MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01 Fall 2012

Problem Set 8 Rigid Body Fixed Axis Rotational Motion

Due: Thursday Nov 8 at 9 pm. Place your solutions in the appropriate section slot in the box outside 26-152. Write your name, section, and table number on the upper right side. Week 9 and 10 Schedule and Reading Assignments: Sections 1-4 (Monday-Wednesday) Oct 31 W09D2 Two-Dimensional Rotational Kinematics Reading Assignment: Suggested: Young and Freedman: 9.1-9.6 8.01 Course Notes: Rigid Body Kinematics: Fixed Axis Rotation http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_kinematics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 9.1-9.6 Tuesday Oct 30 Problem Set 7 Due 9 pm Nov 2 W09D3 Two Dimensional Rotational Dynamics Reading Assignment: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3 Nov 5 W10D1 Two-Dimensional Rotational Dynamics Hour Two, Angular Momentum Reading Assignment Part One: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3 Reading Assignment Part Two: 8.01 Course Notes: W010D1 Angular Momentum http://web.mit.edu/8.01t/www/materials/modules/CN_W10D1_angular_momentum.pdf Suggested: Young and Freedman Chapter 10.5-10.6 Sections 5-7 (Tuesday-Thursday) Oct 30 W09D2 Two-Dimensional Rotational Kinematics 8.01 Course Notes: Rigid Body Kinematics: Fixed Axis Rotation

http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_kinematics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 9.1-9.6 Nov 1 W09D2 Two-Dimensional Rotational Dynamics Reading Assignment: 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3 Nov 2 W09D3 Problem Solving 07: Two Dimensional Rotational Dynamics Reading Assignment: Review 8.01 Course Notes: Rigid Body Dynamics: Fixed Axis Rotation http://web.mit.edu/8.01t/www/materials/modules/CN_W09D1_rigid_body_dynamics_v01.pdf Suggested: Young and Freedman: 1.10 (Vector Product), 10.1-10.2, 10.4, 11.1-11.3 Nov 6 W10D1 Angular Momentum (Longer Presentation) Reading Assignment Part Two: 8.01 Course Notes: W010D1 Angular Momentum http://web.mit.edu/8.01t/www/materials/modules/CN_W10D1_angular_momentum.pdf Suggested: Young and Freedman Chapter 10.5-10.6 All Sections Nov 7/8 W10D2 Experiment 3: Rotational Dynamics; Experiment 4: Angular Momentum 8.01 Course Notes: Thursday Nov 8 Problem Set 9 Due 9 pm Nov 9 W10D3 Problem Solving 08: Conservation of Angular Momentum Reading Assignment: Suggested: Young and Freedman: 10.5-10.6; 8.01 Course Notes:

Problem 1: Physical Pendulum A physical pendulum consists of a disc of radius R and mass md fixed at the end of a rod of mass mr and length l .

a) How far from the pivot point is the center of mass? b) What is the moment of inertia of the physical pendulum about an axis passing

through the pivot perpendicular to the plane of motion of the pendulum.

Solution: The physical pendulum consists of two pieces. A uniform rod of length d and a disk attached at the end of the rod. The moment of inertia about the pivot point P is the sum of the moments of inertia of the two pieces,

IP = IP,rod + IP,disk (1)

We calculated the moment of inertia of a rod about the end point P in class, and found that

IP,rod =

13

mrl2 (2)

We can use the parallel axis theorem to calculate the moment of inertia of the disk about the pivot point P ,

IP,disk = Icm,disk + mdl2 (3) We calculated the moment of inertia of a disk about the center of mass in class, and found that

Icm,disk =

12

md R2 (4)

So the total moment of inertia is

IP =

13

mrl2 + md gl2 +

12

md R2 (5)

Problem 2: Claw Hammer A claw hammer is used to pull a nail out of a board. The nail is an angle ! = 60! to the board. The nail exerts a force on the hammer

!F1 along the line of the nail. The hammer

head contacts the board at point A , which is a distance d from where the nail enters the board. A horizontal force

!F2 is applied to the hammer handle at a distance h above the

board with d < h . The forces !F1 and

!F2 on the hammer are shown in the figure to the

right (the arrows are not in scale with the magnitudes of the forces). You may ignore the weight of the hammer. What is the ratio,

!F1 /!F2 ?

Solution: Let‘s calculate the torque about the point A . The torque static equilibrium condition is that

!! A = !rA,1 "

!F1 +!rA,2 "

!F2 =!0

In the diagram below we show the relevant vectors.

We first calculate the torque about A due to the pulling force

!F2 , where

!rA,2 = xi + hj .

!! A,2 =

!rA,2 "!F2 = (xi + hj) " (#F2 i) = hF2k .

Note that the even though the quantity x was not specified it doesn’t enter into the torque about A . In order to calculate the torque on the hammer due to the force of the nail !F1 , we note that

!rA,1 is perpendicular to

!F1 and the direction of the toque about A is

in the !k direction. From the diagram below we can determine that the length of

!rA,2 = d 3 / 2 .

Therefore the toque about A due to

!F1 is

!! A,1 =

!rA,1 "!F1 =

!rA,1

!F1 (#k) = #

d 32

F1 k

Therefore the torque about A is

!0 = !

d 32

F1 + hF2

"

#$

%

&' k

Setting the above z -component equal to zero yields

d 3

2F1 = hF2 .

Therefore the ratio, F1 / F2 is

F1

F2

=2 3

3hd

.

Check

From the diagram above, we can determine the vector

!rA,1 from the point A to the point

where the force !F1 acts (head of the nail)

!rA,1 =

d 34

i + d34

j

Therefore the toque about A due to

!F1 is

!! A,1 =

!rA,1 "!F1 =

d 34

i + d34

j#

$%

&

'( " (F1 cos(60" )i ) F1 sin(60" ) j)

=d 3

4F1 sin(60" )()k) +

d34

F1 cos(60" )()k)

= )d 3

8F1 )

d3 38

F1

#

$%

&

'( k = )

d 32

F1

#

$%

&

'( k

,

which is in agreement with our earlier calculation.

Problem 3: Rotating Wheel A light flexible cable is wrapped around a cylinder of mass m1 , radius R , and moment of inertia Icm about the center of mass. The cylinder rotates about its axis without friction. The free end of the cable is attached to an object of mass m2 . The object is released from rest at a height h above the floor. You may assume that the cable has negligible mass. Let g be the gravitational constant.

a) Draw free body diagrams of the cylinder and the object, indicating all of the forces acting on each. Also choose appropriate coordinates the linear motion of the object and the angular motion of the cylinder, and indicate the relation between them. b) Find the acceleration of the falling object. c) Find the tension in the cable. d) Find the speed of the falling object just before it hits the floor.

Problem 4: Spinning Sphere

A uniform sphere of mass Ms and radius R rotates about a vertical axis. Assume there is no frictional torque acting along the axis of rotation. The sphere has moment of inertia IS = (2 / 5)MSR

2 for an axis through its center of mass. A light string passes around the equator of the sphere, and over a massive pulley of radius b and moment of inertia IP about its center of mass, and the string is attached to a small object of mass m1 . When the small object is released from rest, the string does not slip around the sphere or pulley, and the small object falls vertically downward. What is the magnitude of the acceleration of the small object after it is released? The downward gravitational acceleration has magnitude g . Express your answer in terms of Ms , m1 , IP , g , b , and R as needed.

Problem 5: Simple Pendulum A simple pendulum consists of a massless string of length l and a point-like object of mass m attached to one end. Suppose the string is fixed at the other end and is initially pulled out at a small angle !0 from the vertical and released from rest. You may assume the small-angle approximation, 0 0sin! !! . The gravitational acceleration is g .

a) Find a differential equation for d

2! / dt2 in terms of l , ! , m and g as needed by calculating the torque about the pivot point. b) What is your differential equation if ! << 1? Solution We shall apply the torque equation about the pivot point. With our choice of a cylindrical coordinate system with origin at the pivot point, the angular acceleration is given by

2

2ˆd k

dt!" =

!. (6)

The force diagram on the pendulum is shown below.

The only forces on the pendulum are the gravitational force on the pendulum bob (the point-like object) and the tension in the string.

The torque about the pivot point is given by

!! P = !rP,Fgrav

" m!g + !rP,T "!T . (7)

Because

!rP,T is anit-parallel to

!T and hence the tension force exerts no torque. The rod is

uniform, therefore the center of mass is a distance d / 2 from the pivot point. So the torque about the pivot point is given by

!! P = !rP,Fgrav

" m!g=lr " mg(# sin$ $ + cos r) = #lmg sin$ k . (8)

The rotational dynamical equation (torque equation) is P PI! "=

!! . (9) The moment inertial of a point-like object about the pivot point is IP = ml2 . (10) Therefore

!lmg sin" k = ml2 d 2"

dt2 k . (11)

We can rewrite this equation as

d 2!dt2 = "

gl

sin! . (12)

When the angle of oscillation is small, then we can use the small angle approximation sin! " ! . (13) Then the torque equation becomes

d 2!dt2 = "

gl! (14)

which is the simple harmonic oscillator equation.

Problem 6: Torsional Oscillator A disk with moment of inertia about the center of mass cmI rotates in a horizontal plane. It is suspended by a thin, massless rod. If the disk is rotated away from its equilibrium position by an angle ! , the rod exerts a restoring torque given by cm! "#= $ .

a) Determine a differential equation that the angular displacement ( )t! satisfies. b) Show that the function !(t) = Acos("0 t) + Bsin("0 t) satisfies the differential

equation you found in part a), where the angular frequency of oscillation is given by !0 = " / Icm

c) Determine an expression for the z -component of the angular velocity.

d) Suppose the disk is released from rest at 0t = , at an angle !(t = 0) = !0 . What

are the constants A and B ? Based on that information determine an expression for ( )t! .

Solution: Choose a coordinate system such that k is pointing upwards, then the angular acceleration is given by

2

2ˆd k

dt!" =

!. (15)

The torque about the center of mass is given in the statement of the problem as a restoring torque ˆ

cm k! "#= $! . (16) Therefore the k -component of the torque equation cm cmI! "=

!! is

!"# = Icm

d 2#dt2 . (17)

b) Show that the function !(t) = Acos("0 t) + Bsin("0 t) (18)

where the angular frequency of oscillation is given by !0 = " / Icm (19) satisfies the differential equation you found in part a).

e) Determine an expression for the z -component of the angular velocity.

d!dt

(t) = "#0 Asin(#0 t) +#0 Bcos(#0 t) . (20)

f) Suppose the disk is released from rest at 0t = , at an angle !(t = 0) = !0 . What

are the constant A and B ? Solution: 0( 0)t A! != = = , and (d! / dt)(t = 0) ="0 B = 0 . Therefore !(t) = !0 cos( " / Icm t) . (21)

Problem 7: Race to the Bottom A uniform stick of length l and mass m is pivoted at one end. The stick initially forms an angle !0 with a table such that cos!0 " 2 / 3 . At the other end of the stick a small block is balanced on top of the stick. The stick is released from rest and hits the table. The block is in free fall and also hits the table.

a) Calculate the speed of a point at the end of stick the instant before the stick hits the table. b) Calculate the speed of the block the instant before the block hits the table. c) Does the ball or stick hit the table first? Solution: We assume the pivot is frictionless and so that energy of the stick is constant.

Choose U (! = 0) = 0 . The initial energy is then

Ei = msg(l / 2)sin!0 . The moment of inertia of the stick about the pivot point is IP = (1 / 3)msl

2 . The final energy of the stick when it just hits the ground is all rotational kinetic energy about the pivot point

E f = (1 / 2)IP! f

2 = (1 / 6)msl2! f

2 . Because energy is constant

msg(l / 2)sin!0 = (1 / 6)msl

2" f2 .

Thus the angular speed of the stick just before it hits the ground

! f =

3g sin"0

l.

If the block fall straight down it hits the ground at a point A , a distance dA = l cos!0 from the pivot point P . Therefore the speed of the point A of the stick just before it hits the ground is

vA, f = dA

3g sin!0

l= l cos!0

3g sin!0

l= cos!0 3gl sin!0 .

b) The energy of the block is also constant. The initial energy is

Ei = mbgl sin!0 . The final energy of the block is

E f = (1 / 2)mbvb, f

2 . Because energy is constant

mbgl sin!0 = (1 / 2)mbvb, f

2 . Hence

vb, f = 2gl sin!0 .

c) Let’s consider the point A on the stick that hits the ground where the block hits. The point A on the stick is traveling in a circular arc which is a shorter distance than the distance traveled by the block.

From the figure above, the distance traveled by the block is db = l sin!0 . Then distance

traveled by the point A is dA = (l cos!0 )!0 . The ratio of these distance is greater than one,

db

dA

=l sin!0

(l cos!0 )!0

=tan!0

!0

> 1.

Therefore is the speed of point A is greater than the speed of the block for fixed ! , point A will hit the ground before the block hits the stick. Using energy arguments for the block we have that at the angle ! ,

mbgl sin!0 " mbgl sin! = (1 / 2)mbvb2 (!) ,

hence

vb(!) = 2gl(sin!0 " sin!) . Using energy arguments for the stick we have that at the angle ! ,

msg(l / 2)sin!0 " msg(l / 2)sin! = (1 / 6)msl2# s

2 (!) . Hence the angular speed of the stick is

! s(") =

1l

3gl(sin"0 # sin") .

The speed of point A is then

vA(!) = (l cos!0 )" s(!) = cos!0 3gl(sin!0 # sin!)

= cos!0

32

2gl(sin!0 # sin!) = cos!0

32

vb(!).

Recall that the angle !0 satisfied the condition that

cos!0 >

23

therefore vA(!) > vb(!) . So the stick hit s the ground before the block hits the stick. We could also argue as follows. The magnitude of the z -component of the end of the stick is given by

vend ,z (!) = (l cos!)" s(!) = cos!0 3gl(sin!0 # sin!) = cos! 3

2vb(!) .

Because cos! " cos!0 > 2 / 3 , the end of the stick is always traveling faster than the

block, so vend ,z (!) > vb(!) .

Problem 8: Static Equilibrium: Back-Bending Exercise

The human vertebral column is shown in the figure. The sacrum is rigidly attached to the pelvis. The fifth lumbar vertebra is separated from the sacrum by a disc. In this problem, approximate the spine as a rigid body hinged at the bottom of the lumbo-sacral disc. The top of the sacrum exerts a force

!Fdisc on the base of the

lumbo-sacral disc at an angle ! relative to the axis of the vertebral column. When a person of mass m = 50 kg bends over or lifts an object, the main muscles that lift the back are the erector spinae (sacrospinal muscles).

These muscles act approximately as a single cord (“rope”) on the vertebral column and exert a force,

!Fmuscle , at a point that is (2 / 3) of the distance from base of the spine (where

the sacrum pushes on base of the lumbo-sacral disc) to the center of gravity of the head and arms. The angle of insertion of this “rope” is about ! = 12! relative to the axis of the vertebral column. Assume that mass of the head and arms is m2 = (1 / 5)m and that the

mass of the trunk is m1 = (1 / 5)m , where m is the mass of the person. The center of mass of the trunk is (1 / 2) of the distance from the base of the spine to the center of gravity of the head and arms. Suppose the person bends over so that the spine makes an angle of ! = 30! with respect to the horizontal.

a) Find the magnitude of the force, Fmuscle =

!Fmuscle , that the muscles exert on the

spine.

b) Find the angle, ! , of the disc force with respect to the spinal axis and the magnitude,

Fdisc =

!Fdisc , of the force on the base of the lumbo-sacral disc.

c) What is the ratio of the magnitude of the force Fdisc to the weight of the body?

d) Suppose you lift a mass equal to the mass of the head and arms with your arms

hanging straight down. Assume you bend bent at the same angle of 30! with the horizontal. By how much does the magnitude of the reaction force increase?

e) Explain why you should bend your knees when you pick up a mass. Base your

analysis on the work done in parts a) through d). Solution:

a) Choose coordinates and a sense for positive rotation as shown in the force diagram shown below.

We first vector decompose each of the forces with respect to the unit vectors shown in the figure above. m1

!g = !m1g sin" i ! m1g cos" j (1.1) m2

!g = !m2g sin" i ! m2g cos" j (1.2)

!Fmuscle = !Fmuscle cos" i + Fmuscle sin" j (1.3)

!Fdisc = Fdisc cos! i " "Fdisc sin! j (1.4)

The two equations for force equilibrium are then

disc 1 2 muscle

disc 1 2 muscle

ˆ : cos sin sin cos 0ˆ : sin cos cos sin 0.

F m g m g F

F m g m g F

! " " #

! " " #

$ $ $ =

$ $ $ + =

i

j (1.5)

Choose the base of the lumbo-sacral disc as the point S about which to calculate torques. Then torque equilibrium becomes

!!S , total =

!!S ,1 +

!!S ,2 +

!!S , muscle =

!0 . (1.6)

We begin with the gravitational force acting at the center of mass of the trunk. The vector from S to the center of mass of the trunk is

!rS ,1 = (1 / 2)l i (1.7)

The torque about S due to the center of mass of the trunk is

!!S ,1 =

!rS ,1 " m1

!g = (1 / 2)l i " (#m1g sin$ i # m1g cos$ j)

= #(1 / 2)l m1g cos$ k. (1.8)

The vector from S to the center of mass of the head and arms is

!rS , 2 = l i (1.9)

The torque about S due to the center of mass of the head and arms is

!!S , 2 =

!rS , 2 " m2

!g = l i " #m2g sin$ i # m2g cos$ j( )= #l m2g cos$ k.

(1.10)

The vector from S to where the muscles act approximately as a single cord (“rope”) on the vertebral column is

!rS , muscle = (2 / 3)l i (1.11)

The torque about S due to the muscular force is

!!S , muscle =

!rS , muscle "!Fmuscle = (2 / 3)l i " #Fmuscle cos$ i + Fmuscle sin$ j( )

= (2 / 3)l Fmuscle sin$ k. (1.12)

The total component of the torque in the k direction must vanish, !(1 / 2)lm1g cos" ! l m2g cos" + (2 / 3)l Fmuscle sin# = 0 . (1.13) We can solve the torque equation (1.13) for the magnitude of the muscular force,

Fmuscle =

((3 / 4)m1 + (3 / 2)m2 )g cos!sin"

. (1.14)

Substituting the given values for the masses of the trunk m1 = (1 / 5)m , and arms and head, m2 = (1 / 5)m , into Eq. (1.14) yields

Fmuscle =(9 / 4)m1g cos!

sin"=

(9 / 20)cos(30! )sin(12! )

mg

= 1.9mg = 9.2 #102 N

. (1.15)

The force in the erector spine muscles is 1.9 times the body weight. b), c) We can solve for the angle the force on the disc makes with respect to the spine, by rewriting the force equations (1.5) as

disc 1 2 muscle

disc 1 2 muscle

cos sin sin cossin cos cos sin .

F m g m g FF m g m g F

! " " #! " " #= + += $ $ +

(1.16)

Dividing these equations yields

disc 1 2 muscle

disc 1 2 muscle

cos sin sin cossin cos cos sin

F m g m g FF m g m g F

! " " #! " " #

+ +=$ $ +

. (1.17)

Substituting Eq. (1.14) into Eq. (1.17) yields

cotan! =m1g sin" + m2g sin" + ((3 / 4)m1 + (3 / 2)m2 )g cos"cotan#$m1g cos" $ m2g cos" + ((3 / 4)m1 + (3 / 2)m2 )g cos"

=m1g sin" + m2g sin" + ((3 / 4)m1 + (3 / 2)m2 )g cos"cotan#

($(1 / 4)m1 + (1 / 2)m2 )g cos"

. (1.18)

We now substitute m1 = m2 , ! = 30! , and ! = 12! , yielding cotan! = 8tan" + 9cotan# = 8tan30! + 9cotan12! = 47 . (1.19) Therefore The magnitude of the disc force cancels and we can take the inverse cotangent to find the angle

! = cot"1 47( ) = 1.2! (1.20)

Finally, we can solve for the magnitude of the disc force by using the first equation in (1.16), with ! = 1.2! and muscleF from the first equation in (1.15);

Fdisc =m1g sin! + m2g sin! + Fmuscle cos"

cos#= m1g sin! + m2g sin! + (9 / 4)m1g cos! cotan"

cos#

=(1 / 5)mg 2sin! + (9 / 4)cos!cotan"( )

cos#

=(1 / 5)(50kg)(9.8m $ s%2 ) 2sin(30! )+(9 / 4)cos(30! )cot(12! )( )

cos(1.2! )

= (1.0 &103 N) " 2.0mg.

(1.21)

d) Suppose you lift a mass equal to the mass of your head and arms with your arms hanging straight down. Assume you bend bent at the same angle of 30! with the horizontal. By how much does the magnitude of the reaction force increase? Lifting an additional weight 2m introduces extra torque at the end of the spine. We assume that lifting the extra weight just doubles the center of mass of the head and arms. Therefore the extra torque,

!!S , weight =

!!S , 2 = "l m2g cos# k . (1.22)

We adjust all the above calculations by introducing this extra factor of two,

Fmuscle =(1 / 2)m1g cos! + 2m2g cos!

(2 / 3)sin"=

m1g((1 / 2)cos! + 2cos!)(2 / 3)sin"

=(3 / 4)mg cos!

sin"

=(3 / 4)(50kg)(9.8m# s$2 )cos(30! )

sin(12! )

= 1.5%103 N " 3.1mg.

(1.23)

! = cot"1 m1g sin# + 2m2g sin# + (3 / 4)mg cos# cot$"m1g cos# " 2m2g cos# + (3 / 4)mg cos#

%

&'(

)*

= cot"1(3 / 5)sin# + (3 / 4)cos# cot$( )

cos#("(3 / 5) + (3 / 4))

%

&'

(

)*

= cot"1 (16 / 3)tan(30! ) + 5cot(12! )( )= 2.15!.

(1.24)

Similarly

Fdisc =m1g sin! + 2m2g sin! + (3 / 4)mg cos! cot"

cos#

=mg (3 / 5)sin! + (3 / 4)cos! cot"( )

cos#

= (50kg)(9.8m $ s%2 )(3 / 5)sin(30! ) + (3 / 4)cos(30! )cot(12! )( )

cos(2.15! )

= (2.1&103N) " 4.3 mg.

(1.25)

The force on the disc has increased by 1.0 !103N when lifting a weight of m2g = (1 / 5)mg = (1 / 5)(50kg)(9.8m ! s"2 ) = 9.8 #101N , (1.26) about ten times the weight lifted. e) The large force found in part d) is in part due to the horizontal forces needed for torque equilibrium. By bending the knees while lifting, the forces are closer to vertical, and the force on the disc will be approximately the weight of the upper body plus the weight lifted.