

The x-component of

On the right hand side above, add and subtract to get

Do the same manipulation for the other components to get

1.19 ( i)

(ii) Body diagonal from the origin to the opposite corner is .

This vector is perpendicular to the vectors AB, AC and BC in the plane, because

its dot product with each one of them vanishes. This shows that the diagonal is

perpendicular to the plane.

6

Another way to see this is to find a vector perpendicular to the plane by taking

cross product of any of the two vectors from AB, AC or BC, and show that it is

parallel to the body diagonal. For example taking the cross product of AB and AC

gives

which is parallel to the body diagonal.

(iii) If the angle between OA and AB is then

This gives . Note that we have taken dot product with the vector BA rather

than AB because we wish to keep less than . In the same manner we get the

angle between OA and AC also to be .

1.20 The problem is to be solved exactly in tha same manner as done in example 1.3 by

replacing the position vector by the velocity vector and the velocity vector by the

acceleration.

1.21 Position vector

Velocity vector

Acceleration

1.22 (i) In reference frame 2, the components of vector at time t are given

in terms of its components in frame 1 by formula (1.10) as

Here are time-independent because vector is constant in frame 1.

Differentiating with respect to time, we get

7

(ii) From (i) it is clear that

1.23 Done in later chapters

1.24 Magnitude of a vector quantity is fixed. This means

Differentiating both sides with respect to time we get

Since is not zero, the equation above implies that and are

perpendicular to each other.

An everyday example is a particle moving in a circle. The magnitude of its position

vector is a constant and therefore its velocity, which is the time-derivative of its

position vector, is perpendicular to the position vector.

1.25

The area of triangle OAB is given as

This is maximum at

1.26 Let the angle between the z axis and the vector be . Then the component OB is

. Thus the magnitude of OB is . However, its direction is

perpendicular to the plane containing the z axis and the vector OA. To get the proper

8

direction we again take cross product of with . The component in the z

direction is given as . Thus the vector OA is . In general

can be replaced by .

9

Chapter 2

2.1 (i)

(ii) Since the element of length y is in equilibrium, we have

Using Taylor series expansion for T(y+y), which gives

And taking limit leads to the differential equation for T(y). The equation is

Solution of this equation is

where C is the integration constant. This is determined by using the fact that at the

lose end (y=L) of the rope, the tension is zero. This gives

and

10

yL

yT(y)

T(y+y)+

2.2 Torque

It is given that and the force has magnitude 50N and acts in the direction of

vector . Thus the force is 50 times the unit vector in the direction of the given

vector. This gives

With this the torque is

2.3 (a)

(b) The centre of the rod is at (3, 2)

The right end is at position and the force at this end is

The left end is at position and the force at this end is

Torque with respect to the origin =

100N

100N

11

Torque with respect to the centre of the rod =

Torque with respect to the left end of the rod =

Torque with respect to the right end of the rod =

(c) Torques about all the points are equal because the net force on the rod is zero.

2.4 (a)

(b) gives

(i)

Net torque about A is zero, which gives

(ii)

Equation (ii) gives

This substituted in equation (i) gives

30N0.5m

A 3m

1m

B

70N 120N

TA TB

12

2.5 Component of force in the plane perpendicular to the axis is at a distance of R

from the axis.

Therefore the torque about the axis is

2.6 Free-body diagram of the block

N1, N2 and W are three forces in a plane. Thus they must pass through one common

point for equilibrium. So the equilibrium conditions are only the force conditions.

gives

gives

Solution of these two equations is

N2

N1

W

13

2.7 Free-body diagram of the plank

Free-body diagram of the block

If the rod makes angle with the horizontal then

and

(a) To get the horizontal force F, we first calculate the normal force N on the rod. To do so,

we calculate the total torque about the hinged end of the plank and

equate it to zero. This gives

Now we balance the horizontal forces on the block to get

(b) Force balance on the plank

N

0.2m

Rx

100N

2m

1mRy

F

Nground

N

14

This gives

Minus sign in front implies that the direction is opposite to that shown in the free-body

diagram above.

2.8 Free-body diagram of the rod

Balancing the vertical forces gives N1 = W = 50N

Balancing the horizontal forces gives N2 = F

Balancing the torque about the centre of gravity gives

leading to

2.9 Free body diagram of the painting

W

N2

F

N1

15

Force balance equations give

and

N1 and N2 are equal because the component of torque perpendicular to the wall must vanish.

This gives

N1=N2=25N

Balancing the component of torque parallel to the wall taken about the lower end of the

painting gives

giving

2.10 We first calculate the forces at the ends of the rod. These forces are applied by the

supports. After finding the forces on the rod, we then calculate the forces and the

torques applied by the wall on the supports.

Free body diagram of the rod

N1+N2

T

W

Fx

W

N1 N2

16

Free body diagrams of the left and the right supports

The forces on the rod satisfy which gives

Taking torque about the left end and using gives

This gives

Now balancing vertical forces and the torque on the supports gives

For the left support F1=20N and

For the right support F2=15N and

2.11

35N

60cm

140cmN1 N2

1

F1

N1

5cm2

F2

N2

5cm

17

To find the force applied by the plastic block, we balance torque about the upper left corner.

This leads to

Balancing the vertical forces gives Ry = 40N

Balancing the horizontal forces gives Rx = 30N

Negative sign means that the direction of Rx is opposite to that assumed in the free-body diagram

above.

Free-body diagram of the pole

Ry

40cm

40N

60cm

Rx

N

18

Balancing the vertical forces on the pole gives N = 40N

There is no net horizontal force and the two horizontal forces give a couple = 300.4 = 12Nm

Balancing the torques on the pole about the ground gives = 300.4 = 12Nm

2.12 Free-body diagram of the table

To find Ny, we balance the torque on the table about its left hand edge to get

30N

40cm

30N

40N

N

j

i

90cm

Nx

NyRy

Rx

20N

19

By balancing the vertical forces, we get . The negative sign tells us that the force is

direction opposite to that shown above.

Free-body diagram of one of the rods

Free body diagram of the entire system

To get Nx, we balance the component of the torque coming out of the paper on the entire system

about the lower hinge. This gives

The negative sign again tells us that the direction of the force is opposite to that shown.

Balancing the horizontal component of the force on the table then gives

Sx

Sy

2Rx 30

2Ry

2Sy

Nx

Ny

30

90cm

20N2Sx

20

Note: The net force on each rod on its upper end is which is along

the rod as it must be for the equilibrium of a rod held at its ends.

Balancing the horizontal and vertical components of forces on each rod gives

Thus the net force on each rod is 10N compressive.

2.13 Free body diagrams of the two side portions and the portion AC over the pulley:

Tension TA and TC at both ends of the portion over the pulley is the same because the torque

about the centre must vanish. This gives

Free body diagrams of the portion AB and BC

TA

TATC

TCN

F

gLML1

gLML2

gL

RM

21

Notice that Torque of the normal reaction about the centre of the cylinder vanishes because for

each small portion of the rope over the cylinder, the normal reaction is radial. Thus T A (or TC)

and TB cannot be equal because they together provide a torque to balance the torque due to the

weight of the rope. Balancing the torque about the centre on AB gives

Thus if the net force by the cylinder on the rope is Neff at an angle from the horizontal then by

force balance

Note that Neff acts at a point different from the centre of BC because on different infinitesimal

portions it is different.

2.14 The support does not apply any torque about the x-axis. All other components and torques

are balanced by the support.

2.15 When forces are applied at two points of the rod, force balance demands that the force be

equal and opposite. However two such forces acting at two different points will give rise to

a couple moment. The couple moment is zero only if the forces point along the rod (see

figure below)

Neffx

TA

TB

gL

RM2

Neffy

Neffx

TC

TB

gL

RM2

Neffy

22

2.16 Let cables OA and OC make angle 1 and OB and OD angle 2 with the vertical. Then

balancing the vertical forces gives

The sine of the angles is easily calculated to be

This gives T=14050N

2.17 The torque direction is given by the direction of cross product , which is

perpendicular to . This implies there is no component of the torque in the direction of .

2.18 The net force on the plate is

Therefore the force that must be applied to the plate to keep it in equilibrium is

. Since there are only three forces acting on the body, they must all pass

through the same point so that their net torque is zero. This is shown in figure below.

Couple moment non-zero Couple moment zero

23

The force is at an angle from the line DC. Thus it

does not pass through O and intersects with side AD and diagonal BD of the square. Therefore:

(i) It is not possible to keep the square in equilibrium by applying the third force at O.

(ii) It is possible to keep the square in equilibrium by applying the third force at a point

on BD.

Equation of BD with O as origin is

Equation of line along which the third force acts is

Solving the two equations gives

This gives the distance of point O = 0.125a

And from B =

(iii) It is clear that for equilibrium, the force can be applied only on AD and BC.

D C

BA

O

24

Chapter 3

3.1 For the three trusses shown, m = 21, j = 12. Thus they all satisfy 2j-3 = m. Thus they are

all simple trusses.

3.2 Showing that pin E is in equilibrium

There are five forces acting on E of which two (FE and ED) are horizontal, two (CE and

the external load) are vertical and one (BE) is at an angle. We wish to check if the

horizontal and vertical forces add up to zero. It is solved in 3.1 that

Since all the forces are tensile, they all pull the pin. In addition there is the external load

of 5000N vertically down.

The net horizontal force is

Similarly the net vertical force is

3.3 (i) The truss has 4 members and 4 joints. Number of force balance equations therefore is

8 (2number of joints). On the other hand, number of forces available is only 7

(3+number of members), which implies that the truss will not be stable and will collapse.

In terms of stability condition which implies that the truss will collapse.

(ii) If we add one more member to the truss, i.e make m = 5, then is satisfied

and the truss becomes stable and a simple truss. Let us add a member across AC.

To find forces in each member we start by first finding the forces applied by the

external supports. The free-body diagram of the truss is as follows:

25

The direction of the forces applied by the external supports has been anticipated as

shown. To find ND, we balance torque about A to get

Now balancing the vertical and horizontal forces on the truss gives

NAx = 0 and NAy = 1875N

The negative sign again tells us that the direction of the force is opposite to that

shown.

We begin to apply the method of joints from point D since at this point there are two

unknown forces FAD and FCD. Assuming these forces to be tensile gives the free body

diagram of joint D as follows

Balancing the vertical forces on point D gives

CB

5000NNAx

NAy

ND

A D

60FAD

FCD

6875N

26

The negative sign shows that the force FCD is compressive and not tensile as assumed.

Balancing the horizontal forces on point D gives

The negative sign again shows that the force FAD is compressive and not tensile as assumed.

Next we go to point A and balance the forces there. The free body diagram of point A is

In drawing the figure above, we have shown the direction of FAD according to it being a

compressive force.

The length of rod AC is = m

So

Now balancing the horizontal forces at A gives

and balancing the vertical forces at A gives

Solving these two equations gives FAB = 0 and FAC = 4390N

Since the sign of FAC is positive it is in the same direction as assumed and therefore tensile.

Now we can easily see that force FBC will be zero because point B is under equilibrium under

only two forces FAB and FBC and FAB has already been determined to be zero. Thus FBC = 0.

Thus all the forces are now determined. They are

FAB

60

3969N

1875N

FAC

27

FAB = 0 FAC = 4390N (tensile) and FBC = 0.

Finally to check our answer we make the forces at point C and see if they all balance. The

free body diagram of point C is

Balancing the horizontal forces at C gives

Balancing the vertical forces at C gives

This indicates that our answers are correct.

Note: We see that FAB and FBC both vanish. This implies that members AB and BC may

not bee needed for the truss. This is true because with just three members AD, AC and

CD (m=3) there are only three joints (j=3) and the truss satisfies the condition

for it to be a stable structure.

3.4 Free body diagram of the truss

7939N

60

5000N4390N

28

Since point C is in equilibrium under one known and two unknown forces, both unknown

forces can be determined easily. The forces on C look as follows

Balancing the vertical forces at C gives

With this implies FAC = 54N (compressive)

Balancing the horizontal forces at C gives

The only force left is at AB. We calculate this by balancing forces acting on pin A, which

look as follows.

Ry

B 30cm

20cm30N

A

C

N

Rx

FBC

30N

FAC

29

This gives

Additionally we can also solve for the normal reaction N and the forces Rx and Ry. These

are

N = 54N, Rx = 45N, and Ry = 30N

3.5 Rod AB provides a vertical force to hold pin A. However if it is removed and the vertical

force is provided by a fixed pin joint, the structure will remain stable because we need

3j=6 forces for equilibrium of 3 joints; two of these are provided by the fixed supports

and two by the two members. The forces in the members remain the same. So do the

forces by the two support except that the fixed point at A also provides a vertical fore of

30N that was earlier provided by member AB.

3.6 Free-body diagram of the truss

FAC

FAB

N

30

Since pin D has only two unknown forces acting on it, we can start our calculations from this

point onwards. The forces on D are

It is immediately clear that

FAD = 0 and FCD = 100 N (tensile)

Next we go to pin C and balance the forces there. The forces acting on C are

C

N

Rx

Ry

100N

A

B

D

FCD

100N

FAD

31

Balancing the vertical forces at C gives

(compressive)

Balancing the horizontal forces at C then gives

(tensile)

Next we go to pin A. The forces there are as follows

Balancing the vertical forces at A gives

(tensile)

Balancing the horizontal forces at A gives

N=100N

Finally balancing forces at pin B will give the external forces

Rx = 100N and Ry = 100N

3.7 Free body diagram of the truss is as follows

FAC

100N

FBC

N

FAB

N2100

32

(i) There are 4 reaction forces at the supporting pins at A and B. In addition the forces

generated by the members of the truss equal 6. This makes the total number of forces

available = 10. The number of joints in the truss is 5 that require exactly 10 number

of forces for equilibrium. Thus the truss is a stable one.

(ii) It is also statically determinate since the number of forces available is equal to the

number of equations to be satisfied for equilibrium.

(iii) First we find Nx by balancing the torque about point B. This gives

We now begin by balancing the forces at point D

Balancing the vertical forces at D gives

(compressive)

DC

A

B

500N

Rx

Ry

Nx

Ny

E

FED

500N

FCD

33

Balancing the horizontal forces at D then gives

(tensile)

Nest we go to pin E because it has two unknown forces acting on it. The forces are as

follows

Balancing forces at E gives

(compressive)

(tensile)

Next we go to pin at A. The forces there are

Balancing the forces gives

(compressive)

and

(tensile)

Finally we go to point C where only one force FBC is unknown. The forces on C are

FAE

N

FCE

FAC

Ny

500N1000N

34

Balancing the horizontal forces at C gives

FBC=1000N

As a final check, the value for FBC gives the horizontal force Rx by pin B on the truss to be

1000N and vertical force Ry to be zero. This is consistent with the overall equilibrium of the

truss when it is treated as a system by itself.

3.8 Since each member of the truss weighs 50N, at each pin we take the load by each pin at

that point to be 25N. The free body diagram of the truss is as follows; here each small

arrow pointing down indicates the weight of the truss member, acting at its centre.

We firs find NE. To do this we balance the torque about B. This gives

FAC

FBC

500N

500N

CB

1000NA

DE

NE

NBxNBy

35

This immediately gives, by balancing forces on the entire truss

NBx = 0 and NBy = 925N

The negative sign showing that the force I opposite to the direction assumed in the figure

above.

We begin at pin D as there are two unknown forces there. The force diagram on pin D is as

follows (there are two members meeting at pin D that give a load of 225=50N there)

Balancing the forces gives

(tensile)

and

(compressive)

Next we go to pin E. The forces acting there are (including 325=75N from 3 members)

Balancing the forces gives

(compressive)

FCD

1050N

FDE

FAE

2275N

FCE +75N

1050N

36

and

(compressive)

Next we go to pin A. The forces acting there are (including 325=75N from 3 members)

Balancing the forces gives

(tensile)

and

(compressive)

Net we move to point B where the forces are (including 225N=50N from two members)

This immediately gives

FAB = 975N (compressive) and FBC = 0

As a final check, one balances all the forces at C and sees that they all balance properly. This

implies that the answers obtained by us are correct.

FAB

1050N

FAC

75N

FAB

FBC

925N

50N

37

3.9 Free body diagram of the truss with the weight of each member included. The free body

diagram is then as follows.

By balancing the torque about A we get

Balancing the vertical and horizontal forces on the truss, this gives

and

For calculating force in members, we take the weight of each member shared equally at each

joint. The forces on A are (including the weight of two members)

Balancing the forces at point A

Next we go to point F. The forces at point F are (including the weight of three members)

F E D

CB

5000N

NA ND

Nx

A

N3

11750

FAF

FAB500N

38

Balancing the forces here gives

Next we go to point B since now there are only two unknown forces there. At point B the forces

look as follows (including the weight of four members)

Balancing the forces

Negative sign above means that the direction of the force is opposite to the one assumed. So

FBC = 5083N (compressive)

N3

10250

FBF

FFE

750N

750N

FBC

4832N

FBE1000N

39

We next consider point C and balance the forces there. The forces at point C are (including

750N form the weight of three members)

Balancing the forces

Next we go to pin D where the normal reaction is N and balance the forces there. The

force diagram there is (including 500N form the weight of two members)

It is easily seen that the vertical forces balance at this point. This points to the correctness of our

calculations so far. Balancing horizontal forces then gives

FED = 5083 N (tensile)

750N

FCE

5083NFCD

500N

7188N

FED

N3

16750

40

As a final check we should check whether all the calculated forces balance at pin E. The forces

at pin E are (including the weight of four members)

All these forces balance as can be seen by calculating the net x and y components of the

forces. Thus our calculations are correct.

3.10 Free body diagram of the truss

Taking torque about B we get

Balancing the horizontal and vertical forces now gives

NBx = 1000 N and NBy = N

Negative sign above means that the direction of the force is opposite to the one assumed.

1000N

2357N

N

4333N

5083N

5000N

E

D

C

B

A

1000N

2000N

NBx

NBy

NE

41

We begin at pin D. The forces there are

In the diagram above

Balancing the vertical forces gives

(compressive)

Balancing the horizontal forces gives

2789N

Next we go to pin E. The forces there are

Balancing the vertical forces gives

(compressive)

Balancing the horizontal forces gives

N (compressive)

1000N

FED

FCD

2000N

FAE

FCE

N

2683N

42

Next we go to pin A. The forces there are

Balancing the vertical forces gives FAC = FAB.

Balancing the horizontal forces gives

This also means that

FAC = 895N (tensile)

Now we go to pin C. The forces there are

The vertical forces are already balanced here. Balancing the horizontal forces gives

To check our answers, we finally balance the forces at pin B and see if they all balance there.

At pin B the forces are as follows

FAC

FAB

1192N

895N

FBC

895N

2789N

43

It is easily seen that all the forces above balance. So our answers are all consistent.

3.11 The weight of the road = volume of the roaddensityg

= 128.3200010

= 576000N

This weight is divided equally between the two trusses on the sides. Thus only

288000N is supported by each truss.

Weight of the members of the truss = 135000

= 65000N

Total weight supported by each truss therefore is = 353000N

Free body diagram of the truss is

From the balance of forces, is clear that

895N

1000N

N

1596N

NAxH G F E

DCB

A

353000N

NAy NE

44

Let us now consider forces at each pin one by one. Each pin has the following forces

acting on it: The weight of the road divided over 5 pins, which is ;

the weight of the members at that pin; and the forces applied by the members.

Let us now balance forces at point E. The forces on pin E are (including the weight of

the members)

In the figure above

Balancing the vertical forces in the figure above gives

(compressive)

Balancing horizontal forces then leads to

(tensile)

Next we consider pin D. The forces on pin D are (including the weight of the

members)

57600N

FEF

176500N

FDE5000N

45

Balancing horizontal forces then leads to

(compressive)

Balancing the vertical forces gives

(tensile)

Next we look at pin F. The forces there are (including the weight of the members)

Balancing the vertical forces in the figure above gives

(compressive)

Balancing horizontal forces then leads to

(tensile)

By symmetry of the problem, forces on the members to the left of member CG will be

exactly the same as on the corresponding members to its right. The only force that we

now have to calculate is on member CG. For this we consider point G. Two horizontal

FCD

142375N

FDF

7500N

57600N

FGF

106400N

FCF10000N

85425N

46

forces at G are by HG and GF and are equal to 114525N each. The forces at point G are

then given as

This gives FCG = 65100N

Finally we check our answer at point C. The forces there (including the weight of 5

members meeting there) are

As is easily seen, the forces at C balance and therefore our calculations have been

consistent throughout.

3.12 Free body diagram of the truss is as follows

57600N

FCG

7500N

114525N114525N

65100N

48500N 48500N

FCF12500N

85425N85425N

47

Here the distances and the angles are

AH = HG = GF = FE = 2.5m BH = DF = 1.25m (similarity of triangles)

Balancing torque about point A gives

Thus NA = 2000N and RA = 0

Now that the external reactions have been determined, we can go about calculating the forces

in the members. We start with pin E because there are only two unknown forces there. The

forces at E are

Balancing the vertical forces in the figure above gives

(compressive)

Balancing the horizontal forces gives

N (tensile)

RA

NA

1000N

E

D

C

B

A FGH

2000N

NE

FEF E

1000N

FDE

48

Nest we go to pin D. The forces there are

Balancing the forces gives

(compressive)

FDF = 0

Next we go to point F where the forces are as shown below.

Balancing the forces gives

FGF = 2000N (tensile) FCF = 0

Pin G is considered next. Since the forces there are only vertical and horizontal, even

without making the forces there, we immediately can write

FCG = 0 and FGH = 2000N (tensile)

Next we consider point A where two members AB and AH meet and therefore there are two

unknown forces. The forces there are

D

2236NFDF

FCD

2000N

FCF

FFGF

0N

49

Balancing the vertical forces gives

(compressive)

Balancing horizontal forces then gives

(tensile)

Next we go to point B. Here there are four forces acting and each pair (FBH and 2000N; and

FBC and 4472N) has two forces in opposite directions. Thus without solving the detailed

force balance equations, we can directly write

FBH = 2000N (compressive) and FBC = 4472N (compressive)

Next we go to point H and balance the forces there. The forces there are as follows

By balancing the forces at H, it becomes clear that (tensile)

Finally the answers are checked at C. The forces at C are as shown below

FAB

FAHA

2000N

45

FCH

4000N2000N

H

2000N

50

As is easily seen, the horizontal a n vertical forces all balance at C. Thus our answers are all

correct.

To calculate the forces by method of sections, we make a cut through the truss so that it

passes the concerned members. In the present case we take the following section of the truss

and show various forces on that section.

In the figure above, FCH is determined easily by taking torque about A since the torque due to

FCB and FGH both vanish about A. This gives

(tensile)

To find FCB, we balance the vertical component of the forces to get

Negative sign here means that the force is opposite to the direction assumed and therefore is

compressive in nature.

Finally, balancing the horizontal forces leads to

1000N

D

2236N4472N

N22000

2000N

FCB

B

A H

2000N

FCH

FGH

51

(tensile)

3.13 The free-body diagram of the truss on one side is as follows (Notice that the weight of the

truck is equally divided between the two trusses)

We first calculate NE by balancing torque about A. This gives

This gives NA = 12.5 kN and RA = 0

To find forces in members CD and DG, we make a cut through CD, DG and GF. This looks

like the following

RA

NA

NE

50kN

H G F

E

DCB

A

FGF

37.5 kN

50kN

FGD

F

E

DFCD

52

To find FGF, we balance torque about point D about which the torques due to FCD and FGD

vanish. This gives

(tensile)

To obtain FCD, we take torque about point where FDG and FGF intersect, which is point G.

This leads to

(compressive)

Now we balance the horizontal and vertical forces on the truss. Balancing horizontal forces

gives

Negative sign here means that the force is opposite to the direction assumed and therefore is

compressive in nature.

To find the forces in the members BC and BG, we make a cut through the members BC, BG

and HG as follows and then calculate the forces.

To obtain FBC, we take torque about point where FGH and FBG intersect, which is point G.

This leads to

(compressive)

Next we find FGH by taking torque about B. We get

(tensile)

Finally we get FBG by balancing vertical and horizontal forces. Horizontal force balance

gives

(tensile)

12.5 kN

HFGH

FBCB

A

FBG

53

Finally to find FCG, we make the following cut through the truss

Since the vertical forces all balance, this implies FCG = 0.

Further, the horizontal forces are also balanced.

12.5kN37.5kN

50kN

H G F

E

DFCGB

A

54

Chapter 4

4.1

4.2 Since the block is in equilibrium under three forces, the three forces must pass through

the same point. Thus the normal reaction will be at the point where the arrow showing

the weight meets the inclined plane. This is shown below.

4.3 The free body diagram of the block is as follows

Applied force

Frictional force

sN

Fmax

f

mg

θ

N

55

Balancing the horizontal forces gives

Balancing the vertical forces gives

Since the maximum frictional force , for equilibrium we should have

4.4 We consider two different situations when the weight on the table is about to move to the

left or to the right. When it is about to move to the left, its free body diagram will look

as follows

By equilibrium conditions, we have

N = 50g

f + mg = 10g

f

N

mg

F

10g

N

50g

f

mg

56

Since

We have

Thus the minimum value of m is 5kg when the frictional force is at its maximum pointing to the

right. As m is increased above 5kg, frictional force becomes less and less, eventually changing

direction and attaining its maximum value pointing left. In that situation, the free body diagram

of the block on the table is as follows.

In this situation, the equation for horizontal force balance is

f + 10g = mg

This coupled with leads to

Thus

4.5 Taking the x axis along the plane and the y axis perpendicular to the plane, the free-body

diagram of the block looks as follows.

50g

f

10g

N

mg

57

The equations describing equilibrium in the X and the Y directions are

The first equation implies that

Taking g = 9.8ms2, the value of F for different values of F is

4.6 Free body diagram of the box when it is about to move (i.e. the frictional force is at its

maximum) is shown below

N

F

F

30º

30º

XY

100g

N

a

bh

F N

mg

58

When the box is about to move, the friction is at its maximum and is equal to N. The force F

also equals N at this point. This creates a couple that is counterbalanced by the couple formed

by the weight of the box mg and N (=mg). This is the reason that N shifts towards the direction

of the push. However, the maximum couple moment that can be created by mg and N is .

Thus for the box not to topple, the couple created by F and the friction should remain less than

. Thus implies

4.7 suppose each break show makes an angle at the centre as shown below

The force F is assumed distributed uniformly over the shoe. Then the torque due to the

frictional force will be

With two shoes therefore, the torque would be

4.8 It is given that mass M is balanced by mass m. The contact angle is π. Since each time the

string is wound once more around the rod, the mass M that can be balanced by m becomes

twice as large, we have

This gives = 0.11

59

4.9 Neglecting the length of the rope passing over the pulley, we have mass on one

side of the pulley that is balanced by mass on the other side. Thus we have

4.10 As the weight is put, it has a tendency to move down. Hence

the frictional force will be in the counterclockwise direction. Thus if the tension in the rope

on the spring balance side is T1 and that on the weight side is T2 then

Now it is given that T1 = 5g and T2 = mg. Thus we get

An interesting possibility exists if a person had pulled the weight down and then slowly

brought it to equilibrium. In that case the tension will work in the other direction and

However we have not considered this possibility.

4.11 There is a range of M2 that exists because frictional force can act with its maximum value

in one direction to the maximum in the other direction. Largest value of M2 is when the

mass M1 is about to slide up the plane. The free body diagram of M1 in that case is as

follows

f

N

M1g

T

60

When the mass M1 is about to slide up, we have

The contact angle between the rope and the pulley is

Since the rope has a tendency to move clockwise, the frictional force due to the pulley will be

acting counterclockwise. Thus we have

The other extreme is when the mass M1 is about to slide down the plane. In that case the free

body diagram of M1 is

Thus we have

Now the rope has a tendency to move counterclockwise, the frictional force due to the pulley

will be acting clockwise. Thus we have

4.12 Free body diagram of the tire when it is loaded and is about to roll is as follows

f

N

M1g

T

61

Balancing the vertical forces gives

Balancing the horizontal forces gives F = f

Balancing torque about the centre of the wheel gives

NW

F

f

62

Chapter 5

5.1 Consider a composite surface of total area A made up of N different surfaces. Then the

coordinates of the centroid satisfy

If the area of each surface is then

Now in the definition of the centroid, the integrals can be performed separately over each surface

so that we can write

This immediately gives

5.2 By symmetry it is clear that XC = 2. We are nevertheless going to prove it below. We first

calculate the area of the surface. It is

Substituting we get

To calculate Xc, we take vertical strips of width dx on the

surface at distance x from the origin and then calculate

63

O X

Y

x

Thus

Similarly to calculate YC, we take horizontal strips of width dy at height y and calculate

At height y, the strip extends from x1 to x2. These

points are given by the equation

Therefore

We thus have

Substituting so that , we get

This gives

Thus the centroid is at

5.3 One curve (call it curve 1) in this problem is the same as that in the problem

above. The other curve (curve 2) is

64

x1 x2OX

Y

y

The y-axis of curve 2 is thus 4 times curve 1. The area of curve 2 is therefore . Similarly

the x coordinate the centroid of curve will remain at 2 but the y coordinate will be .

Thus we have

The area A for which we wish to obtain the centroid is obtained by removing surface

formed by curve 1 from the surface formed by curve 2. We thus have

5.4 Trapezoidal loading is shown in the figure below

The total force on the beam will be equal to the area under the curve. Thus the total force is

equal to

This load will be acting at the centroid of the area. Thus it acts at

X

f(x)

X1 X2

w2

w1

65

Adding and subtracting and in the numerator we get

5.5 From figure 5.14, for a plate of width w

Similarly, from figure 5.15

66

This gives from formula 5.9

which is equivalent to the depth given by formula (5.17)

5.6 Loading on the tank door is triangular as shown below

The average pressure is the pressure of water at the centroid of the submerged part. Thus the

average pressure will be

Thus the total force due to the water pressure is

This force acts at the centroid of the loading that is triangular in this case. Thus it is at a distance

below point A.

We now apply equilibrium conditions to the door. This leads to

0.25m

NB

NA

N1

153.125N

19.6N

0.5m

67

5.7 The rectangular surface looks as follows

To find Ixx, we take a strip (see figure above) of width dy parallel to the x-axis and calculate

Similarly to find Iyy, we take a strip (see figure above) of width dx parallel to the x-axis and

calculate

To find Ixy, we take a small square (see figure above) of size dxdy parallel and calculate

by symmetry of the inegrand.

b

a

X

Y

68

5.8

From the figure

From the formula for transformation of area moments (taking X and Y axis as in the problem

above) we get

Similary

b

Y’

X’

a

O

69

and

5.9

To calculate IXX we take a horizontal strip of width dy at y (see figure) for dA and calculate

dyybbaydAyI

b

bXX

2222 2

Taking , we get

Similarly for IYY, we take a vertical strip at x for dA and calculate

Taking , we get

b

a

X

Y

70

And by symmetry

We now calculate the moments and product of inertia about a set of axes rotated by an angle

with respect to the original one and with the same origin. Thus

Similarly

Product of inertia is calculated using the formula

5.10

71

To calculate IXX we take a horizontal of width dy strip at y (see figure) for dA and calculate

To evaluate the integral, we substitute so the integral is transformed to

Now substituting , we get

Similarly for IYY, we take a vertical strip of width dx at x for dA (see figure) and calculate

X

Y

O x1 x2

y

X

Y

O

x

72

Taking we get

Now , and . This gives

5.11 Product of area about the origin O is given as

If the centroid is at O’ which has the coordinates and the coordinates of a point with

respect to O’ are then

Therefore

O

(x0,y0)

X

Y Y’

X’

73

However, by definition of the centroid

Thus

5.12 Consider the moments and product of inertia of a square about a set of axes parallel to its

sides and passing through its centre.

For this set of axes

Now by the formula

Ix’y’ will always remain zero because of the equality of Ixx and Iyy irrespective of the angle of

rotation of the new set of axes x’y’. Thus any set of axes passing through the centre is the

principal set of axes.

5.13 The formulae for the moments of inertia in rotated frames are

a

Y

X

74

Taking the second derivative of these expressions with respect to , we get

Thus the two derivatives have opposite signs. This implies if one of them is a maximum, the

other one will be a minimum.

75

Chapter 6

6.1 (i) (a) and (d) are the virtual displacements because these are the only ones consistent with

the constraint that the block can move only in the vertical direction.

(ii) Suppose the strech is y0. In that situation, the forces on the block are as shown

Now a virtual displacement gives a displacement of in the vertical direction. Taking it in

the vertically up direction gives the virtual work to be

Equating this to zero gives

6.2 Figure below shows the students and the plank on a wedge and a possible virtual

displacement of the system.

It is clear that as long as the point on the wedge does not move, the only possible displacement is

the rotation of the plank about this point and the system has only one degree of freedom. For the

ky0

mg

x (3-x)

40g50g

30g

76

virtual displacement shown, the displacement and the virtual work done by various forces is as

follows:

Since the net virtual work must vanish for equilibrium, we have

6.3 (i) Since the number of parameters required to describe the system is 1, the number of

degrees of freedom is 1. We choose it to be the angle , the rod makes from the vertical. A

virtual displacement will be to change by .

(ii) To apply the principle of virtual work, we need to calculate the virtual work done by various

force when is changed to +. For this we first write the coordinates (x, y) of the tip of the

rod and yCG of the centre of gravity of the rod with respect to the pivot point. These are

As is changed to +, these coordinates change and the changes are given by

Thus the total virtual work done by the external forces – 2g at the CG, 20g at the tip, both in the

positive y direction, and T at the tip in the positive x direction – is

20kg

T2kg

1.5m

77

Equating this to zero gives

6.4 To find the forces applied by the bricks, we treat these forces as external. For only vertical

motion, there are two degrees of freedom. One is the vertical displacement of the centre of

gravity and the other the rotation of the plank about the CG. Equivalently, we can take the

vertical displacements of the ends A and B as the virtual displacements. We choose the

second option because this is related directly to the forces applied by the bricks. The plank

in equilibrium and virtually displaced is shown below

Now the vertical virtual displacement of different points is

Point A :

Point B :

Centre of gravity of the plank :

Athlete :

Thus the total virtual work done is

500N

NA

2m100N NB

y2

y11m

1.5m

78

Equating the virtual work to zero and therefore the coefficients of each independent

displacement (y1 and y2) to zero gives

NA=175N and NB = 425N

6.5 (i) Constraints on the system: length of the strings holding the masses is fixed

(ii) There is only one degree of freedom. This can be understood as follows. There are

three variables needed to describe the system: The distance of two masses and one

pulley from the ground. However there are two constraints: To of the strings have

fixed lengths. Thus only one variable is left to change freely.

(iii) In terms of the lengths shown in the figure

The constraint that the longer string has fixed length is expressed as

The constraint that the shorter string has fixed length is expressed as

(iv) The constraint forces are the tension in the two strings. It is by these tensions that

the constraints are maintained.

(v) To apply the method of virtual work, we make a virtual displacement of mass 1.

The corresponding displacement of mass 2 is then . However, since there is only

y2

M2

M1h1

y1

h2

79

1 degree of freedom, we will finally express in terms of to apply the method

of virtual work. The virtual work done in these processes is

Notice that the two virtual displacements are not independent. Therefore we should not

conclude that M1=M2=0 for equilibrium. To apply the method, we first express the virtual

work in terms of only .

From the first constraint equation, since h1 if constant,

From the second constraint we have

Substituting these in the expression for the virtual work gives

Equating this to zero then gives, for equilibrium

6.6 (i) Since the two blocks are free to move in one dimension without any constraint, the

number of degrees of freedom for the system is 2.

(ii) The system in equilibrium is shown below. At equilibrium spring on the left is

stretched by x1 and the total stretch of the two springs together is by x2. Thus spring

on the right is stretched by (x2x1). Thus force on mass m1 is to the left and

to the right. These two forces must be equal for equilibrium but we wish

to obtain this by applying the principle of virtual work. Similarly the two force

acting on m2 are F and to the left. Again these two forces must be equal

for equilibrium and we will obtain this by applying the principle of virtual work.

80

As the mass m2 is moved by a virtual displacement , let us assume that mass m1 moves

by . Then the virtual work done will be

Equating this to zero and therefore the coefficients of and to zero gives

and

This gives and .

6.7 (i) There is only one degree of freedom. Although the piston moves in the vertical direction

and the wedge in the horizontal direction, their movements are connected because the piston

moves on the surface of the wedge.

(ii) Normal reactions on all the surfaces are the constraint forces. In the present context, the

constraint force specific to the piston’s movement on the surface of the wedge is the normal

reaction of the wedge surface on the piston.

(iii) if the distance of the middle line of the piston is at a distance a from the origin, its

height is y and the distance of the left edge of the wedge is x (see figure) then the

constraint is expressed as

F

k1 k2m1 m2

x1 x2

81

Now the method of virtual work is applied by considering x as the free variable, varying it by x,

and equating the total virtual work to zero. The virtual work is

Equating the coefficient of x to zero gives

6.8 When the equilibrium angle is , the distances of various points (see figure) , taking A as the

origin are as follows

The force on the two points B and D due to the spring is

to the left on B and to the right on D

F

m

a

yx

82

It is clear that we need only one parameter to specify the system. Thus there is only one

degree of freedom. Now let us make a virtual displacement by changing by . In that

situation

Thus the total virtual work done as is changed by is

Vanishing of the virtual work then implies

Or equivalently

If then the solution is . For very small value of therefore we have ,

where x is very small. Substituting in the equation above, we get

W

A

B D

C

x

y

83

Since is small, we get by Taylor series expansion

Thus the equation for equilibrium is

Or to a good approximation

Thus

6.9 (i) Even if we consider only one dimensional motion, we would require two variables, one

the angle that the bar makes with the vertical and the other describing the position of the

mass. However, the two are connected by a rope. Hence the two variables cannot vary

independently since the length of the rope is foxed (constraint). Therefore there is only one

degree of freedom in the system.

(ii) The constraint is the length of the rope remaining constant. It is enforced by the

tension that develops in the rope. This tension makes the movement of the bar and

the mass restricted. Thus it is the tension in the rope that is the force of constraint.

(iii) A virtual displacement would be to displace the bar by an angle from its

equilibrium position. Since the length of the rope is fixed, the midpoint of the bar

moves by the same distance as the mass connected to the rope. Therefore the tension

does positive work at one end of the rope and exactly equal but negative work at the

other end. The sum of the work done by the tension then vanishes.

(iv) Let us say we make a virtual displacement of the bar by turning it counterclockwise

by an angle from the vertical. Then the end of the rod moves by l in the

84

direction of the force. At the same time, the virtual displacement of the mass is

equal to the movement of the midpoint of the bar. Thus the mass moves by

opposite to the gravitational force (see figure)

The net virtual wok therefore is

Equating this to zero then gives

6.10 Initially the bars are at 90. The weight has a tendency to fall down so the torque applied is

such that it tends to pull the weight up. Thus is a virtual displacement of angle is made

in the direction of the torque, the weight will be lifted up by the corresponding distance y.

The corresponding virtual work done by the torque is and the corresponding work by

the gravity is Wy. By the method of virtual work then we have

This gives

m

F

85

Next we calculate the relationship betweeny and . As the rod is rotated by angle , the

length of the horizontal diagonal decreases by . If the corresponding angle at the

vertical corner changes from 90 to 90+α, then we have (see figure)

Change in the length of the horizontal diagonal

=

This should equal so we have

As the angle changes the length of the vertical diagonal changes by

Thus we have

This gives

6.11 (a) For the motion in a plane, the orientation of the rod can be described by the

displacement of its centre of mass and the angle it has rotated by about its CM. Or

W

(90+α)

86

equivalently the displacement of its two ends is sufficient to describe its orientation. Thus

the degrees of freedom is 2.

(b) We take the vertically down displacement of the two ends as the virtual displacement as

shown in the figure.

Let the centre of mass be at a distance from the left hand end of the rod. In that case,

the centre of mass moves down by . The virtual work

done by the springs in such a virtual displacement is and respectively while

that by the weight of the rod is . Thus equating the net virtual work to

zero gives

Now equating the coefficient of each independent displacement to zero gives

y2

y1

yCM

87

Chapter 7

7.1 For Cartesian coordinates (x,y), the planar polar coordinates are and

. If y<0 and x<0, 270>>180 and if y<0 and x>0, >270.

Unit vectors are given by

and

7.2 At each point the velocity is given as . Thus for (iv) for a

particle at (-2,-3), the velocity is

7.3 The two points with polar coordinates and and the corresponding

vectors are shown in the figure below

(i) As is clear from the figure, the angle between the vectors and is . This also the

angle between unit vectors and and unit vectors and . Therefore

2

1

(21)

88

Similarly, angle between and is and that between and is .

Therefore

(ii) Similarly, from the figure

(iii)

7.4 The trajectory of the projectile is shown schematically in the figure below. The

horizontal distance from the origin to the point of highest elevation (height 1.25m) is

. The vertical component of the velocity at this point vanishes while the

horizontal component is . Similarly, the vertical component of the

acceleration is 10ms-2 vertically down. Also shown in the figure are unit vectors in the

radial and the tangential directions at the highest point and on the ground.

From the figure it is clear that for the point of highest elevation

Therefore the radial and tangential components of the velocity at the highest point are

Similarly the radial and tangential components of the acceleration are

1.25m

89

On the ground, the radial unit vector points towards positive x direction and the

tangential unit vector is in the negative y direction. Thus the radial component of the

velocity is its x component and the tangential component is its y component. Therefore

Similarly, since the gravitational acceleration on the ground is in the negative y direction,

its radial and tangential components are

7.5 The position of the particle at time t is shown in the figure below.

The polar coordinates of the particle at time t are given as

Therefore

This gives the velocity in polar coordinates as

Differentiating the velocity vector gives

2t

1

2ms-1

90

Now substituting and , we get the acceleration above to be

zero.

7.6 (i) Kepler’s second law states that the rate of the area swept by the radius vector is

constant. This can be expressed as (see figure for the symbols used)

constant

Now differentiating the equation above with respect to time gives

The expression on the right is the tangential acceleration. Thus Kepler’s second law gives

tangential acceleration to be zero.

(ii) Since the tangential acceleration is zero, the force is only in the radial direction.

Acceleration in the radial direction is

We differentiate the orbit equation with respect to time to get

Sun

r

91

Since is constant, differentiating the equation above once more with respect to time, we get

This gives

This shows that the force is proportional to r2.

7.7 It is given that

which gives

Now write and differentiate it twice with respect to t to get

This gives

Substitute this in to get

Multiplying and rearranging terms gives

The equation does not appear to be easily ingrable.

7.8 It is given that and

92

(i) Since , integrating the equation for gives

or equivalently

(ii)

If the velocity vector is at 45 to the radius vector, we have

This gives

Squaring both sides gives

OR

This is equivalent to

(iii) Radial acceleration zero implies

which leads to

giving

This also gives the distance from the origin to be

7.9 Free body diagram of the bead when its radius vector is making an angle (increasing as

the particle slides down) from the vertical is shown below

93

Taking the components of the forces in the radial and tangential directions and equating these

to mass times the radial and tangential components of the acceleration, respectively, gives

In the radial direction

In the tangential direction

Since the particle moves on a path of constant radius , we have

When substituted in the equations above, this gives

Now using , we get from the second equation above

which gives upon integration

When substitutes in the equation , this gives

7.10 The free body diagram of the bead at equilibrium is shown in the figure below.

R

mg

N

94

The horizontal components of the normal reaction N provides the centripetal force while the

vertical component balances the weight of the bead. Thus

Dividing the first equation by the second one gives

The slope of the curve is also equal to . Thus

This gives

and

7.11 Since , we have

;

Thus and

7.12 We know that

Therefore

mg

N

95

Similarly

7.13 Since

We get from the solution

From , we get

Now from we get

And from we get

7.14

96

(i) The tension in the outermost string provides the centripetal force for the outermost bead.

Let the tension in this string be T1. Then

Similarly, centripetal force for the second bead from outer end is provided by the difference

in the tension T2 in the second string and tension T1 in the first string. Thus

Extending further

and so on. In general we can write for the ith string from the outside,

This is then easily seen to be

(ii) Now we generalize the result of part (i) to a rope of length L and mass per unit length

λ. To make the transformation from the problem in part (i) to this problem, we

consider the mass of each bead to be distributed over the connecting string whose

length we take to be vanishingly small i.e. . Thus we have

and

Substituting this in the expression for Ti above, we get

ml

T1T2

97

On taking limit , we then get

To get this answer by considering the rope directly, we take a small portion of the rope of

length at a distance x from the centre O. The centripetal force to it is provided by the

difference in the tension at its two ends, as shown in the diagram below

Then

This gives the differential equation

The solution of this equation is

where C is the integration constant. With the condition that the tension vanishes at x=L, we

then get

7.15 Consider a small section of the rope making a small angle at the centre of the

loop. The force at its two ends due to the tension in the rope is shown in the diagram

below

T(x) T(x+x)

98

For small angle the forces at the ends give a net force towards the centre which is equal to

This provides the required centripetal force for the segment. Therefore

7.16 The relevant coordinates and the free body diagrams of the two masses are shown

below

We treat the mass m using the polar coordinates as shown in the figure. The total number of

unknowns in the problem are : y coordinate of mass M, polar coordinates of mass m and

tension T in the string. The corresponding equations are

TT

/2

Tsin(/2)

r

y

M

m

Mg

T

mg

T

99

And the constraint equation

7.17 (i) After time t, the end of the rod that was at the origin has moves by a distance .

Thus the relationship between the x and y coordinates will be

(ii) Free body diagram of the bead is shown below

Thus the equations of motion are

From the constraint equation

Thus

which gives

Thus

Or

θθ

N

100

(iii) Integrating the equation above with the condition gives

Thus the bead will take time to reach the lower end.

7.18 Suppose the lower corner with angle is at the origin at t=0. Then if the position of

the mass is (x, y) at time t, the relationship between these coordinates is

if the acceleration of the wedge is A. This implies

(i) Now if the mass falls vertically down, which gives i.e. the

wedge accelerates to the right with acceleration . The interpretation is

quite simple: when the wedge moves by horizontal distance , the mass

should move vertically down by and the relationship between the two at

minimum acceleration is .

(ii) For the particle not to move with respect to the wedge, we have

which implies

.

The free body diagram of the mass is as follows

101

Thus we have by the conditions above

which gives

7.19 From example 7.9

This gives (assuming )

Similarly

This gives

Y

θX

N

102

Thus if mass m starts from height h, it will take time

And at this time the speed of the wedge will be

7.20 The coordinates used in solving the problem are shown in the figure below

Let the tension in the string be T. The equations of motion for the two masses are

And since the pulley P3 is massless, we have

Thus we have four unknowns y1, y2, y3 and T but only three equations. One more equation is

provided by the constraint equation. The constraint is that the length of the string is a constant.

If the heights of the two fixed pulley are h1 (for P1) and h2 (for P2), the constraint is expressed as

P1

P2

P3m

my1

y2

y3

T TT

Th1

h2

103

or equivalently as

Now adding the equations of motion for the two masses gives

On substituting and from the equations above, we get

This gives

The general solution of the equation above is the sum of its homogeneous solution yh(t) and the

particular solution yp(t). We have

Here A and B are two constants to be determined by the initial conditions and . Thus

the general solution is

Now the initial condition is

These give

Thus

This immediately gives through the equation

104

This gives

This is easily integrated. Since it is gives that and , we get

In exactly tha same manner we get

7.21 (i) Wedge m3 is free to move only in the horizontal direction; m1 and m2 move both

horizontally as well as vertically. Thus we would have had 5 degree of freedom.

However, there are three constraints: The length of the string is fixed, mass m1 moves

only in the vertical shaft and mass m2 moves on the plane of the wedge. These

constraints reduce the degrees of freedom to 2. Thus there are only two degrees of

freedom.

(ii) The origin and the coordinate axes chosen to describe the motion are given in the figure

below. Also shown are the free-body diagrams of the three masses

m2

m1

m3

x1

y1

x2

x3

y2

105

We see that there are in total 8 unknowns: x1, y1, x2, y2, x3, N1, N2, N3 and T. The equations of

motion are:

These are the equations of motion. In addition there are three constraint equations.

In the above, h is the height of the wedge. There are a total of nine variables and nine equations.

Of these equation (vi) is not relevant for the motion since the wedge moves only in the horizontal

direction.

Equations (i), (iii), (iv) and (ix) give

which gives

This is the equation expressing momentum conservation in the horizontal direction.

Similarly equation (iv) and (v) along with (ii) give

m2gm1g

m3N1 N1

N2 N2

N3

TT

TT

m2g

106

Now substituting for from (viii) and from (vii) and using (ix), we get

Which is equivalent to

Now using , we eliminate from the equation above to get

This gives

And using , we then get

Now using equation (viii) we get

Using equation (vii), we get

As , we get

7.22 Consider two parts of the rope x and (L-x) in length. The free body diagrams (showing

only the horizontal forces) of the rope and these two parts are shown below

107

Since the rope is moving with constant speed, there is no net force on the rope or any part of

it. Thus from the free body diagram of the rope

From the free body diagram of the left portion of the rope

Or from the free body diagram of the right portion of the rope

It is also instructive to solve the problem by considering a small portion of length dx at

distance x from the left and balancing the forces there to get a differential equation for T(x).

The free body diagram of such a portion is as given below

With the condition that T(L) = 0, the equation above can be integrated to get

xF

Friction

F

Friction

T(x)

T(x)

Friction

T(x)

Friction

T(x)+dT(x)

dx

108

7.23 The force applied should be such that the frictional force on mass m is sufficient to balance

its weight. Free body diagrams of the two blocks are shown below

If the entire system is moving with acceleration a then

This gives

If mass m is not falling then

Thus

For m=16kg, M=88kg, =0.38 and g=9.8ms-2, we get

mgM

Ffriction

NN

friction

Mg Normal reaction

109

7.24 The forces on the bead are: Its weight, normal reaction NV in the vertical direction and

normal reaction NH in the horizontal direction. Thus the free body diagram of the bead is

as follows (assuming the rotating arm is going into the page)

These force provide the radial and the tangential acceleration given by

Since the rod is rotating with a constant angular speed , we have

(i) If the bead is stationary at r=R, we have

This gives

Since , we get for stationary bead

(ii) If and negligible weight of the bead, we have

The minus sign for the friction shows that since the bead slides outwards, the frictional force in

inwards. The equation describing the motion of the bead is then

Assuming a solution of the form and substituting it in the equation above we get

Solution of this equation gives

mg

NV

NH

friction

110

Thus the general solution is

Here A and B are to be determined by the initial conditions that .

Thus

The solution is

Thus the distance of the bead from the center is given as

7.25 The solution for the distance travelled by the particle is

For k=0, we can’t substitute this directly in the formula since we are dividing by k in the

expression above. Thus we take the limit . In this limit, the first nonzero term we get is

This gives

7.26 The equation of motion for the particle is

111

The solution of the homogeneous equation

is

Here A is a constant to be determined by the initial conditions. The particular solution is

Therefore the full solution is

Now the initial condition is . This gives

Thus

This is easily integrated to get y(t) also. Thus

When the ball reaches the highest point, its speed is zero. If this time is tup, then

This gives the height h to be

112

If the total time of flight is T then T=tup+tdn, where tdn is the time taken to come down from

the highest point. At time T, y(T)=0. Therefore

Thus tdn will be given by solving

To understand whether tup is larger or tdn is larger, let us see the time change in the limit of

very small k. In that case (up to order k)

And tdn is given by solving

This gives

Therefore

113

A comparison shows that tup is smaller than tdn. This makes sense because the while coming

down, the average speed is smaller since the particle has lost energy due to viscosity and

continues to do so.

This also gives the total time of flight to be (up to order k)

However, this approximation will be valid only if

5.27 It is given that and . Therefore . It is also

given that .

(i) Height = Range =

(ii) When , from the expression derived in the problem above, we get with the initial

vertical speed

Height =

Substituting the values , we get

For k=0.1, Height = 202m

For k=0.2, Height = 172.3m

To find the range, we first fine the total time of flight and then use formula derived in example

7.13 to find the horizontal distance travelled. From the solution of the previous problem, we

know that the time of flight T is given by solving

For k=0.1, this gives

114

Since the time of flight without drag is , and as the result of the problem

shows the time of flight becomes smaller for , we tabulate T and the right hand side of the

equation above to find T for

It is clear from the Table above that the expression on the right was smaller than T till T=13s and

becomes larger at 12.5s. Thus the time of flight will be between 13 and 12.5s. A little more

tabulation gives T=12.8s.

Substituting this in , we get Range = 669m

Since , we get tdn= 6.75s. This confirms numerically that time

taken to come down is greater.

For k=0.2, the equation to determine the time of flight is

Making a table like above gives T=11.8s and Range T

14 13.62

13.5 13.29

13 12.93

12.5 12.58

115

(iii)

(iv) When drag is introduced, it is the range that is affected much more than the height.

(7.28) In the problem above, we have already found the range for . Let us now take the

case of k=0.1 and find the range for and .

For , the equation

beomes

or

and gives

T=11.7s.

This gives a range of

116

k=0

k=0.1

k=0.2

= 678m.

Thus for , the range increases.

For , the equation

beomes

or

and gives

T=13.8s.

This gives a range of

= 641m.

Thus for the range decreases.

The two calculations above show that for maximum range, the projectile should be launched at

an angle less than .

The reason for this is as follows. Since the horizontal speed reduces as the projectile moves, it

should cover a larger distance in the initial part of the flight. For this it is better to have a

relatively larger horizontal component of the velocity compared to the case when there is no

drag. Thus the angle should be smaller than .

(7.29) In this case the drag force is proportional to the square of the speed. So the equation of

motion will be given as follows for the motion up and motion down (taking vertically up

direction to be the positive y direction)

Motion up

Motion down

Since we are only interested in height, we change to to get the speed as a

function of the vertical distance of the ball from the ground. The first equation in that case is

117

The solution of this equation is the sum of the solution of the homogeneous equation

and a particular solution . These solutions are

and

Here A is a constant to be determined from the initial conditions. The full solution therefore is

The initial condition is that . This gives

This the speed of the ball as it moves up is

At the maximum height h, the speed becomes zero. Therefore

This gives

Now we consider the motion for downward motion. This can be rewritten as

Again the solution of this equation is the sum of the solution of the homogeneous

equation and a particular solution . These solutions are

and . Thus the complete solution is

Now the initial conditions are . This gives, with

118

This gives

If the final speed is vf, then

Note that of k=0 then The answer can also be written as

119

Chapter 8

8.1 Since there is no external force on the system in the horizontal direction, the total momentum

in the horizontal direction is conserved.

Initial momentum in the horizontal direction = momentum of the carriage + momentum of rain

=

=

Final momentum of the system after time t =

Here vf is the final velocity. Equating the two moment gives

8.2 Since the water leaking out of the carriage still has a horizontal velocity equal to the velocity

of the carriage, total momentum of water after it came out for time t is =

If the initial amount of water in the carriage was m0, then the initial momentum of the system

(carriage + water in it) =

If the aped of carriage (with left over water in it) after time t is vf, then by momentum

conservation

8.3 Exactly like in problem 8.2, there will be no change in the speeds of the two bicyclists. This

is easily done by considering the momentum of the two friends before and after the books

are given by one of them to the other person. Consider the person giving the books. Her

momentum before transferring the books is . After she gives the books, let her speed by

vf. Then by momentum conservation

Similarly, for the person receiving the books

8.4 Conserve momentum after the first bullet has been fired. Initial momentum is 0. Let the

velocity (since the motion is one dimensional, we write only the symbol for it, the direction

is taken care of by the sign) of the gun after the bullet is fired be v1. Since the relative

velocity of the bullet when it leaves the gun is u, and the bullet leaves the gun when the gun

120

is already moving with v, bullet’s speed ug with respect to the ground is calculated as

follows:

Therefore momentum conservation gives

Now the momentum of the gun and bullets in it is

Now let the speed of the gun after the second bullet is fired be v2. Then momentum

conservation gives

Similarly one can now show that if the speed after the third bullet is fired is v3 then

Generalizing this we get after N bullets have been fired

8.5 (i) Momentum of the system = sum of the momentum of each particle

= kg ms1

(ii) velocity of the centre of mass = total momentum/total mass

8.6 (i) acceleration of the CM

121

(ii) No, the acceleration is not in the same direction as the momentum of the CM.

8.7 If the base of the cylinder is in the xy plane, as shown in the figure, the x and y coordinates

of the CM are (0, L/2). We thus have to calculate the z coordinate of the CM.

To obtain the z coordinate of the CM, consider a rectangular sheet of thickness dz at height z, as

shown in the figure below.

If the density of the material that the cylinder is made of is , the z coordinate of the CM, by

definition, is

L

R

x

y

z

z R

122

To evaluate the integral, we substitute and . This gives

8.8 (i) CM of a cone shown in the figure below

The CM is on the axis of the cone by symmetry. To calculate its height, we take a thin disc of

thickness dz at height z. By similarity of triangles, it radius r is given by

If the density of the material that the cone is made of is , then the position of the CM is given

by

It is reasonable that the location of the CM is more towards the base since larger mass of the

cone is concentrated there.

R2

h

zr

123

(ii) Hemispherical bowl of radius R is shown in the figure below.

The CM will be on the line passing through the centre of the base. To calculate its height zCM, we

take a ring of height dz at height z. According to the figure

and ,

If the mass per unit area for the shell is , then the mass dm of the ring is

Thus

8.9 Given N particles of masses mi (i=1-N) with total mass at positions (i=1-N),

position of their CM is given as

r

R z

124

Now let us make m subsystems of these masses with number of particles N1, N2, N3……Nm in

them. Then we have the mass of each subsystem as . The position of the CM can

then be written as

By definition of the CM we have for the position of each subsystem .

Thus the expression above can be written as

This shows that the CM of the system can be calculated by treating each susbsystem as a point

particle of mass Mi located at the CM of each subsystem.

8.10 To find the CM, we will treat the cone and the hemisphere as two subsystems. It is also

clear by symmetry that the CM will be on the extended axis of the cone. Taking the axis

as the z direction with z = 0 at the base of the cone, we have

Assuming the entire system is made of a material of uniform density, we get

8.11 Since there is no external force on the system in the horizontal direction, the position of the

CM will remain unchanged as the small block moves from one side to the other. Taking

horizontal direction to be the x-direction, let the position of the CM when the block in on

125

the left be X1 and let it be X2 when the block is on the right. Then the poison of the CM of

the block is (X1R) and (X2+R), respectively. Since the CM does not move, we have

This immediately gives

8.12 When the ball is compressed, it looks like shown in the picture below

The radius r of the circular area of contact for x<<R is

Thus, if the pressure in the ball remains unchanged, the force that the ball applies on the wall is

8.12 As a photon hits the surface, it gives it an impulse proportional to its momentum. If it gets

absorbed, the impulse and if it is reflected then . And the force by the

stream of photons hitting the surface will be where n is the number of particles hitting the

surface per second. If the cross-sectional area of the parallel beam of light is a, then

Thus the pressure P

R

Rx

126

(i) when the light is completely absorbed

(ii) when light is perfectly reflected

8.14 By equation (8.42b), the force on the planar surface will be equal to momentum

transfer per unit time. On hitting the surface, the component of momentum perpendicular

to the plane becomes zero while that parallel to the plane remains unchanged. Thus all the

momentum that water stream carries perpendicular to the surface is transferred to it. The

momentum carried by the stream of water per second is

Its component perpendicular to the surface is

When the water stream hits the surface, it makes an elliptical cross sectional area on the

surface because the surface is slanted. The major axis of the ellipse is =

And the minor axis remains the same as

Thus the cross-sectional are of the ellipse is =

Thus the pressure on the surface =

8.15 At steady state flow let the mass flow rate from the upper portion of the hour-glass be

. If the height through which it falls before hitting the lower surface is h then the

amount of mass in the air is the rate at which the mass is falling and the time it takes

for it to reach the bottom. Thus it is and its weight is . Thus the hour

glass should have weighed less by this amount. However as the sand hits the bottom,

it transfers momentum to the hour glass that exactly compensates for the weight in the

127

air. This is shown as follows. As the mass hits the bottom, its speed is . Thus

the momentum it transfers to the bottom per second is .

8.16 Consider equation (8.43) for the rocket.

Now since the mass coming out leaves the rocket with , we have

This is because when the mass leaves the rocket, it has already achieved velocity

. This gives in the equation above

8.17 Example (8.9) using equation derived above. The example is solved exactly as done

in the text except that in applying the equation derived above, each

time the bullet is fired. This immediately leads to equation (8.50) and the rest is the

same as done in the example.

8.18 (i) Force needed to hold the chain is equal to the force required to hold the part of the

chain hanging vertically. This force is =

(ii)If the chain is to be pulled at a constant speed v, its mass increases at the rate of .

This gives momentum change per unit time = . This is the additional force required

to provide momentum. Thus the total force is . This is also seen easily by the

rocket equation

Now it is given that . This substituted in the rocket equation

immediately gives the result derived above.

(iii) The rocket equation, with is

128

If at a time t, the length of the chain on the table is x then and

. Thus the rocket equation can be rewritten as

This is integrated as

Upon integration this gives

Upon solving this gives

8.19 Since the peg is frictionless and the length of the portion of chain passing over the

peg is negligible, the other portion has length (Lx) and the tension in the chain is the

same throughout. Taking the tension to be T, the equation of motion for the two

portions is (see figure)

129

Adding the two equations to eliminate T gives

This equation is also obtained by direct application. Since the total mass being moved is

M, the net force is and the acceleration is .

The solution for the equation above is a sum of the solution of the homogeneous equation

and the particular solution . The solution of the homogeneous part is

Where A and B are two constants to be determined by the initial conditions. The full

solution is

x(t)

TT

130

The initial conditions are . This gives . This give

the solution for x(t) to be

8.20 (i) Since the pressure inside the box is p, and the force is unbalanced over an area S,

the force on the box will be pS.

(ii) The acceleration of the box will be

(iii) In the simplest calculation, the rate at which the molecules are coming out in one

second will be those contained in a cylinder of height , where is the average

speed in the x direction in the rms sense. Thus the rate at which the gas will be

leaking out is where n is the number density of molecules and m the mass

of each molecule.

(iv) Equation (8.45)

In our case so

by equation (8.40)

8.21 By the rocket equation . If the mass of the fuel is M, we have

, Mf = M. Thus we have

131

This gives

8.22 In this case the rocket equation becomes (assuming vertically up direction to be

positive)

Solution to the equation above is given by the sum of the solution for the homogeneous

part and the particular solution. This gives

A is determined by the initial condition . This gives . Thus

132

Chapter 9

9.1 Consider a frame the origin O and another frame with origin O’. Frame O’ is moving

with velocity V in the x-direction with respect to frame O. Let there be a force F act

on a particle; the force is the same in the two frames. Then by the work energy

theorem in frame O, we have

Now in the frame O’, the corresponding velocities are related to the velocities in frame O

as follows

and

and

Substituting this in the work energy theorem gives

Now is noting but the momentum change , which is the

same in both the frames. Thus . This, when substituted in the

equation above leads to

Which is the work energy theorem in frame O’.

9.2 Kinetic energy of each particle is

Thus the total kinetic energy of the system is = 0.45J

133

Momentum of the system is =

Velocity of the CM therefore is =

Thus the kinetic energy of the CM is =

Thus the kinetic energy of then particles in the CM frame is = 0.450.417 = 0.033J

This can also be checked by calculating the kinetic energy of each particle in the CM

frame, which is

9.3 Since the momentum of the system is zero, the kinetic energy of the CM = 0

9.4 Masses, velocities, moment and kinetic energies o feach particle are shown in the table

below

mi(kg) vi(ms1) pi(kgms1) KEi(J)

1 1 1 0.5

2 2 4 4.0

3 3 9 13.5

4 4 16 32

5 5 25 62.5

Thus the total mass is = 15kg

Total momentum is = 15kgms1

Velocity of the CM =

Total kinetic energy = 112.5J

Kinetic energy of CM =

Kinetic energy about the CM = 105J

134

This is easily checked by calculating the kinetic energy of each particle about the CM and

adding them all up. Thus

KE about the CM =

= 105J

9.5 From equation (9.37), we have

Thus the kinetic energy of the system

Since , we get

9.6 (a) We will be using to calculate the force. Thus ,

where C is a constant chosen according to the reference point x0, where .

(i)

(ii) To calculate the potential energy, substitute so that

to get

(iii)

135

(iv)

(b) The plots of force and potential in each of the cases above are as follows. The force is

shown on the left and the corresponding potential on the right in each case.

(i) We have taken k = 1.5. We have chosen the constant such that U(0) = 0.

(ii) The plots are drawn for a = 5

(iii) We have chosen the constant such that U(0) = max.

136

(iv)We have chosen the constant such that U(0) = 0. Notice that the average force is always

positive, i.e. pointing in the positive x-direction. Therefore the potential energy curve keeps

going down as x increase.

9.7 We use and keep the potential continuous everywhere. In the

plots the force is shown on the left and the corresponding potential on the right in each

case.

(i)

If we choose

For the plots drawn below we have chosen k = 2.

137

(ii)

If we choose U(x) = 0, C=0.

For the plots below, we have chosen k=2.

(iii)

The nature of force implies that the potential is a constant for

138

Choosing U(0) = 0 gives and

For the plots below, we have chosen k=2 and a =15.

(iv)

The nature of force implies that the potential is a constant for

Choosing U(0) = 0 gives

Thus

For the plots below, we have chosen C=250 and a =20.

139

9.8 We use . This gives

(i)

(ii)

(iii)

(iv)

(b) Plot for part (i) is straightforward; it is a constant force and linearly varying potential

Plots for part (ii) are similar to that of part (i) in problem 9.7. This has been given above.

(iii) We have chosen a = 5. The force is shown on the left and the corresponding potential

on the right.

140

Notice that the force is negative for negative x and positive for positive x. Thus it is a force

that pushes a particle away from x = 0 and the position at x = 0 is an unstable equilibrium

point. Thus when a particle is displaced slightly from this point, it will runaway to infinity.

(iv) We have chosen a = 5. The force is shown on the left and the corresponding

potential on the right.

Notice that the force is positive for negative x and negative for positive x. Thus it is a

restoring force and the position at x = 0 is a stable equilibrium point. Further the force is

varying almost linearly with x near x = 0. Thus when a particle is displaced slightly from this

point, it will perform simple harmonic oscillation.

9.9 Since all the potentials except (i) are time-dependent, only (i) is conservative.

9.10 For the block not to fall off the track, its speed v at the top of the loop should be such that

Now by the conservation of energy we have

9.11 For the potential energy we have the potential energy curve (with

parameters C = 0.5 and a =2)

141

It is clear from the curve that a particle will always experience a force in the negative direction

except at a = 2. From this it is clear that the point x = a is a point of unstable equilibrium.

For the potential energy we have the potential energy curve (with parameters C

= 0.5 and a =2)

This potential energy curve gives force opposite to the displacement, and the force is zero at x =

a. Thus this point is a point of stable equilibrium. Notice that unlike the potential energy for a

142

spring , this potential energy is quite flat near the equilibrium point (because

for . For larger displacements from x = a, the curve rises much faster

than the spring potential energy.

9.12 Let the velocities of the balls be v1 and v2 after the collision (since this is a one

dimensional motion, we are not putting vector signs on top of the velocities). Then by

momentum conservation

The first equation gives

This and the second equation implies that

Thus either v1 is zero, i.e. the first ball is not moving. This is the situation after collision.

Or v2 is zero, i.e. the second ball is not moving. This is the situation before collision.

9.13 According to the situation give, the two balls are undergoing an elastic collision (no loss

in energy) when they are moving in the opposite directions with equal speed V. Let

the mass of the lighter ball be m and that of the heavier one be M. If the velocity of

smaller ball is v1 after the collision, then by equation 9.36 (taking vertically up

direction to be positive)

In the extreme limit, when M>>m, we have

If the balls are dropped from a height h then

If after the collision the smaller ball goes to height H then

These two equations give

9.14 By momentum conservation we have

143

Totla initial energy =

Final energy =

As is clear, the final energy is less than the initial energy. Therefore some energy is lost in

the process.

Now consider a pile of beads, each of mass m, connected with strings of length l between

them. As the first bead falls over the edge, its initial speed is zero bu by the time the string

connecting it to the second bead is taut, it gains a velocity of . However, as soon

as it pulls the second bead, by momentum conservation at that instant (neglecting the

impulse due to gravity) the speed of two beads is

Now the two beads fall together and as the string connecting the second bead to the third

bead becomes taut, the energy of hse two beads is

This gives the speed of the two beads to be

Thus when the third bead is pulled over, momentum conservation at that instant gives the

speed of the three beads to be =

One can go on like this and build up the solution up to p th bead falling by recognizing a

pattern. We can also do the problem in the following way.

Suppose when the (p-1)th bead has just fallen of the edge, the speed of the beads becomes

. Then when these beads fall and the string between the (p-1) th and pth string becomes

taut, the energy of the system will be

Thus the speed of these beads just before the pth bead falls off the edge is given by

144

By momentum conservation, therefore, we get the speed when the pth bead falls off as

follows

We know that i.e. when the first bead just falls off the edge of the table, its speed is

zero. Now we can write

And so on so that continuing in this manner we get ( )

This gives

9.15 From problem 8.18, the total force required to move the chain is . Thus the

power delivered by the person pulling the chain is =

On the other hand, the energy of the chain is = kinetic energy + potential energy

If the length of the chain on the table is x then and

Kinetic energy = potential energy =

Thus Total energy E =

This gives

Thus the rate of change of the energy of the chain is not equal to the power delivered. As

shown in the previous problem, the energy is lost in the inelastic collision.

One may ask a question: what if instead of the chain, it s a rope or a strin that is being

pulled? Where does the energy go in that case? The answer is that the energy difference is

then used in stretching the rope to generate the required force.

145

9.16Let the length of the chain be L. If its length y is hanging from the table, the force

pulling it down is =

Taking the vertically down direction to be positive, the coordinate of the end of the hanging

portion of the chain, when its length is y , is also y.

Thus the equation of motion of the chain is

The equation can also be derived by considering the two protions, the one hanging from the

table and the other on the table, separately.

Writing we get from the equation above

This gives

The second term in the equation above is the change in the potential energy of the chain as

the length of its hanging portion changes from y1 to y. Thus the total energy of the chain

remains unchanged as it slips off or the energy is conserved. This can also be seen as

follows

When length y1 is hanging,

kinetic energy = 0 potential energy = Total energy =

When length y is hanging,

kinetic energy = potential energy =

Total energy =

It is clear from the above that the total energy remains unchanged.

9.17As the chain unfolds, the mass of the hanging (and moving) portion keeps on

changing. Thus the problem is like the variable mass problem. We take the mass per

146

unit length of the chain to be λ. Taking the vertically down direction to be positive,

the coordinate of the end of the hanging portion of the chain, when its length is y , is

also y. The external force on the chain is , , and the relative velocity of

the mass that is added to it is . Thus the equation of motion of

the chain is

Using we write this equation as

We now solve this equation for as the sum of the solution of the homogeneous

part and the particular solution to get

Here A is a constant. Using the condition that , we get and

therefore

The same result can be obtained by letting and such that in the solution for

a coiled set of beads in problem 9.14.

Now let us compare the energy of the chain at the beginning o fthe motion and agter it has

slipped so that y of its length is hanging.

At the beginning, when length y1 is hanging,

kinetic energy = 0 potential energy = Total energy =

When length y is hanging,

Kinetic energy =

147

Potential energy =

Total energy =

Thus we see that the energy is lost during the motion. This happens due to the inelastic

collision between the chain links as each new link is pulled by the moving chain (see

problem 9.14).

148

Chapter 10

10.1 (a) The fields are shown on the axses. They are the same all over the place.

(i)

(ii)

(b) If we think of the force arrows as velocity of a fluid, we see that a matchstick thrown in

that fluid will tend to rotate clockwise in (i) and counterclockwise in (ii). Thus the curl of

both the fields is not zero; it is negatve for (i) and positive for (ii).

(c) curlf of force field (i)

Curl of force field (ii)

149

x

y

x

y

10.2

(i) Force field .

Path 1: Work done is zero as the particle moves along the x-axis. Similarly as it moves from

(2,0) to (2,1), work done is again zero since the particle is moving perpendicular to the force

field. Thus alongh path 1, the net work is zero.

Path 2: Work done is zero as the particle moves along the y axis. However, as it moves from

(0,1) to (2,1), it is under a constant force of 1 unit in the x direction. So the work done by the

field is 2 units.

Path 3: Along path 3 . Thus . The displacement along path 3 is given as

. Thus

unit

(ii) Force field

Path 1: Work done is zero as the particle moves along the x-axis since the particle is moving

perpendicular to the force field. Similarly as it moves from (2,0) to (2,1), Field is constant and

equal to 2 units. Thus the work done by the field will be 2 units. Thus alongh path 2, the net

work is 2 units.

Path 2: Work done is zero as the particle moves along the y axis because the field is zero. As it

moves from (0,1) to (2,1), it is is moving perpendicular to the force field so the work done is zero

again. So the net work done by the field is zero.

O X

Y

(2, 1)

150

Path 3: Along path 3 . Thus . The displacement along path 3 is given as

. Thus

unit

In both the cases the work done depends on the path. It is therefore consistent with the curl of

both the fields not being zero.

10.3 Force field is . The paths are shown below

We will calculate the work done along path 1 in two parts: first when the particle mones

along the x-axis and secondly when it is moving from (1,0) to (1,1). Along the x-axis, the

force is and the displacement is . Thus the work done is zero when the particle

moves along the x-axis. When the particle moves from (1,0) to (1,1), the force is

and displacement is . Thus the total work done is

This then is the total work done along path 1.

Along path 2 y = x so that dy = dx. The work done is given by the integral

Since the work done along two paths is different, the force field is NOT conservative.

10.4 The figure for the path si shown below. The way we have set up the transformations

Path2

O X

Y (1, 1)

path1

151

, the corresponding angle is as shown in the figure and

varies from π to 2π. Thus the integral from A to B over the

semicircular path is

The integral can be carried out easily and gives

10.5 . The force field will be conservative if its curl vanishes. The curls

of it is

If the curl has to vanish, we should have .

For the force field it is clear by inspection that the potential should be

. However, to derive it formally we put

This gives

Y

X

C

BA

(0,1)

(2,1)

152

Which is integrated to

Differentiating this with respect to y and writing gives

By the condition , the const. is zero. Thus

10.6 (i) Curl of the field

is not zero. Therefore the field is not conservative.

(ii) The curl of the field

is not zero. Therefore the field is not conservative.

10.7 (i) The function will be a constant where its

argument is a constant. Thus the contours are given by

These are concentric circles with centre at .

(ii) The gradient of the function is given by

10.8 Kinetic energy of the system of N particles with masses and positions

is

153

Let the position of the CM be so that . If the positions and velocities of these

particles with respect to the CM are given respectively by and then

and

Substituting this in the equation for kinetic energy above, we get

Denoting the total mass as and using the property of the CM that , we

get

This is the desired result.

10.9 When some energy is released during collision, the equations during the collision are

(momentum conservation)

(energy conservation)

Here the unprimed quatities are before collision and the primed quantities are after collision.

Further . The easiest way to prove that in the bove situation, the paricles cannot move

stuck together is to write all quatities in the with respect to the CM. In that case (referring to

quantities in the CM frame with subscript c)

And using the splitting of the kinetic energy as in the problem above and the fact that

Now if the particles are moving together, their momenta after the collision with respect to the

CM will be zero. This is also seen mathematically as follows. By momentum conservation

154

And if they are moving together, then

The above two equation give

Putting this in the energy conservation equation gives

However that is not possible because two positive quantities cannot add up to zero. Thus the

particles cannot move stuck together.

10.10 The picture of the carom coin and the striker is shown below

The coin will move in the direction of the impulse that it receives from the striker. This

direction is perpendicular to their common surfaces and theefore along the line joining their

centres. Thus the coin moves at an angle such that

We will work in cgs units. Initial momentum of the system is that equal to the initial

momentum of the striker which is .

Y

X

2cm

2ms-1

2.05cm 1.55cm

Direction of impulse on coin

155

If after the coliision, the velocity of the striker is and the speed of the coin is v

then its velocity would be and the momentum

conservation will lead to

Since the collision is elastic, the total energy is conserved and we get

These equations simplify to

These are 3 equations for 3 unknowns so all the answers can be obtained from these

equations. Substituting

In the energy conservation equation gives

This gives after collision. The trivial answer of course refers

to the situation before collision.

This then leads to . This gives the angle of deflection

of the striker to be below the x-axis.

(ii) KE o fthe CM remains unchanged during the collision. It is therefore .

(iii) Velocity of the CM =

Therefore the striker is moving with velocity 0.5 ms1 and the coin with -1.5 ms1 along the x-

axis in the CM frame. The magnitude of momentum of each body is 0.0075 kgms1 in

156

directions opposite to each other (see figure). In the CM frame the magnitude of velicities

does not change during an elastic collision. Thus the momentum and the velocity vectors of

the striker and the coin just rotate. Further, the impulse J is in the direction (see figure) such

that it makes and angle

from the x-axis. As is clear from the figure, the magnitude of J must be such that the final

momentum has the same magnitude as the initial momentum. This is seaily done by drawing

a circle of radius equal to the magnitude of the momentum with its centre at the tail of the

initial momentum vector. We then draw the impulse vector from the head of the initial

momentum vector at the angle and take its length such that its head touches the circle. The

vector from the centre to this point gives the final momentum. The construction

immediately shows that

It is also clear from the figure that

It is also clear from the figure that the angular momenta vector srotate by

33.75

J

0.0075 kgms1

0.0075 kgms1

157

As a check we now calculate the velocities of the striker and the coin in the lab frame from

the information above and compare our answer. As is clear from the figure, the x component

of the striker’s velocity in the CM frame is

This then gives Vxlab as follows (keep in mind that the CM is moving in x direction only)

And we get from the figure (the CM is moving in x direction only)

Thus

Similarly for the coin (see figure)

This gives

and we get from the figure

This gives

These match with the previously obtained answers.

10.11 As in the problem above, the stationary balls will move in the direction of impulse

after the collision. This is going to be along the line joining the centre of the striking ball

and the centres of the stationary balls. At the time of colliding, the three balls will look as

shown below.

158

Since the net impulse (due to the two stationary ball) will be along the line of its initial

motion by symmetry (see figure), it will continue to move along its original direction. This is

also seen by noticing that the statinary balls have a net momentum in the direction of v0.

Thus there will be no component of the momentum to be balanced in the direction

perpendicular to that. Thus the striking ball will continue to move along its original

direction. Let the speed of the striking call be v1 after the collision and the speeds (equal by

symmetry) of the other balls be v. Then by momentum conservation

Since the collision is elastic, we have by energy conservation

There are two unknowns v1 and v and two equations so we can get their values. Squaring the

momentum conservation equation and subtracting it from the energy conservation equation

gives

One of its solutions is the trivial solution referring to the situation before collision.

Theother solution is

Substituting this in the the momentum conservation equation gives

30

Direction of impulse on ball

Direction of impulse on ball

159

Chapter 11

11.1 Disc rotating about a point on its periphery. Consider first a rotation about a point on

the periphery by an angle. This is shown on the left in the figure below. The dashed

circle shows the initial position of the disc while the solid circle shows the position

after rotation.

On the right we show the same rotation carried out by translating the CM of the disc first to

its new position, as shown by the straight arrow and then carrying out the rotation about the

CM. We have made the disc in this position by dotted circle. As is evident from the figure,

the magnitude and the sense of rotation is the same as in the figure on the left.

11.2 Now we generalize the result of the problem above. On the right in the figure below,

we show the same rotation as in the figure on the left carried out by translating the opposite

end of the diameter of the disc first to its new position, as shown by the straight arrow and

then carrying out the rotation about the this point. We have made the disc in this

intermediate position by dotted circle. As is evident from the figure, the magnitude and the

sense of rotation is the same as in the figure on the left. Similarly, it can be shown about any

other point.

160

11.3 The angular momentum about the origin is given by . Thus the answers in

different cases are

(i) (ii) (iii)

(iv)

11.4 The position of the wheel and the stone stuck to its periphery is shown in the figure

below.

Vt

t

X

Y

x(t)

y(t)

161

Thus we have for the position , velocity (obtained by differentiating with

respect to time once) and acceleration (obtained by differentiating with respect to

time twice) of the stone

As is evident, the acceleration is towards the centre of the wheel. It is the centripetal

acceleration. The force for this is provided by the force that keeps the stone stuck to the

wheel. This force is

The angular momentum of the stone with respect to the origin is

This gives

The angular momentum changes due to the torque provided by the force that keeps the stone

stuck to the wheel and is equal to the centripetal force. Thus the torque

Thus it is equal to the rate of change of the angular momentum.

11.5 We assume that at time t = 0, the paricle is on the x axis. Thus after time t, its position

vector is give as (see figure)

162

Thus its velocity and acceleration are (obtained by differentiating with respect to time

once for the velocity and twice for the acceleration)

As expected, the acceleration is the centriprtal acceleration. It is provided by the external

force towards the centre. The angular momentum of particle with respect to the origin is

Its time derivative is

The torque due to the external force is

This is the same as the rate of change of angular momentum.

11.6 The angular momentum for a collection of particles about an origin O is

Now let is choose a different origin O’ such that the position vector of O’ from O is (see

figure) so that

and

X

Y

R

Ot

163

Substituting the expressions above in the formula for , we get

where we have used the fact that .

Now the angular momentum about O’ is

Thus if the total momentum of the system is zero, then

and

11.7 At the maximum distance Rmax and the minimum distance Rmin from the sun, the

velocity vector is perpendicular to the radius vector. If the speed of the earth at these

distances is Vmax and Vmin, respectively, then the angular momentum of the earth at

these two points is MearthRmaxVmax and MearthRminVmin , respectively. By the conservation

of angular momrntum we have

MearthRmaxVmax = MearthRminVmin

This gives

O

O’

X

X’Y

Y’

ZZ’

164

11.8 Initial momentum , the final mpmentum and the change in the momentum

are shown in the figure below.

As is evident from the figure, the magnitude of is where and it is

in the direction bisecting the angle between the incident and the scattered direction.

Mathematically it can be seen as follows. Choose the direction of incoming particle to be the

x direction. Then

and

Thus the magnitude of is and it is in the direction .

(ii) Since the paticle is moving parallel to the x axis at a distance d, its angular momentum is

mvd going into the plane of the paper. This can be seen as follows. The position and

velocity vectors of the particle are

and

This gives for the angular momentum

Further, since the force is central, the angular momentum about the origin is a constant.

X

Y

165

(iv) Since , we have . Therefore is in the same direction as

.

To calculate , we change the integration over time to integration over the angle by

using the angular momentum conservation. Using polar coordinates, we write the angular

momentum of a particle moving in the xy plane as . Since in the present

problem, it is a constant equal to , we have

The force between the two charges is in the radial direction and is equal to

Thus

Thus the magnitude of is . Comparison of this expression with that for

the momentum change shows that the two are in the same direction and

Or

This is the Rutherford formula.

166

11.9 (i) At angle the potential energy of the girl is with

respect to the equilibrium point, i.e. when the swing is in the vertical position. The

kinetic energy of the system at the lowest point , if the angular speed is , is

. By energy conservation

(ii) Whe the child stands up at the lowest point, all the external forces on the her are

passing through the pivot so her angular momentum remains unchanged. However

as she stands up her moment of inertia about the pivot is where ICM

is her moment ofinertia about her CM. Neglecting it gives the moment of inertia to

be . Then by conservation of angular momentum

keeping only linear terms in since d<<L.

(iii) The final kinetic energy of the girl is

keeping only linear terms in since d<<L. Thus we see that the kinetic energy of the

system has increased in comparison with the original energy. Threfore the swing would go

higher on the other side.

(iv) Suppose the swing goes up to angle on the othwr side, we have by energy

conservation and the fact that

167

Now we have the approximation d<<L, which gives

keeping only linear terms in . Thus

This gives the increase in the angle of swing in half a period. While going back, the new

amplitude by the time the swing reached its starting point will be.

In the full period, the net increase in the amplitude will therefore be

168

Chapter 12

12.1 Moment of inertia of a ring. Let the mass of the ring be m and its radius R. Since all

the points on the radius are at the same distance from the axis, each small mass on the

periphery gives the same contribution to the moment of inertia. Thus the total

moment of inertia is

A disc can be thought of made of many rings of width at different radii (see figure).

The moment of inertia of each ring is . Thus

the total moment of inertia is

For the moment of inertia about the diameter of a ring, consider a small portion of it of

extent at and angle It moment of inertia about the diameter is .

Thus the total moment of inertia is

12.2 Moment of inertia of a rectangular sheet. The sheet is shown in the figure below

169

We first calculate its moment of inertia about the x axis. For this, let us take a thin strip of

width dy parallel to the x axis and at a distance y from it. Its mass is .

Since the perpendicular distance of all its points is y from the x-axis, its moment of inertia

about the x axis is . Thus the total moment of inertia about the x-axis is

Similarly by taking a thin strip parallel to the y axis, we will get the moment of inertia Iy

about the y axis to be

Let us now calculate the moment of inertia about the z axis. Consider again the strip

parallel to the x axis. By parallel axis theorm the moment of inertia of this strip about the

z axis is = (moment of inertia of its CM with respect to the z-axis + its moment of inrtia

about the axis passing through its CM and parallel to the z-axis)

which upon integration leads to

Z

Y

b

a

X

y

170

12.3 The disc and a strip on it perpendicular to the axis of rotation (y-axis) is shown in the

figure below. The strip is of width dy and at a distance y fron the centre.

Length of this strip is and its mass is . Thus its moment of inertia

about the y-axis is

The integration is easily carried out by substituting and gives the resuly

12.4 This problem can be done exactly in the same manner as above. We now divide the

shell inti thinn rings at an angle (see figure).

X

Y

yR

171

The radius of the ring is and therefore its mass is and

its moment of inertia about the rotation axis is

Thus the total moment of inertia is

12.5 In this problem, we divide the sphere into thin discs and add the moments of inertia of all

discs. Thus for a disc of thickness dy at height y (see figure), its radius is , its

mass is

.

Therefore, the moment of inertia of the disc is

.

Integrating it from R to R gives the total moment of inertia

X

Y

yR

172

12.6 For the moment of inertia about the y-axis, the problem is done exactly in the same

way as in 12.5 by taking a disc at height y. The only difference is that the mass M is

now distributed over half the volume and the y integration runs from 0 to R. Thus the

mass of the disc is

And its moment of inertia is

Thus the total moment of inertia is given as

Similarly, to calculate the moment of inertia about the x axis, we make a cylindrical shell as we

did in example 12.3 (see figure)

Thus the moment of inertia will be calculated exactly like in example 12.3. The mass of this

shell is given by

Y

R y

X

173

Th factor of comes because there is only half the shell in a hemisphere. Therefore the moment

of inertia of the hemisphere is

12.7 A conical shell is shown below. To calculate the moment of inertia, we take a ring at a

distance x from the base extending from x to x+dx. Then by similarity fo triangles, its

radius is . Similarly the length of its side is

Thus the mass of the ring is . Its moment of

inertia therefore is

The moment of inertia of the shell is then given by integrating the expression above over x from

0 to h and gives

Now we calculate the moment of inertia about the y-axis. To do this we use the parallel axis

theorem and write the moment of inertia of the ring considered above as that of its CM plus its

moment of inertia about the axis parallel to the y axis and passing through the CM. Thus

X

Y

h

r

x

174

Integrating this we get

We now do the calculations for a solid cone. The calculations proceed in exactly the same

manner as for the shell bu now instead of a ring, we take a disc at distance x. The mass of the

disc is . Thus its moment of inertia is

Thus the total moment of inertia is

For the moment of inertia about the y axis, we again consider the same disc and use the parallel

axis theorem to write the moment of inertia of the ring considered above as that of its CM plus

its moment of inertia about theaxis parallel to the y axis and passing through the CM. So

Now the integration over x from 0 to h gives

12.8 (i) The total angular momentum = angular momrntum of large disc + angular momentum

of small disc

I2 is calculated using the parallel axis theorem and its value is

Thus

(ii) Let the large disc start rotating with angular speed in the opoosite direction.

Then by conservation of angular momentum

175

(iii) For and , we have

This keeps the total angular momentum zero since the large disc rotates in the opposite

direction with exactly the same angular speed as given to the small disc.

(iv) In a helicopter also, as the large rotor starts, the body of the helicopter will tend to

rotate in the opposite direction. To prevent this, a tail rotor is fitted that provides

the counterbalancing torque.

12.9 The position of the person (represented by the filled circle) after time t is shown on

the platform in the figure below.

As the person moves on the platform, the platform starts rotating clockwise with angular

speed so that the total angula momentum remains zero. As the platform rotates, the

person also moves with it. The net velocity of the person with respect to the ground is

equal to (velocity of the person with respect to the platform + rotational velocity of

the person due to the rotation of the platform). This is expressed as

Thus the angular momentum of the person is

u

A

B

utO

176

Here is the position vector of the person from the origin O on the axis. The position,

various distances and the velocities, as seen from the top, are shown in the figure below.

It is clear from the figure that

Here the sign takes care of the ddirection of angular momentum along the aixs of rotation;

it is positive for counterclockwise rotation and negative for clockwise rotation. Thus the

angular momentum of the perso is

The angular momentum of the platform is

Now equation the total angular momentum to zero (by conservation of angular momentum)

gives

O

A

B

ut

177

This gives

Total time that the person takes to move from A to B is

Thus the angle through which the platform rotates by the time the person reaches B is

This integration can be done by substituting

So we get

Now substitute which gives to get

where . This gives

178

If m = 0, we have and . Thus although on the platform, the person has moved

from A to B but overall he remains at his original position since the platform has rotated

back by .

For , we have

Thus , where 1 . We can therefore write

Thus

This gives

12.10 This is similar to the variable mass problem except that in this

problem we will get the answer by applying the principle of conservation of angular

momentum because this is an extended body and if we try to apply Newton’s second

law to each point particle, it is impossible to solve the problem because of a large

number of internal forces involved.

Let a small mass come out at a given time t during the time interval from t to

t+t from one of the containers. Let the angular speed of the display be at t. Then

the speed of gases coming out with respect to the ground is , as shown in the

figure

179

If in time the angular speed of the display becomes , then by conservation of

angular momentum

Here factor of 4 comes because of 4 containers. Here is the mass of each container at

time t. Obviously . Neglecting the second order term , we get

Thus

This equation can not yet be integrated because should be written in terms of for

that. Obviously . This gives

Integrating this gives, starting from = 0 at t = 0

Thus after the entire gubpowder is exhausted, we have

u-l

u-l

180

12.11 This problem is exactly the same as the problem above except

that the moment of inertia of the display is different because of different positioning

of the containers of gunpowder. Conservation of angular momentum now gives

Neglecting second-order terms and simplifying leads to

Following the steps in the solution of problem 12.10, this leads to

and

12.12 As the water comes out of the sprinkler pipes, it carries angular momentum with it. If

the opposing torque were not there, the system will rotate in the opposite direction with

increasing angular speed to conserve the angular momentum. However it does not happen

because of the opposing torque which is the external torque on the system. Taking the

direction of rotation of the rods to be positive, we have

Here external torque is with the minus sign indicating that it is in the direction opposite

to the rotation. Since the mass of the rod remains unchanged, if the mass of water coming

out from each end in time interval is , we have

and

Thus we have

181

This gives

Since the area of each hole is S, we have , where is the density of water. In

steady state, the sprinker system rotates with constant angular speed so that .

Substituting and in the equation above gives

Assuming that at the beginning of turning the spinker on, all pipes are filled, we also solve

how the angular speed changes with time until it approaches the above vaue. The

differential equation for can be written as

This equation has the solution

Since , we get

This gives

12.13 Let the angle between the unit vector and vector be (see

figure).

182

It is clear from the figure that the magnitude of is and it is in the direction of

. Since the magnitude of is also , we get .

12.14 As the disc hits the rough surface, let the surface apply an

impulse J on it (see figure).

As a consequence the disc moves with a smaller velocity and also starts rotating because

the impule applies a torque on it about its CM. By rolling condition we have .

By the equation for the CM, we get

And by the torque equation

This gives

r

n

rO

JV

V1

183

12.15 Free body diagram of the disc is given below.

Here f is the frictional force on the disc. By the equation for the motion of its CM we get

Taking torque about the CM of the disc we get

For pure rolling we have . Substituting this in the above two equations we get

12.16 (i) Free body diagram of the cylinder is shown below.

Here Jf is the frictional impuse on the disc. By the equation for the motion of its CM we

get

Similarly the angular momentum equation about the CM gives

Where the second equality follows from the rolling condition . Solving the above

two equations leads to

fF

Jf

Jd

184

(ii) For a sphere the equations are the same as for a cylinder except that its I about the

CM is different. Thus the equations become

Their solution gives

.

12.17 The rod in example 12.11 bounces back with angular speed

and speed V. Before it impacts the ground, its angular momentum about the point of

impact is equal to the angular momentum of its CM because it is not rotating, and it is

pointing into the page (see figure).

Thus

After the impact, the rotation about CM gives angular momentum coming out of the page

and the translational motion gives angular momentum going into the page. Thus

h

V

l/2

185

By conservation of angular momentum, we have

By energy conservation we have

These are two equations for two unknowns. Substituing from the

first equation into the second equation, we get the equation

This is a quadratic equation in V, and its solution is

The plus sighn in front of 1 above gives the trivial solution that the velocity of the rod

before impact is downward. And after the impact its velocity is gives by the second

root which is

From , this gives

It is evident from the equations above, for , the rod rebounds without any rotation.

12.18 This problem is similar to the problem 12.17 . However, the

CM and the moment of inertia about the CM are different in this problem. In this

case the CM is at distance from mass m. Thus the moment of inertia about

the CM

Like the previous problem, before the system impacts the ground, its angular momentum

about the point of impact is equal to the angular momentum of its CM because it is not

rotating, and it is pointing into the page. Its value is

186

Let the system bounce back with angular speed and speed V after the impact. Then

Thus angular momentum conservation about the point of impact gives

Similarly energy conservation gives

Again we substitute in the second equation to get

Its solution gives one trivial solution and for speed after impact

and

12.19 The figure below shows the rod and the stone just before the

impact.

187

(i) Right after the impact the angular momentum about the pivot is conserved. Thus,

if the rod and stone stuch with it move with angular speed after the impct, we

have (moment of inertia of the rod and the stone together is )

With , we get

(ii) The kinetic energy of the rod and the stone system after the impact is

If the rod rises by an angle , the potential energy of the system at this point is

Equating the KE and PE, we get

(iii) The impulse by the pivot. As soon as the stone hits the rod and gets stuck with it,

the horizontal momentum of the system changes. This change is brought about by

l43l

188

the impulse that the pivot applies on the rod. Thus by calculating the change in the

momentum of the system, we obtain the impulse that the pivot applies on the rod.

Initial momentum of the system =

Final momentum of the system =

Thus after the stone gets stuck, the momentum of the system changes by

in the same direction that the stone was moving initially. Thus the impulse given by the

pivot is

in the same direction that the stone was moving initially.

It is also instructive to solve this problem by angular momentum consideration by

calculating the change in the angular momentum of the system and attributing it to the

torque provided by the impulse. For this we apply the equation

Here CM is the centre of mass of the entire system (rod+sone). Note that this equation can

be applied only about the CM of the system and NOT about the CM of the rod alone.

Distance of the CM from the pivot point =

Thus the CM of the rod is above the system CM. Similarly the stone is

below the system CM. The velocity of the CM is in the same

189

directon as the stone. Thus befor the stone strikes the rod, it is moving with respect to the

system CM with speed in the same direction and the CM of the rod is moving with

speeed (in the opposite direction). This is shown in the figure below in the system

CM frame.

Thus the angular momentum of the system in its CM before the impact is

where we have used the fact that . The sense of rotation is counter clockwise.

Immediately after the stone hits the rod, rod starts moving with speed

With respect to the CM of the system, the CM of the rod and the stone are moving with

speed (keep in mind that is the same about any point and we are thinking of the motion

of the system as the translation of the system CM plus rotation about the system CM. Thus

with respect to the system CM, the motion is roational)

and

190

Thus the angulat momentum of the system about the system CM after the impact

where we have used the fact that . The sense of rotation is counter clockwise. Thus

the impuse gives a torque equal to

The negative sign here shows that the change is in clockwise direction. Its magnitude is the

torque impulse which is equal to the impulse J provided by the pivot times the distance fo

the pivot from the system CM. For clockwise sense, J is in the same direction as the

firction of stone’s velocity. Thus

12.20 Let the distance of the sweet spot be ls. The ball hitting the bat and its swing is

shown in the figure below.

ls

0

F

191

The point where the bat is held is the pivot point of the bat. Further, let the ball come

horizontally with speed vi and retun alonh the same line with speed vf. Considering

clockwise rotation to be represented by the positive direction, we have:

Initial angular momentum

Final angular momentum

Since there is no external torque about the pivot, we have

or

Finally the momentum change of the system is caused by the force applied on the bat.

Initial momentum is the sum of the momentum of the bat and that of the ball. The final

momentum is that of the ball only. This gives

Since for the sweet spot force is zero, we have

This combined with the equation gives

(i) For M=2 kg, LC=40 cm and I=0.3kgm2, we get

(ii) From the momentum equation above, we get

(iii) With the wrapping of the tape, the mass the moment of inertia of the bat and its CM

all change. We have

and

This gives

192

Thus the sweet spot is lowered by 1.1 cm.

12.21 The problem is solved as example 15.2 in chapter 15.

12.22 For generality we first take the masses on the dumbel to be of

mass M each and the mass striking to have mass m. Since there are no external

forces or torques acting on the system, its momentum and angular momentum

remains the same before and after the collision. For convenience we take angular

momentum about the origin. We show various distances in the figure below.

The momentum components of the mass m are as follows

Initial angular momentum about the origin arises only from the motion of mass m. When

particle is on the x axis, only the y component of its momentum gives the angular momentum

about the origin and it is

clockwise

After the mass m gets stuck with M after hitting it, the CM of the system is at

al/2

M

l/2

Ov

M

mX

Y

193

immediately after the impact.

By momentum conservation the momentum of the system after the imact is the same as before it.

Thus the angular momentum of the CM of the system after the impact is (see figure below)

Since angular momentum is conserved, the rest of the angular momentum comes from the

angular momentum about the CM of the system. Thus the angular momentum about the CM is

This then gives the rate of rotation through the equation

For this we first calculate . It is

Thus

With these we easily calculate the velocity of mass (M+m) and mass M right after the impact.

These will be given by adding the velocity of the CM to their velocity due to rotation about the

CM. Thus

Y

px

M

O

yCM

M

(M+m)

X

py

194

For these answers become

Angular momentum of the system CM about the origin after the impact =

Angular momentum of the system about its CM

and

Chapter 13

13.1 (i)

Thus

(ii) To find the principal axes, one rotates the frame with respect to the z-axis by an

angle so that the coordinates for each particle transform by the formulae

and the resulting coordinates make . Numbering the masses as shown in the figure,

we get

195

Substituting these in the formula for and equating the resulting expression to zero

gives

Substitution of new (x’, y’) also gives

and

13.2 The principal axes (1,2) of the rectangle passing through its CM are shown in the

figure when it is in the plane of the paper. Axis (3) is coming out of the paper.

Y

w

lX

1 2

34X’

Y’

196

(i) Components of the angular velocity are

(ii) Components of angular momentum along the principal axes are

Thus if the unit vectors along the principal axes are denoted as and then

(iii) Kinetic energy of the rectangle

13.3 The system and its principal axes (1,2) are shown in the figure below. Principal axis

(3) is coming out of the paper.

a a

b

12

197

We have

Rate of change of angular momrntum can be found easily by either the Euler’s equations or

by taking components of angular momentum in directions parallel to and perpendicular to

the direction of .

By Euler’s equations: Since the components of in the direction of the principal axes

remain unchanged, we have for the change in angular momentum

With the components calculated above, we get

Thus the vector is always in the direction opposite to that of principal axis . Thus at

the instant shown, it is pointing into the paper.

1

2

a

b

L

198

By taking components of angular momentum in directions parallel to and

perpendicular to the direction of : If we decompose the angular momentum

components L|| in the direction parallel to and L perpendicular to , then as the body

rotates, L|| remains unchanged and L rotates with angulae speed and changes at the rate

and that gives the rate of change of angular momentum. From the figure above, it is

clea that

This is also shown in the figure above. It is then clear that at the instant shown this vector

is going to rotate into the plane of the paper. Thus the direction of change of angular

momentum is into the paper and its magnitude is

13.4 The picture of the system of eight particles is shown below. We also show the

diagonal about which we consider it rotating.

(i) To show that the coordinate system chosen is is also the principal axes system,

it is sufficient to show that all the off diagonal elements of its moment of inertia

tensor vanish. This is easily shown. We do it for Ixy here.

X

Y

Z

199

Thus the set of axes chosen is the principal set of axes.

(ii) For the diagonal shown, the unit vector is =

Thus the angular velocity is given as

To calculate the angular momentum, we also need the moment of inertia about the principal

axes. These are

Thus we have

The kinetic energy

13.5 As the rod rotates, its angular momentum also changes. The required torque is

provided by the gravitational force pulling it down. Thus the rod can be in

equilibrium at two positions. Absolutely vertical or at an angle from the vertical.

The first solution is trivial. For the second solution, we find by equating the

angular momentum change about the pivot (a staionary point) to the torque. Later

we will check the answer by applying the torque equation about the centre of mass

also.

The rotating rod in the plane of the paper is shown in the figure along with its principal

axes at the pivot; axis (3) is coming out of the plane fop the paper. Also shown is the free

200

body diagram of the rod. The pivot applies a vertical force N to balance the weight mg of

the rod and a horizontal force F to provide the required centripetal accelerartion to the CM

of the rod.

The moment of inertia of the rod about the principal axes at the pivot are

The angular velocity components are

Thus the angular momentum components are

Since the angular velocity and its components along the principal axes are constant, we have for

the torque

Thus

Here the negative sign shows that the torque is pointing into the paper. This is proided by the

weight of the rod. The torque of the weight about the pivot is pointing into the paper and its

magnitude is

1

2 N

F

mg

201

Thus we should have

The same answer can also be obtained by taling horizontal (perpendicular to ) LH and

vertical (parallel to ) LV components of the angular momentum and then calculating the

rate of change of angular momentum as LH.

Now we calculate F. This provideds the centripetal force so we have

Check: To check our answers, we apply the torque equation about the CM and see if the forces

calculated by us satisfy this. About the CM we have

Further, for the principal axes at the CM (parallel to those at the pivot)

So that (angular velocity components are the same)

Thus we have, with

Let us see if , N and F calculated by us give the same torque. This is in the direction

perpendicular to the paper and its magnitude is

Substituting we get the torque as

202

Thus our answers also satisfy the torque equation about the CM and are therefore correct.

13.6 (i) Since the wheel is rolling without slipping, it rotates about the horizontal axis

with an angular speed

Further, it rotates about the vertical axis with an angular speed . This can also be

seen easily as follows. Since the wheel rotates a full radians in time , its angular

speed about the vertical is in counterclockwise direction looking at it

from the top.

(ii) Angular momentum about O = angular momentum of the CM + angular momentum about

the CM

Angular momentum of the CM about O = MLV in the vertically up direction

Angular momentum about the CM has both the horizontal and the vertical components.

Angular momentum about the CM = in the vertically up direction + in

the horizontal direction pointing towards O.

Thus total angular momentum about O is

Horizontal component LH = in the horizontal direction pointing towards O.

Vertical components LV= in the vertically up direction.

(iii) As the wheel rotates, the vertical component of the angular momentum remains

unchanged while the horizontal component rotates. Its rate of change is

therefoe equal to

If seen from the top, its direction is as shown in the figure below

203

(iv) The free body diagram of the wheel, when it is going into the page is shown below.

As the wheel rotates, the change in the angular momentum is such that at the position

shown above, it will point out of the paper. On the other hand, torque due to the weight

about O is pointing into the paper. Thus to generate torque coming out of the paper, the

normal reaction of the ground has to be more than the weight of the wheel so that

By balancing the vertical forces, we get

Thus when the wheel is moving, it is pulled down by the vertical shaft at the centre. In

turn, the shaft is pulled up. Thus with time it will tend to come out of the ground.

The horizontal force applied by the shaft provided the centripetal acceleration. Thus

OLH

LH

mgO

N

N1

LF

204

The explanation for why the wheel presses theground harder has been given above.

Another way to think about it is as follows. If there were no gravity, the horizontal shaft of

the wheel will tend to turn clockwise about a horizontal axis as the wheel moves so that the

change in the torque is zero. In turn the wheel presses the ground and generated enough

torque so that the rate of change of its angular momentum is equal to the torque generated.

13.7 This problem is like the previous problem except for position of the centre of mass

and the moment of inertia of a cone. We again obtain the horizontal component of the

angular momentum of the cone and multiply it with the vertical component of the angular

velocity of the cone to get the rate of change of the angular momentum. The horizontal and

the vertical components of the angular velocity and the free body diagram of the cone are

shown below.

Since the speed of the centre of cone’s base is V and its radius is R, . Similarly

.

The principal axes of the cone at its vertex are its axis and two axes perpendicular to it.

The moment of inertia about the axis of the cone is . Thus the horizontal moment

of inertia is

And its rate of change is

H

V N1

mgN

F

205

Coming out of the plane of the paper at the position of the cone shown. This shoud be

equal to the torque about the pivot. Thus

And

Similarly, the force F provides the centripetal force. Thus

13.8 This is a problem where we take the components of the angular velocity according

to the convenience of applying the rolling condition. So we split the angular velocity so

that it has component about the vertical and about its axis. We then find the

relationship between them by demanding that the speed of all the points on the ground

vanishes so that the cone rolls without slipping.

(i) Now a point at distance x from the vertex (pivot point) on the line touching the ground

moves with speed due to the rotation about the vertical axis through the vertex and

speed (see figure below) in the opposite direction due to rotation about the axis of

the cone.

Thus for rolling we get

Now if the centre of the base moves with speed V, we have

x

206

This also gives

As is clear from the components, the net angular velocity of the cone is parallel to

the line touching the ground an in direction shown in the figure. Thus

(ii) To calculate the angular momentum of the cone, we calculate the angular momenta

along the principal axes shown in the figure below.

From the figure it is clear that

In this case, the angular momentum changes because its vertical component rotates with

angular speed . Thus the rate of change of angular momentum is given by

and it is in the direction of axis 3. Its sign depends on the relation between h and R.

Check: The answer can be checked easily by applying the Euler equations that give

12

207

which is the same answer as obtained above.

13.9 In the figure below, we show the horizontal angular momentum LH of the rotating

disc, its change LH as the platform rotates and its free body diagram.

We have

If the torque applied on the disc is zero, i.e., N1 and N2 are equal, there cannot be any

change in the angular momentum. Hence the disc will tend to rotate clockwise in the

position shown so that N1 will become larger than N2. Finally, N1 and N2 will be such that

the torque is equal to change in the angular momentum. This gives

Solving these two equations gives

13.10: It has been worked out in section 13.5.3 (see figure 13.18). Carry it out further for

each axis step by step.

13.11 We show in the figure below the principal axes for the system

at the pivot point and the forces that act on the vertical rod at the bearings. Principal

mg

N2

N1

LH

LH

208

axis 1 is along the rod and 2 and 3 are perpendicular to the rod with axis 3 coming out

of the plane of the paper at the instant the rod is shown. In the free body diagram of

the rod, we have not shown the vertical forces. The cetre of mass of the system is

also indicated in the figure.

The moment of inertia about the principal axes are as follows

The angular velocity components are

Thus the angular momentum components are

The torque required to keep the rod rotating can be calculated either by Euler’s equations or

by the horizontal component LH of the angular momentum and multiplying it by .

Euler equations give

This is the same answer as obtained through LH as is easily checkd.

N2

d

12

d

LH

N1

209

The torque is provided by the reactions of the bearings. The difference in these reactions

also provides the centripetal force as the CM of then system is moving in a circle of radius

. Thus

Solving these two equations gives

13.12 In the figure below we show the principal axes for all the

three rigid bodies. Axes (1, 2) are in the plane of the paper and axis 3 is coming out of

the the paper in all three cases.

In all the three cases we have

and

Thus we have

(b)

1

2

(c)

1

2

(a)

12

210

Now in case (a)

This gives

In case (b)

. This gives

In case (c)

. This gives

13.13 As the ring rotates, the tension in the string does two three things: it balances the

weight of the ring, it provides the centripetal acceleration and it provides the torque

for the angular momentum change of the ring. The figure below shows the principal

axes of the ring at its CM and the free body diagram of the ring. Axes (1, 2) are in

the plane of the paper while axis 3 is coming out of the paper.

211

mgB

AT

O2

B

A1

The components of the angular velocity are

Therefore components of the angular momentum along the principal axes at the CM are

Thus the change in the angular momentum is times the horizontal component of the

angular momentum =

And it points in the direction of axis (3) i.e. coming out of the page at the instant shown.

Euler equations also give the same answer through

However, at the position shown, the tension and the weight of the ring give no torque about

the CM of the ring. Thus the internal forces will move the ring so that there is an opposing

change in the angular momentum. This ring moves so that its B end moves up. This

implies that the ring moves towards position 1. However, as it moves up, the tension

gets misaligned with the CM and starts giving a torque about it and finally the ring stops at

an angle such that the torque equals the angular momentum change.

Now balancing the forces gives

These equations give

If the ring moves up by an angle , The perpendicular distance of the tension and the CM

is about . This gives a torque

Equating this to the rate of change of angular momentum gives

212

This gives

13.14 Following the notation of example 13.7, we have (using parallel axis theorem for

calculating )

Given

(i) The presseion angular frequency will then be

Seeing the change in the angular momentum for the given sense of rotation, the top will be

coming out of the paper.

(ii) We see that . Therefore for the nutation motion we can write

with

The maximum difference between and 0 is therefore . Substituting all the numbers,

we get

(iii) The frictional force is needed to provide the centripetal force. Thus

13.15 The soulution of the Euler equations governing the motion

is the same as in

example 13.7 but the initial conditions are different. Thus we have

213

Here . Let the initial angular speed given about the vertical be . Then

the initial conditions for the subsequent motion are

These give

Thus we get

If the precession velocity of the top is then

This gives

And

214

Upon integration, the equation for gives

The variation of with time is similar to that shown in figure 13.23.

To get the angle through which the top has precessed, we substitute the value of in the

equation for and integrate it numerically. Finally to see the motion of the tip of the

axis of the top, is plotted against .

215

Chapter 14

14.1 For this spring rad s1

(i) giving

(ii) leading to

(iii)

(iv)

(v)

(vi)

14.2 It is given that . Thus the velocity is and the

acceleration is . This is shown below for different values of .

Displacement curve is the one with smallest amplitude of 5, the velocity curve has

216

intermediate amplitude of 10 and the acceleration curve has the largest amplitude of

20. Notice that phase difference of π and π give identical curves.

217

218

14.3 Equilibrium points of the potential are given by .

For , this gives

The roots of this equation are at x = 2 and x = 1. The second derivative of the potential at

these points is

Thus the potential is minimum at x = 2. The corresponding spring constant is

. This gives .

14.4

Equating the derivative to zero gives x = x0. At this point the second derivative

is positive and therefore at this point the potential is minimum.

(i) The corresponding spring constant .

(ii) To plot the potential and harmonic approximation to it, we have taken C=2, x0=1.0

and two values of a: a=0.4 and a=1.2

a = 0.4 a = 1.2

219

(iii)

(iv) As is clear from the figure, depending on the value of a the potential becomes softer

or harder for larger displacements from x = x0. Thus the frequency will become

smaller and time-period larger for a=1.2 and the frequency will become larger and

time-period smaller for a=0.4.

14.5

Equating the derivative to zero gives . At this point the second derivative

Thus the corresponding spring constant is and the frequency of small oscillations is

14.6 The surface charge density gives a electric field that pulls the

displaced electrons back, i.e. it applies a restoring force. We consider volume V of

the displaced block, the force on it is .

The total mass of the block is = number of electrons in volume V electron mass

220

=

So the equation of motion for the displaced block of electrons is (writing )

This gives the plasma frequency (here n is the number density of electrons)

14.7 In this problem

(i)

(ii) Velocity of platform and person on it is calculated by principle of linear momentum

conservation and gives

(iv) If we take the equilibrium point of (person+platform) as y=0, then the displacement

without the person on the platform is = . Thus this is a problem

with the initial conditions

Representin the motion as , we have

Thus the displacement is given by

14.8 In this case, it is the motion of a rigid body about a pivot point. Thus the equation of

motion is written in terms of the moment of inertia, angular acceleration and the

restoring torque. If the square is displaced by an angle about the vertical (see

figure) the torque is

221

The moment of inertia about the pivot is calculated by the parallel axis theorem and is

The quation of motion is

This can be rewritten as

This gives the frequency of oscillation to be

14.9 These two pendulums with their displacements are shown in the figure below.

a

mg

222

For visualizing the force applied by the spring, let us take . Then the bob on the left

feels a force to the right due to the spring and to the left due to the weight of the bob. The

spring force is in the opposite direction on the pendulum on the right.

(i) Since the rods are rigid, and therefre generate internal forces, we use the angular-

momentum torque equation for describing the motion of the pendulums. Thus the

equations of motion for the two pendulums are

(ii) The equations above are solved by adding them to get the equation for

and for . These are

These give

Combination of these two gives and . There are four unknowns that will be given

in terms of four initial conditions.

M

L

1

K

2

223

14.10 Angular displacement of the rod by and angle and the

corresponding angle that the strings make with the vertical are shown in the figure

below. Also shown are the tensions in the strings; these are equal by symmetry.

Since we are dealing with a rigid rod, the correct equation of motion to be used is the

angular momentum-torque equation. There are three unknowns in the problem: , and

T. We thus need three equations. These are

In the small angle approximation we have

Then

And the equation of motion becomes

Thus the frequency of oscillation is .

14.11 We show in the figure below a displacement of the rod when

the left spring is stretched by y1 and the right one by y2.

l

L

TT

224

The equation of motion for the CM is

with .

And the equation for rotation about the CM is

where we mesure counterclockwise as positive. Then

Thus we have for the CM

And for rotation about the CM

With the initial conditions and the rod held stationary in the beginning,

we have

This then gives the answer

L

y1y2yCM

225

14.12 Since we are writing the displacement as , the projection of the

phasor on the y-axis gives the displacement. The phase angle is measured from the x-axis.

In the following we show the position of he phasor at t = 0. After that it rotates

counterclockwise.

14.13 Assuming that the swing is performing simple harmonic

motion, the period of the swing is

Therefore the speed of the person on the swing as the swing passes through its equilibrium

point is =

(i) When the swing is at the extremes and the child is handed over, the amplitude will

not change. It is like strating a swing, ireespective of its mass, from a distance A

from the equilibrium with zero initial speed. Energetically it can be seen as

follows:

When only the man is on the swing the total energy is =

When the child is handed over to the person at the extreme, the child brings in additional

potential energy (with respect to the equilibrium point)

= Mchildgthe height of the extreme point with respect to the equilibrium point

(i) (ii) (iii)

(iv) (v) (vi)

226

Thus the total energy that the swing posseses after the child is handed over is

If the new amplitude is Anew, this should equal (keep in mind that remains unchanged)

A comparison in the total energies calculated in two ways immediately gives .

(ii) When the child is handed over at the equilibrium point, the speed os the swing decreses

by the conservation of linear momentum. It is

And the total energy of the system is now

If the new amplitude is Anew, then (keep in mind that remains unchanged)

This gives .

(iii) As is clear from the calculations above, energy is conserved in case (i) only.

14.14 The equation of motion now will be

Its general solution is

With the initial conditions , the solution is

Since the velocity of the block is

227

It becomes zero at , At this time

14.15 (i) Since this is the motion of a rigid body pivoted at a point

under an external force, its motion is described by the angular momentum-torque equation.

The torque on the body is applied by the weight acting at its CG, and it is a restoring torque

(see figure). The value of the torque about the pivot point is

The moment of inertia of the rod about the pivot point is

Thus the equation of motion for the rod is

For small angles this equation bocomes

(ii) As the pendulum swings about, the frictional torque causes it to lose energy and

therefore the amplitude of the pendulum decreases slowly. The equation of motion

mg

228

can be written and solved exactly as in the problem above. When the pendulum in

the figure above is moving clockwise, the equation of motion is

With the initial conditions , the solution of the equation above is

(just like in the problem of spring-mass system with friction)

Like in the problem above, this then leads to a reduction in the amplitude by by the

time the pendulum reaches the left extreme. Thus in one cycle the amplitude reduces by

. We will now derive this result from energy considerations also.

Suppose at a certain instant the pendulum starts with the maximum displacement of and

goes to the maximum angle on the other side. Then the total angle covered by it is

and the work done against the frictional torque is therefore . The energy

loss during the motion is . Equating this to the work done against

friction gives

Thus in one cycle the amplitude reduces by .

Now if the pendulum starts with an angle and completes N cycles, it will stop swinging

when it is at angle such that . Thus we have

14.16 As the mass leaks out, it carries with it the energy at the rate

. If at time t, the amplitude is A(t) and the frequency (t), then

229

and the average rate of loss of energy E(t) of the

oscillator is

In time averaging A(t) and (t) come out of the integral because they vary slowly over a

few oscillations (that is precisely why they can be defined). The negative sign shows that

the energy is being lost with time.

And if amplitude is A(t), the energy and the rate of change of energy

. Combining all this gives

If the amplitude at t = 0 is A0, this equation is integrated as

This gives

14.17 The energy loss is given by the formula

Thus the average energy loss is

If the damping factor of the system is , then where the energy E is

. Thus for the current problem we get

230

14.18 We consider an oscillator with initial conditions so that its

motion is given by

For undamped oscillatot the siplacement is maximum when . For the damped

oscillator, it will be given at time when

Since Q is very high, the maximum occurs when where is very small. Since

the maximum occurs before , it is advanced by phase angle . Thus

For small , we have

This gives .

14.19 If the person gives an impulse J, the energy gained by him is

Jv, where v is the speed at the equilibrium point. Since , we have

Energy gained per cycle =

This will equal the energy loss per cycle due to damping.

Energy loss per cycle = 2πenergy loss per radian =

Here E is the energy of the swing. This follows from the definition of the quality factor.

231

Thus we get

Putting in all the numbers gives

14.20 It is given that and the mass of the platform is

50kg. Since the system is critically damped for 500kg put on the platform, we have

So the coefficient of drag force

Thus when a 200kg weight is on the platform, the value of

And the angular frequency

Thus , which makes the system hevily damped.

The solution then is

with

So

Here A and B are to be determined by the initial conditions.

Initial conditions:

Since we are taking the equilibrium position of the (load+platform) as y = 0, the initial

displacement from that position is .

(i) The speed of the load when it lands on the platform is .

Then by momentum conservation, the load+platform will move with the speed

.

232

(ii) From the above, the initial conditions are .

Thus we have from the solution above

Thses give

The solution therefore is

14.21 For a forces oscillation the angle by which the displacement

lags the applied force is given by equations

For a very heavily damped oscillator . Thus we have

This gives

233

Chapter 15

15.1 We perform our calculations in a noninertial frame attached with the box and take the

origin (x = 0) at the point where the spring is connected to the box. In this frame,

there is a pseudo force ma acting in the negative x direction. Thus the system in the

noninertial frame looks as shown below.

The equilibrium point x0 is given by

Thus the equation of motion for the mass in this frame is

At , the initial conditions are

We change the variables to and write the above equation as

with the initial conditions This gives the

soulution

15.2 This problem is very easy to solve in the accelerating frame attached with the car.

The free body diagram of the door in the car frame with pseudo force included is

shown below.

ma x = 0

l

234

In the car frame therefore the door is being pulled at its CM by a force Ma and rotates

about the hinges. Thus the problem becomes like examples 12.7 and 15.3 where a

ruler/rod, pivoted at one of its ends’ rotates about an axis at that point when the pivot point

starts accelerating with an acceleration. Then by energy conservation this gives the answer

for the angular speed of the door when it is about to close

15.3 This problem is similar to problem 12.7 and the problem above and is solved in

exactly the same manner.

15.4 We substitute in

to get

Now from equation 15.8 and so we have

Since , the above equation is

Again substituting from equation 15.8 and ,

we get

car

Ma w

235

which gives

15.5 In the rotating frame the rod is stationary and there is a pseudo force – the centrifugal

force – acting on it. Thus the free-body diagram of the rod in the rotating frame is as

shown below. Notice that we are not taking the centrifugal force to be acting at the centre

of mass because it is different for different portions of the rod; it increases with the distance

of fom the axis.

(i) It is clear from the figure above that the rod is pulled out horizontally by the

centrifugal force (in the rotating frame) and would therefore move out when

displaced from the vertical position. Since we are dealing with a rigid body, we

should be working in terms of totques. Thus at the given position, the centrifugal

force applies a counterclockwise torque on the rod but at the vertical position,

neither its weight nor the force by the pivot apply any torque. Thus the rod will

tend to swing away from the axis. It will be clearer mathematically in part (ii).

(ii) When the rod is at an angle from the vertical, the component of force on a portion

of length ds at distance s from the pivot is (see figure).

Centrifugal force

mg

236

Thus the torque due to the centrifugal force is

At the same time there is an opposing torque due to gravity and its value is

Balancing these two gives

Thus as long as , the rod will be at equilibrium at the angle whose

cosine is given above. Thus when disturbed from the vertical position it will move out to

this equilibrium position (with very small friction damping the motion but not affecting the

equilibrium position). For angular speeds such that , the rod will

not move out because the torque due to the centrifugal force will not be able to overcome

the torque due to the rod’s weight.

15.6 This problem is similar to the problem above except that there

is no gravitational force involved here. If looked at from the top along CD, the free

body diagram of the rod in the rotating frame, when it is at an angle from the frame,

looks as shown below.

mg

s

237

If we take a strip of width ds at a distance s from the axis CD, the mass of the strip will be

and the centrifugal force on it = . The net torque on it is

therefore

This shows that both and are equilibrium positions of the sheet.

(i) Thus when disturbed slightly , the torque on the sheet is

and tends to take it away from that position. The equation of motion for slight

disturbance from the position parallel to the frame is therefore

This gives

Therefore the equilibrium position at is an unstable equilibrium position.

(ii) If we call the angle from the perpendicular position of the sheet then .

The expression for torque in terms of remains

However, as the sheet is disturbed from this position, the torque has restoring nature. For

small therefore the equation of motion is

A Bs

238

Thus the motion about this position is oscillatory and is given by

or equivalently by

Therefore the equilibrium position at is a stable equilibrium position.

15.7 We follow the convention for the directions as in the main

text. There is no force in the horizontal plane. The equations of motion in the northern

hemisphere are therefore

If , vx and vy would remain zero for a particle thrown up. Thus correct to the linear

order in , the equations above are

and

This gives

and

Upon integration, the second equation gives

Straightforward approach to solve for the deflection by the time the particle reaches the earth

again would be to find the time for it to come back to the ground and substitute it in the equation

for x(t). We have

,

Time taken to reach the highest point

239

And the total time of flight = 2T = 9.04s.

Thus the total deflection by the time the stone comes back to the ground is

cm

The negative sing in front indicates that the deflection is towards the west. This is an interesting

result. One’s immediate answer would normally have been that the stone deflects one way while

going up and the other way while coming down, since the sign of vz changes, the net deflection

would have been zero. This does not happen because by the time the stone reaches its highest

point, it already has a westward velocity. To see its effect, let us calculate the distances in two

steps: first when the stone goes up and the second when it comes down.

While going up, the deflection of the particle by the time it reaches the highest point is

At this point it has the horizontal velocity

cm s1

The negative sing in front indicates that the velocity is towards the west. Now, as shown in

example 15.9, if a stone is dropped from a height of 100m, it deflects to the east by cm.

However, now the stone has an initial westward speed of cm s1. Thus the nest

deflection fo th stone while it is coming down will be

This gives a total deflection of during the entire flight.

15.8 The initial components of the velocity in the east, north and vertical direction are

Equations of motion in are

240

Since is very small, we will do our calculations correct to the first order in . This means that

for vx and vy, we substitute their initial values and for vz we substitute in the

equations above. This gives

Integration of the last equation, with proper initial conditions , gives

For time of flight T calculation, we substitute z = 0 and get

Thus the time of flight of the projectile increases. Now we integrate the equation for motion in

the x direction, with the initial conditions to get

Similarly, integration of the equation for the motion in the y direction gives

241

Thus the deflection of the projectile in the east and north direction are known. As a check, we

take a projectile thrown staright up and find that the answer matches with that obtained

in problem 15.7. To know the deflection in the direction of launch and perpendicular

to it, we transform these as (see figure)

This transformation gives

To calculate the range and deflection from the path during the flight, we substitute the time of

flight T for t. To the first order in , we then have

X

Y

X’

Y’

α

O

242

Therefor the range, up to the first order in is

So the change in the range due to the coriolis force is .

The sideways deflection is given by

Thus the sideways deflection is .

This shows that a projectile fired eastward will deflect towards the south and that fired

westward will deflect to the north. If we look at the equations of motion in the

northern and the southern hemisphere, they differ by the sign of the terms containing ,

which is equivalent to changing the sign of λ to λ. From the expression derived above, it is then

clear that the change in the range will not change but the sideways deflection will change to

in the southern hemisphere.

15.9 Exact solution of the equations of motion on the surface of the earth. Thse are

243

To get the exact solutions, we differentiate the first equation once and get

Substituting for the derivatives of the velocity components appearing on the right, we get

The solution of this differential equation is given by the sum of the solution for its homogeneous

part and the particular solution. Thus it is

Here A and B are determined by the initial conditions. Subztituting this solution in the equation

for vy gives

Here C is another constant. Similarly from the last equation we get

Here D is another constant.

No wlet us determine the constants appearing in the equations above for a stone dropped from

height h. Since we have solved a second-order differential equation for vx , we need two initial

condtions for it. These are . Then we have

244

This gives

For the y and the z components of the velocity we have . These give

C = 0 and D = 0. Thus the complete solution for the velocity is

In the limit of this gives, to the first order in ,

This leads to the same answers as in example 15.9.

15.10 Cyclone building up in the Bay of Bengal. If only the motion in the horizontal plane is

considered, the equations of motion are

We choose the coordinate system such that its origin is where the storm starts from. Thus the

initial conditions are . The second

equation above can be immediately integrated to obtain

245

where C is a constant. Since , we get C = 0. Thus

We substitute this in the first equation above to get

This gives

With the initial conditions , this gives

Thus

With the initial conditions we get . Thus finally we have

for the coordinates of the eye of the cyclone

and

(i) This gives the equation for the trajectory as

Since , the above equation can also be written in

kilometers as

246

Thus the trajectory is a circlular one with the centre at km.

(ii) Thus the sense of the trajectory is clockwise as seen from above. On the other hand, since

we are in the northern hemisphere, the winds around the eye move in counterclockwise direction.

(iii) To find out where the cyclone hits the coast, we substitute in the equation above

and get and . The second coordinate is for the return

path. Thus the cyclone hits the coast 18km north of where it started from.

15.11 The solution here is similar to the solution above except for the initial conditions. We

take the origin to be where the striker is played from. The x and the y axes are shown in the

figure below.

For the problem . Thus the equation of motion in the xy plane are

1 2

A

B

x

y

247

Note that the signs on th right hand side are opposite to what they are in the equations of motion

in the orther hemisphere of the earth because the direction of the angular velocity is opposite.

Assuming that the striker is played at an angle from the x axis, the initial conditions are

The second equation above is integrated to give

With the initial condition , we get . Thus

Substituting this in the first equation of motion, we get

The solution for the equation is then obtained as the sum of the solution for the homogeneous

part and the particular solution. Thus

With the initial condition , we get . Thus

When substituted in , this gives

Solved with the initial condition y(0) = 0, this gives

Thus the trajectory is described by the coordinates

These can be combined together to give the equation of trajectory as

248

Thus the trajectory is a circle with centre at and the sence of direction is

therefore counterclockwise. This implies that the striker should be played in direction 2. Now

we want the trajectory to be such that it pass through point B, i.e. through coordinates .

When substituted in the equation for the trajectory, this gives

If the angle with AB is α, then

The striker is played in direction 2 making an angle from AB.

(ii) We first solve this problem for correct to first order in in an easy way. We then

solve the problem exactly.

Easy approximate solution correct to first order in :

It is given that initially and . Thus as we integrate the equations above and

therefore . So up to order , vy can be taken to be a constant equal to v. Thus

and the integration of the first equation above gives

Since , we get C = 0. This gives

because x(t=0)=0.

Therefore by the time the striker reaches the other end, time taken by it is and its has

moved a distance and has x-component of the velocity . Now the

249

striker starts its journey back after hitting then board. It again takes time to reach the side

it started from. Now the initial x-componet of the velocity is and the displacement

. Now if we integrate the equation with these conditions, then

This gives

Keep in mind that t is being measured from T onwatds. Thus t = 0 inplies when the striker starts

its journey back. Integrating this with the initial condition , we get

For this gives

This is the final displacement when the striker comes back. One would have thought that the

striker will come back to its original position; that does not happen because when it starts its

journey back, it has an x-component of velocity that makes it drift. Had the x-component been

zero when it started back, the striker would have reached its original position.

Exact solution:

If the striker is played along line AB then . The trajectory of the striker is then

and with time x(t) and y(t) vary as

Thus for y = L, we have from the trajectory equation

250

Thus the striker gets displaced in x direction as it reached point B. One of the two answers

(smaller magnitude ) is when the particle is moving upward. The other

one (larger magnitude ) is when the striker would have returned had it

not hit the other side of the board. Further, when the striker reaches the other end, it would have

taken time T such that y(T) = L. This gives

The x and y components of the velocity at the other end are therefore

From the above, one sees that the speed remains unchanged during the motion. This is

expected since the coriolis force is perpendicular to the velocity and therefore does no work on

the striker. As the striker hits the opposite side of the carom and returns back, its x velocity

remains unchanged and the y component of the velocity changes direction and it traverses a new

trajectory. Taking the time when it starts back to be t = 0, the new trajectory is thus determined

by the initial conditions

One could solve this problem with these initial conditions. However, we can make use of the

equations already derived for the trajectory if we describe the journey back with new coordinate

system with the origin at the point where the striker starts back. This is shown in the figure

below. In this coordinate system the initial conditions are

So that we have initial speed v at an angle such that

251

Thus the x and y displacement of the striker in the new coordinate system is

These can be combined together to give the equation of trajectory as

For y = L this then gives

While cutting the y = L line for the first time, the value of x will be

This is in addition to the previous displacement of the origin in journey. Thus the total

displacement is

x

y

v

241

2 2

22 vv

Lvx

A

B

252

It is easily seen that for this goes over to the approximate answer obtained earlier.

253

Appendix A

A.1 In this problem there are four variables. So we make one dimensionless variable and equate

it to a constant. The variables of the problem and the corresponding dimensions are

Let the dimensionless variable be . Since this is a dimensionless variable we have,

on equating the dimensions of the right hand side in the equation above to zero

Since we want an expression for p, we take a = 1. Then the equations above give

and . Therefore the dimensionless variable is

Equating it to a constant k, we get

A.2 This problem has three variables. Therefore to apply Buckingham’s π theorem we form one

dimensionless variable and equate it to a constant. The variables of the problem and the

corresponding dimensions are

Let the dimensionless variable be . On equating the dimensions of the right hand

side in the equation above to zero we get

Taking a = 1, since we want an expression for p, we get and . Therefore the

dimensionless variable is

Equating it to a constant k, we get

254

A.3 This problem has four variables. Therefore to apply Buckingham’s π theorem we form one

dimensionless variable and equate it to a constant. The variables of the problem and the

corresponding dimensions are

If the dimensionless variable is , on equating the dimensions of the right hand side

in the equation above to zero we get

Since we want an expression for E, we take a = 1. Then the equations above give

and . Therefore the dimensionless variable is

Equating it to a constant k, we get

A.4 This problem has four variables. Therefore we form one dimensionless variable and equate

it to a constant. The variables of the problem and the corresponding dimensions are

If the dimensionless variable is , on equating the dimensions of the right hand side

in the equation above to zero we get

Since we want an expression for , we take a = 1. Then the equations above give and

. Therefore the dimensionless variable is

Equating it to a constant k, we get

255

A.5 This problem has five variables. Therefore to apply Buckingham’s π theorem we form

twodimensionless variables and write one of them as a function of the other. The variables of the

problem and the corresponding dimensions are

If the dimensionless variables are expressed as , on equating the

dimensions of the right hand side in the equation above to zero we get

These give and . Since we want an expression for , we form one of

the dimensioless variables with a = 1. Then the equations above give and

. Taking d = 0 gives giving one dimensionless variable as

For the other dimensionless variable, we take a = 0. This gives b = c = 0 and . Taking e =

1, the second dimensionaless variable is

Thus the functional relationship between the variables can be expressed as

Now we have several choices for the function f. If we take it to be a constant, we get

But this is not consistent with observations. We next take where k is a constant. Then

we get

This gives answer consistent with observations. For corrections to higher orders, we can take

higher powers of and construct higher order functional relationships.

256