physics - 1 (lecture - 2) · section1 newton’s law of motion santabrata das (iitg) physics - 1...

PHYSICS - 1 (Lecture - 2)

Santabrata Das

Indian Institute of Technology Guwahati

[email protected]

August 6, 2014

Santabrata Das (IITG) PHYSICS - 1 (Lecture - 2) August 6, 2014 1 / 56

Section 1

Newton’s Law of Motion


First Law of Motion

Isolated Bodies Move With Uniform Velocity.

Common Notion: All moving bodies come to a halt.Isolated BodiesInertial Frames


Second Law of Motion

F = ddt (mv) = ma

It is a law, not a definition!Mass is constant and additiveForce is in the direction of change of velocity

Force: a spring force

L

Force = L − L0

L’

Force = L’ − L0



Mass is a measure of inertiaExternal agency that alters the state of motionForce is a vector quantity F =

∑iFi

��

F2

F1

F1 + F2

Arises out of interaction between systems.



Therefore, isolated bodies ⇒ No interaction ⇒ No force ⇒ motion withuniform velocity.

Question: Is there laws for the forces themselves?Answer: YES, in some cases.

GravityElectrostaticMass on a spring

But, for comlex systems, it is difficult to have a complete idea aboutlaws of the forces

three body motioncollision of two automobilesmolecular motion of a gas


Third Law of Motion

Every action has equal and opposite reaction

Describes the nature of forcesValidates the concept that force is necessarily the result ofinteraction between systems.Is it always true?There is some problem with electromagnetic forces.


Fundamental Forces

Gravitational ForcesElectromagnetic ForcesWeak Nuclear ForcesStrong Nuclear Forces


Gravitational Forces

Tycho Brahe and KeplerNewton’s Law of Gravitation

F = Gmm′

r2r

Force is along the line of centres and is attractiveInertial or gravitational Mass?Explains planetary Motion (Large distances)Cavendish Experiment (On Earth)


Electrical and Magnetic Forces

Coulomb’s Law (static charges)

F =qq′

r2r

Magnetic Force (moving charges)Fields E and B

Lorentz ForceF = q (E+ v ×B)


Nuclear Forces

No known law for nuclear forcesRange of Nuclear Forces ∼ 10−13cmScattering ExperimentsNewtonian or Quantum Mechanics?

Quantum mechanical laws are valid with such small particles at atiny distance.


Applications of Newton’s Laws

Mentally divide the system into smaller systems, each of which canbe treated as a point massDraw a force diagram for each massFix the coordinate systemUse third law whenever necessaryFind out constraints and construct the relevent equationUse second lawIdentify the number of unknown quantities. There must be enoughnumber of equations ( Equations of motion + third law pairs +constraint equations) to solve for all the unknown quantities.


Everyday Forces

Tension in the ropesNormal ForcesFrictional ForcesViscous ForcesSpringsAtomic Forces


Forces

Normal ForceForce exerted by a surface on a body in contact with it can beresolved into two componentsComponent perpendicular to the surface is called Normal forceN→ equal & opposite to the resultant of all other forcesperpendicular to the surface

Frictional forceFor bodies not in relative motion, 0 ≤ f ≤ µN ; f opposes themotion that would occur in it’s absenceFor bodies in relative motion, f = µN ; f is directed opposite torelative velocity.


Constraints: Example IA block moves on a wedge which in turn moves on a horizontal table.The wedge angle is θ. How are the accelerations of the block and wedgerelated?

hx

Xy

θ

θ

x − X

h − y

Constraint - I: Vertical accelerationis zero.Constraint - II:

x−X = (h− y) cot θ

x− X = −y cot θ

Coordinates are fixed on groundEquations are independent of forces


Example 2The two blocks M1 and M2 shown in the sketch are connected by astring of negligible mass. If the system is released from rest, find howfar block M1 slides in time t. Neglect friction.

M1

M2

g

x

Equations of motion:

M1x1 = T

andM2x2 =W2 − T

Constraint equations:

L− x1 + x2 = const⇒ x1 = x2

Unknowns: x1, x2, T and three equations.


Example 2

Solve three equations simulteneously and obtain the solution as,

M2g = (M1 +M2)x1

x1 =M2g

M1 +M2

After integration,

x1 =M2gt

2

2(M1 +M2)


Example 3

A particle of mass m slides without friction on the inside of a cone. Theaxis of the cone is vertical, and gravity is directed downward. The apexhalf-angle of the cone is θ.

mυ0

θ

The path of the particle happens to be a circle in a horizontal plane.The speed of the particle is v0. Draw a force diagram and find theradius of the circular path in terms of v0, g, and θ.


Example 3

Force diagram is self-explanatory.Equations of motion:

N sin θ = W

N cos θ = mrω2

Dividing them yields

tan θ =W

mrω2=

mg

mrω2=

g

rω2

Speed of mass m is v0 = rω

tan θ =gr

v20⇒ r =

v20 tan θ

g


Example 4

Masses M1 and M2 are connected to a system of strings and pulleys asshown. The strings are massless and inextensible, and the pulleys aremassless and frictionless. Find the acceleration of M1.

M1

M2


Example 4

The fixed length of the string is a constraint.


M1x1 = T −M1g

M2x2 = T ′ −M2g

Unknowns: x1, x2, T, T ′ and we require four equations to solve theproblem.


Example 4

Net force is 0⇒ 2T ′ = T

Then, from Eqns of motion, we have

T = M1x1 +M1g

M2x2 =T

2−M2g

Constraint: Fixed length of the string is a constraint.

x1 + l1 + l′1 +(x2 + l2 + l′2)

2= constant

x2 = −2x1


Example 4

Upon solving, we get

−4M2x1 = T − 2M2g = M1x1 +M1g − 2M2g

x1 =(2M2 −M1) g

(4M2 +M1)

Acceleration of M2 is twice the rate of M1 and the weight of M1 iscounterbalanced by twice the weight of M2.When

M1 �M2 ⇒ x1 ≈ −g and x2 ≈ −2g

and ifM2 �M1 ⇒ x1 ≈

g

2and x2 ≈ −g


Example 5

A mass m is connected to a vertical revolving axle by two strings oflength l, each making an angle of 45o with the axle. Both axle and massare revolving with angular velocity ω. Gravity is directed downward.Find the tension in the upper string, Tup, and lower string, Tlow.

ω

l45°

45°l

m

Radial distance of m from the axle is l/√2 as cos 45o = sin 45o = 1/

√2


Example 5


Tup√2

= mg +Tlow√

2→ V ertical

mlω2

√2

=Tup + Tlow√

2→ Radial

Unknows: Tup, Tlow and two equations.Solving we get,

Tup =mlω2

2+mg√2

Tlow =mlω2

2− mg√

2


Example 6A disk rotates with constant angular velocity ω. Two masses, mA andmB, slide without friction in a groove passing through the center of thedisk. They are connected by a light string of length l, and are initiallyheld in position by a catch, with mass mA at distance rA from thecenter. Neglect gravity. At t = 0 the catch is removed and the massesare free to slide. Find rA immediately after the catch is removed, interms of mA, mB, l, rA, and ω.

mB

mA

w

lra


Example 6Blocks are constrained by the grooveTangential motion plays no dynamical role, neglected in force diag.Force T on each mass is radially inward.


−T = mA

(rA − rAω2

)−T = mB

(rB − rBω2

)Constraints: rA + rB = l⇒ rA = −rB

Solving,

mA

(rA − rAω2

)= mB

(rB − rBω2

)= mB

[−rA − (l − rA)ω2

]rA = rAω

2 − mBlω2

(mA +mB)


Example 7: Whirling blockA horizontal frictionless table has a small hole in its center. Block A onthe table is connected to block B hanging beneath by a string ofnegligible mass which passes through the hole. Initially, B is heldstationary and A rotates at constant radius r0 with steady angularvelocity ω0. If B is released at t = 0, what is its accelerationimmediately afterward?

A

B

z

T

T

MA

MB

WB

θ

z

r

Two movable bodies and their free body diagram


Example continued ...Equations of motion:

−T = MA(r − rθ2) Radial0 = MA(rθ + 2rθ) Tangential

WB − T = MB z Vertical

Constrain equations: r + z = `⇒ r = −z.Unknowns: ar, aθ, z, T . Four unknowns and four Eqs.

z =WB −MArθ

2

MA +MB

Immediately after B is released, r = r0 and θ = ω0. Hence,

z(0) =WB −MAr0θ0

2

MA +MB


Example 8: Block on wedge with frictionA block of mass m rests on a fixed wedge of angle θ. The coefficient offriction is µ. Find the value of θ at which the block starts to slide.

fN

m

W

y

x

θ

θ

Coordinate system is fixed onthe wedge.

Constraint:

my = 0

Eqns of motion:

mx = Wsinθ − fmy = N −Wcosθ

= 0


Example 8: Block on wedge with friction

When sliding is about to start: f = fmax = µN ; x = 0.Eqns of motion:

W sin θmax = µN

W cos θmax = N

tan θmax = µ

As the wedge angle is gradually increased from zero, the friction forcegrows in magnitude from zero to its maximum value µN , since beforethe block begins to slide we have

f =W sin θ θ ≤ θmax


Example 8: Block on wedge with frictionNow, the wedge is given horizontal acceleration a. Assuming thattan θ > µ, find the minimum acceleration for the block to remain on thewedge without sliding.

aθ

g

Horizontal acceleration of the block is

mamin = N cos θ − f sin θ; f ≤ µN (1)

In the limit f = µN , we have

mamin = N (cos θ − µ sin θ)

Block has zero Vertical acceleration

N sin θ + f cos θ −mg = 0

N (sin θ + µ cos θ) = mg (2)


Example 8: Block on wedge with frictionEleminating normal force N from Eqns (1) and (2) we get

amin =

(cos θ − µ sin θsin θ + µ cos θ

)g

We can estimate the maximum acceleration also.

Horzontal acceleration of the block

mamax = N (cos θ + µ sin θ) ; f = µN

Vertical acceleration of the block

N sin θ − f cos θ −mg = 0

As earlier,

amax =

(cos θ + µ sin θ

sin θ − µ cos θ

)g


Examples 9

A 4 Kg block rests on top of a 6 Kg block, which rests on a frictionlesstable. Coefficient of friction between blocks is 0.25. A force F = 10N isapplied to the lower block.

F

4 Kg

6 Kg


Examples 9: Force Diagrams

4 Kg

6 KgF

F1

N

Wa N

F1

N’

Wb

N’


Examples 9: Coordinate System andConstraints

Fix the coordinate system to the table.

4 Kg

6 KgA

xB

x

yA

yB

yA = constyB = constxA = xB + const


Examples 9: EquationsEquations of Motion in Y direction.

mAyA = N ′ −WA −NmB yB = N −WB

Constraints

yA = 0

yB = 0Solution

N ′ = WA +WB

N = WB


Examples 9: Equations

Equations of Motion in X direction.

mAxA = F − F1

mBxB = F1Constraints

xA = xBSolution

xA = xB =F

mA +MB= 1 m/s2

F1 = mBxB = 4N


Examples 9: Example Continued ...

The force F1 < µN = 10 N, the maximum frictional force betweenthe blocks. Hence the solution is consistent with assumption.What would be the motion if F = 40 N?If the blocks move together then xB = 4 m/s and F1 = 16 N! Morethan the maximum frictional force!


Examples 9: EquationsEquations of Motion in X direction.

mA xA = F − F1

mB xB = F1

But,

F1 = µN

Solution

xA =F − µNmA

= 5 m/s2

xB =µN

mB= 2.5 m/s2


Example 10

A block of mass m slides on a frictionless table. It is constrained tomove move inside a ring of radius l fixed to the table. At t = 0 theblock is touching the ring and has a velocity v0 in tangential direction.

��

��

��

��

m

l


Example 10: Equations

Constraint Equation is r = l, that is r = r = 0.Equations of Motion

m(r − rθ2

)= −mlθ2 = −N

m(rθ − 2rθ

)= mrθ = −f

Eliminating N , we get

θ = −µθ2

v(t) = lθ

=v0

1 + µv0t/l


Tension in a string

A string consists of long chains of atoms characteristic of the particularmaterial found in the string. When a string is pulled, we say it is undertension. The long chains of molecules are stretched, and inter-atomicforces between atoms in the molecules prevent the molecules frombreaking apart. A detailed microscopic description would be difficultand unnecessary for our purpose. Instead, a macroscopic model wouldbe developed for the behavior of strings under tension.


Block and StringA string of mass m attached to a block of mass M is pulled with forceF . What is the force F1 that the string “transmits” to the block?Neglect gravity.

Equations of motion are:

F1 =MaM ; F − F1 = mas

String is inextensible: as = aMBy Newton’s third law: F1 = F ′1

a = as = aM =F

M +m

and

F1 = F ′1 =M

M +mF

Mm

aMaS

F1 F ′1 F

F

M


Block and String

Mm

aMaS

F1 F ′1 F

F

M

The force on the block is lessthan FThe string does not transmitthe full applied force.If the mass of the string isnegligible compared with theblock, F1 = F to goodapproximation.


Continued...

A string is composed of short sections interacting by contact forces.Each section pulls the sections to either side of it, and by Newton’sthird law, it is pulled by the adjacent sections. The magnitude of theforce acting between adjacent sections is called tension. There is nodirection associated with tension. In the sketch, the tension at A is Fand the tension at B is F ′.

F F F ′ F ′

A

A

B

B

Although a string may be under considerable tension, for example astring on a guitar, the net string force on each small section is zero ifthe tension is uniform, and the section remains at rest unless externalforces act on it. If there are external forces on the section, or if thestring is accelerating, the tension varies along the string.


Dangling Rope

A uniform rope of mass M and length L hangs from the limb of a tree.Find the tension in the rope at distance x from the bottom.

The section is pulled up by a forceof magnitude T (x), where T (x) isthe tension at x. The downwardforce on the section is its weightW =Mg(x/L). The total force onthe section is zero since it is at rest.

T(x)

W

L

x

x

Hence T (x) = MgL x.

At the bottom of the rope the tension is zero, while at the top wherex = L the tension equals the total weight of the rope Mg.


The Spinning TerrorThe Spinning Terror is an amusement park ride—a large vertical drumthat spins so fast that everyone inside stays pinned against the wallwhen the floor drops away. What is the minimum steady angularvelocity ω that allows the floor to be dropped safely?

R

ω

Radial Equ. N =MRω2.

By the law of static friction,f ≤ µN = µMRω2

Since we require M to be invertical equilibrium, f =Mg.

M N

f

ω

W = Mg

Therefore, Mg ≤ µMRω2 ⇒ ω2 ≥ gµR

⇒ ωmin =√

gµR


Example

Consider the force as a function of time, position, or velocity, and solvethe differential eq.

md2x

dt2= F (x, t, v)

to find the position, x(t), as function of time

Since we need to solve second order differential equations, we must havetwo initial conditions, usually x0 ≡ x(t0) and v0 ≡ v(t0), to obtain thefinal solutions.

Three special cases: F is function of t only, function of x only, andfunction of v only.


Example

F is a function of x only: F = F (x)

We will use

a =dv

dt=dx

dt

dv

dt= v

dv

dt

Following

F = ma ⇒ mvdv

dx= F (x)

As example,

F = −kx or F = −k/x3


Example Continued ...

Now separate the variables and integrate both sides to obtain

m

ˆ v(x)

v0

v′dv′ =

ˆ x

x0

F (x′)dx′

Separating variables in dx/dt ≡ v(x), yieldsˆ x(t)

x0

dx′

v(x′)=

ˆ t

t0

dt′.


Inverse-cube force field

A particle of mass m starts from rest at x0(> 0) in an attractiveinverse-cube force field F = −k/x3 (k is a positive constant). Calculatethe time taken to reach the origin.The equation of motion is,

mdv

dt= mv

dv

dx= − k

x3

m

ˆ v(x)

0vdv = −k

ˆ x

x0

dx

x3

v(x) =dx

dt= −

√k

mx30

√x20 − x2x

‘-’ sign for motion in ‘-’ x-direction.


Inverse-cube force fieldIntegrating the velocity Eq. with respect to t,

ˆ x(t)

x0

xdx√x20 − x2

= −

√k

mx20

ˆ t

0dt

x(t) = x0

√1− k

mx40t2

‘+’ root as particle is to the right of x > 0.

For t = T0, x = 0, then T0 =

√mx40k and

v(t) = − k

mx30t

(1− k

mx40t2)−1/2


Free motion in viscous mediumA body of mass m released with velocity v0 in a viscous fluid is retardedby a force Cv. Find the motion, supposing that no other forces act.

F = −Cv ⇒ −Cv = mdv

dtdv

v= −C

mdt

ˆ v

v0

dv

v= −

ˆ t

0

C

mdt

logv

v0= −C

mt

v = v0exp(−C/m)t

Let τ = m/C, then v = v0e− t/τ . τ is a characteristic time for thesystem; it is the time for the velocity to drop to e− 1 ≈ 0.37 of itsoriginal velocity.


Free motion in viscous mediumTherefore,

dx

dt= v0e

−t/τ

Hence,

x =

ˆ t

0

v0e−t/τdt

x = v0τ(1− e−t/τ

)As τ →∞, C → 0, means no resistance. Then x = v0t.[

e−x = 1− x+ 1

2x2 − 1

6x3 + ...

]Santabrata Das (IITG) PHYSICS - 1 (Lecture - 2) August 6, 2014 55 / 56

physics - 1 (lecture - 2) · section1 newton’s law of motion santabrata das (iitg) physics - 1...

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