1

Chapter 11 Rotational Dynamics

Introduction Let’s consider a rigid rod thrown in the air (Figure 11.1) so that it is spinning as its center of mass moves with velocity

!v

cm. (Recall in the Introduction to Chapter 4 that a body is

called a rigid body if the distance between any two points in the body does not change in time.) Rigid bodies, unlike point-like objects, can have forces applied at different points in the body. We have explored the physics of translational motion; now, we wish to investigate the properties of rotation exhibited in the rod’s motion, beginning with the notion that every particle is rotating about the center of mass with the same angular (rotational) velocity.

Figure 11.1 The center of mass of a thrown rigid rod follows a parabolic trajectory while the rod rotates about the center of mass.

We can use Newton’s Second Law to predict how the center of mass will move. Since the only external force on the rod is gravitation (neglecting the action of air resistance), the center of mass of the body will move in a parabolic trajectory. How was the rod induced to rotate? In order to spin the rod, we applied a torque with our fingers and wrist to one end of the rod as the rod was released. The applied torque is proportional to the angular acceleration. The constant of proportionality is called the moment of inertia. When external forces and torques are present, the motion of a rigid body can be extremely complicated while it is translating and rotating in space. We shall begin our study of rotating objects by considering the simplest example of rigid body motion, rotation about a fixed axis. 11.1 Fixed Axis Rotation: Rotational Kinematics Fixed Axis Rotation When we studied static equilibrium, we demonstrated the need for two conditions: the total force acting on an object is zero, as is the total torque acting on the object. If the total torque is non-zero, the object will start to rotate.

2

A simple example is the motion of a compact disc in a CD player, which is driven by a motor inside the player. This motor creates a torque on the disc that produces angular acceleration, causing the disc to spin. As it is set in motion, a frictional torque is produced that opposes the applied torque. When these two torques are equal in magnitude, the disc no longer has any angular acceleration, and it now spins at a constant angular velocity. Throughout this process, the CD rotates about an axis passing through the center of the disc, and is perpendicular to the plane of the disc (see Figure 11.2). This type of motion is called fixed axis rotation.

Figure 11.2 Rotation of a compact disc about a fixed axis rotation. When we ride a bicycle forward, the wheels rotate about an axis passing through the center of the wheel and perpendicular to the plane of the wheel (Figure 11.3). As long as the bicycle does not turn, this axis keeps pointing in the same direction. This motion is more complicated than our spinning CD because the wheel is both moving (translating) with some center of mass velocity,

!v

cm, and rotating.

Figure 11.3 Fixed axis rotation and center of mass translation for a bicycle wheel. When we turn the bicycle’s handlebars, we change the bike’s trajectory and the axis of rotation changes direction. Other examples of non-fixed axis rotation are the motion of a spinning top, or a gyroscope. This type of motion is much harder to analyze, so we will restrict ourselves to considering fixed axis rotation with and without translation.

3

Angular Velocity and Angular Acceleration When we considered the rotational motion of a point-like object in Chapter 7, we introduced an angle coordinate ! , and then defined angular velocity (Eq. 7.2.7)

! "d#

dt, (11.1.1)

and angular acceleration (Eq. 7.3.4),

! "d

2#

dt2

. (11.1.2)

For a rigid body undergoing fixed axis rotation, we can divide the body up into small volume elements with mass

!m

i. Each of these volume elements is moving in a

circle of radius r! ,i

about the axis of rotation (Figure 11.4).

Figure 11.4 Coordinate system for fixed axis rotation. Because the body is rigid, all the volume elements will have the same angular velocity ! and the same angular acceleration, ! . Sign Convention: Angular Velocity and Angular Acceleration

Suppose we choose ! increasing in the counterclockwise direction as shown in Figure 11.5.

4

Figure 11.5 Sign conventions for rotational motion. If the rigid body rotates in the counterclockwise direction, then the angular velocity is positive,

! " d# dt > 0 . If the rigid body rotates in the clockwise

direction, then the angular velocity is negative, ! " d# dt < 0 .

If the rigid body increases its rate of rotation in the counterclockwise direction then the angular acceleration is positive,

! " d

2# dt

2> 0 . If the rigid body

decreases its rate of rotation in the counterclockwise direction then the angular acceleration is negative,

! " d

2# dt

2< 0 . Also, if the rigid body increases its

rate of rotation in the clockwise direction then the angular acceleration is negative, and if the rigid body decreases its rate of rotation in the clockwise direction then the angular acceleration is positive.

Tangential Velocity and Tangential Acceleration Since the small volume element is moving in a circle with constant angular acceleration, ! , it has a tangential velocity component

v

tan, i= r

! , i" . (11.1.3)

If the magnitude of the tangential velocity is changing, the volume element undergoes a tangential acceleration given by

a

tan, i= r

! , i" . (11.1.4)

Recall from Chapter 7.3 Equation (7.3.14) that the volume element is always accelerating inward with magnitude

arad, i

=

vtan, i

2

r! , i

= r! , i"

2 . (11.1.5)

5

Example 1: Turntable A turntable is a uniform disc of mass 1.2 kg and a radius 1.3!101 cm . The turntable is spinning initially at a constant rate of

f

0= 33 cycles !min"1 ( 33 rpm ). The motor is

turned off and the turntable slows to a stop in 8.0 s . Assume that the angular acceleration is constant.

a) What is the initial angular velocity of the turntable?

b) What is the angular acceleration of the turntable? Answer: Initially, the disc is spinning with a frequency

f0= 33

cycles

min

!"#

$%&

1min

60s

!"#

$%&= 0.55 cycles ' s(1

= 0.55 Hz , (11.1.6)

so the initial angular velocity is

!0= 2" f

0= 2"

radian

cycle

#$%

&'(

0.55cycles

s

#$%

&'(= 3.5 rad ) s*1 . (11.1.7)

The final angular velocity is zero, so the angular acceleration is

! ="#

"t=

#f$#

0

tf$ t

0

=$3.5 rad % s

$1

0.8 s= $4.3 rad % s

$2 . (11.1.8)

Because the angular acceleration is negative, the disc is slowing down. To phrase this more generally, if ! and ! have the same sign, the body is speeding up; if opposite signs, the body is slowing down. This general result is independent of the choice of positive direction of rotation. 11.2 Torque, Angular Acceleration, and Moment of Inertia For fixed axis rotation, there is a direct relation between the component of the torque along the axis of rotation and angular acceleration. Choose the z-axis to lie along the axis of rotation. Divide the body into volume elements of mass

!m

i. Let the point S denote

any point along the axis of rotation (Figure 11.6). Each volume element undergoes a tangential acceleration as it moves in a circular orbit of radius

r! , i

about the fixed axis.

6

Figure 11.6: Volume element undergoing fixed axis rotation about the z-axis. The vector

!r

S , i= z

ik̂ + r

! ,ir̂ from the point S to the volume element is given by

!r

S , i= z

ik̂ + r

! ,ir̂ (11.2.1)

where

z

i is the distance along the axis of rotation between the point S and the volume

element. Now the torque about S due to the force

!F

i acting on the volume element is

given

!!

S , i=!r

S , i"!F

i. (11.2.2)

Substitute Eq. (11.2.1) into Eq. (11.2.2),

!!

S , i= (z

ik̂ + r

" ,ir̂) #!F

i (11.2.3)

We are only interested in the z-component of the torque so that must arise from the term

(!

S , i)

z= (r

" ,ir̂ #!F

i)

z (11.2.4)

since the cross product of

z

ik̂ !!F

i must point perpendicular to the plane formed by the

vectors k̂ and

!F

i, hence perpendicular to the z-axis. The total force acting on the volume

element has both a radial and tangential component

!Fi= F

r ,ir̂ + F

tan,i!̂ (11.2.5)

7

(since the acceleration in the z-direction is zero, there is no z-component of the force). Substitute Eq, (11.2.5) into Eq. (11.2.4),

(!

S , i)

z= (r

" ,ir̂ # (F

r ,ir̂ + F

tan,i$̂))

z (11.2.6)

The radial force does not contribute to the torque about the z-axis since

r! ,i

r̂ " Fr ,i

r̂ =!0 (11.2.7)

So we are interested in the contribution due to torque about the z-axis due to the tangential component of the force on the volume element (Figure 11.7). Thus the component of the torque about the z-axis is given by

(!

S , i)

z= (r

" ,ir̂ # F

tan,i$̂))

z= r

" ,iF

tan,i (11.2.8)

The z-component of the torque points out of the page in Figure 11.7.

Figure 11.7 Tangential force acting on a volume element. Newton’s Second Law in the tangential direction is

F

tan, i= !m

ia

tan, i. (11.2.9)

Using our result for tangential acceleration we have that

F

tan, i= !m

ir" , i# . (11.2.10)

From Eq. (11.2.8), the component of the torque about the z-axis is given by

(!

S , i)

z= r

" , iF

tan, i= #m

i(r

" , i)2$ . (11.2.11)

8

The total component of the torque about the z-axis is the summation of the torques on all the volume elements,

(!S

total )z= (!

S ,1)

z+ (!

S ,1)

z+ " " " = (!

S , i)

z

i=1

i=N

# = r$ , i

Ftan, i

i=1

i=N

# = %mi(r

$ , i)2&

i=1

i=N

# .(11.2.12)

Since each element has the same angular acceleration, ! , the summation becomes

(!S

total )z= "m

i(r# , i

)2

i=1

i=N

$%

&'(

)*+ . (11.2.13)

Definition: Moment of Inertia about a Fixed Axis

The quantity

IS= !m

i(r

" , i)2

i=1

i=N

# . (11.2.14)

is called the moment of inertia of the rigid body about a fixed axis passing through the point S , and is a physical property of the body. The SI units for moment of inertia are[kg !m2] .

Thus Equation (11.2.13) shows that the z-component of the torque is proportional to the angular acceleration,

(!

S

total )z= I

S" , (11.2.15)

and the moment of inertia,

I

S, is the constant of proportionality.

This is very similar to Newton’s Second Law: the total force is proportional to the acceleration,

!F

total= m

total!a . (11.2.16)

and the total mass, m total , is the constant of proportionality. For a continuous mass distribution, the summation becomes an integral over the body

IS= dm(r

!)2

body

" , (11.2.17)

9

which will be explored in detail in the next section.

10

Example 2: Turntable A turntable of mass 1.2 kg and radius 1.3!101 cm has a moment of inertia about an axis pointing through the center of the disc,1.01!10"2 kg #m2 . The turntable is spinning at an initial constant frequency of

f

0= 33 cycles !min"1 . The motor is turned off and the

turntable slows to a stop in 8.0 s due to frictional torque. Assume that the angular acceleration is constant. What is the magnitude of the frictional torque acting on the disc? Answer: We already calculated the angular acceleration of the disc in Example 1, where we found that the angular acceleration is

! ="#

"t=

#f$#

0

tf$ t

0

=$3.5 rad % s

$1

0.8 s= $4.3 rad % s

$2 . (11.2.18)

So the magnitude of the frictional torque is

!

friction

total= I

S" = (1.0 #10$2 kg %m2 )(4.3 rad % s$2 ) = 4.3#10$2 N %m (11.2.19)

Example 3: Moment of Inertia of a rod of uniform mass density Consider a thin uniform rod of length L and mass m . In this problem, we will calculate the moment of inertia about an axis perpendicular to the rod that passes through the center of mass of the rod. A sketch of the rod, volume element, and axis is shown in Figure 11.7.

Figure 11.7 Moment of inertia of a uniform rod about center of mass. Choose Cartesian coordinates, with the origin at the center of mass of the rod, which is midway between the endpoints since the rod is uniform. Choose the x -axis to lie along the length of the rod. Choose an infinitesimal mass element

dm = ! dx , located a distance x from the pivot

point, where the mass per unit length ! = m / L is a constant, as we have assumed the rod to be uniform.

11

When the rod rotates about an axis perpendicular to the rod that passes through the center of mass of the rod, the element traces out a circle of radius

r!= x .

The limits of the integral for the moment of inertia about the center of the rod are

Icm

= (r!

)2dm

body

" = # (x2 )

x=$ L 2

x= L 2

" dx . (11.2.20)

We are adding together the contributions from each infinitesimal element as we go from

x = !L 2 to

x = L 2 .

The integral now becomes

Icm

= (r!

)2dm

body

" = # (x2 )

x=$ L 2

x= L 2

" dx = #x

3

3x=$ L 2

x= L 2

=m

L

(L 2)3

3$

m

L

($L 2)3

3=

1

12m L

2 . (11.2.21)

Since the rod is uniform, there is no variation in the z-direction. We also assume that the width is the rod is negligible. (Technically we should treat the rod as a rectangle in the x-y plane. The calculation of the moment under this assumption would be more complicated.) 11.3 Parallel Axis Theorem Consider a rigid body of mass m , undergoing fixed axis rotation. Consider two axis parallel to the axis of rotation. The first axis passes through the center of mass of the body. The second axis, parallel to the first, passes through some other point S in the body. Let

d

S , cm denote the perpendicular distance between the two parallel axes (Figure

11.8).

Figure 11.8 Geometry of the parallel axis theorem.

12

Then the moment of inertia,

I

S, about an axis passing through a point S is related to the

moment of inertia about an axis passing through the center of mass, I

cm, according to

I

S= I

cm+ md

S , cm

2 . (11.3.1) Proof of Parallel Axis Theorem Identify an infinitesimal volume element of mass dm . Choose a coordinate system for the rigid body such that the vector from the point S to the mass element is

!r

S , dm, the vector

from the center of mass to the mass element is

!r

cm, dm, and the vector from the point S to

the center of mass is

!r

S , cm. From Figure 11.8, we see that

!r

S , dm=!r

S , cm+!r

cm, dm. (11.3.2)

Note that

(r

S , dm)2=!r

S , dm!!r

S , dm= (!r

S , cm+!r

cm,dm) ! (!r

S ,cm+!r

cm,dm) = (r

S ,cm)2+ (r

cm,dm)2+ 2!r

S , cm!!r

cm, dm.(11.3.3)

The moment of inertia about the point S is

IS= dm(r

S , dm)2

body

! . (11.3.4)

Thus we have for the moment of inertia about S ,

IS= dm(r

S , cm)2

body

! + dm(rcm, dm

)2

body

! + 2 dm(!r

S , cm"!r

cm, dm)

body

! . (11.3.5)

In the first integral above,

r

S , cm= d

S , cm, is the distance between the parallel axes and is a

constant and can be pulled out of the integral. Thus

dm(rS , cm

)2

body

! = m(dS , cm

)2 . (11.3.6)

The second term is the moment of inertia about the center of mass,

Icm

= dm(rcm, dm

)2

body

! . (11.3.7)

The third integral is zero,

13

2 dm(!r

S , cm!!r

cm, dm)

body

" = 0 . (11.3.8)

To see this we can pull the constant term

!r

S , cm out of the integral,

2 dm(!r

S , cm!!r

cm, dm)

body

" =!r

S , cm!2 dm(

!r

cm, dm)

body

" . (11.3.9)

The integral

dm(!r

cm, dm)

body

! = 0 (11.3.10)

because this is the definition of the center of mass vector with respect to the center of mass; hence the integral is zero. So, the moment of inertia about S is just the sum of the first two integrals,

I

S= I

cm+ md

S , cm

2 . (11.3.11) Example 4: Uniform Rod Let point S be the endpoint of the rod. Then the distance from the center of mass to the end of the rod is

d

S , cm= L / 2 . The moment of inertial about an axis passing through the

endpoint, I

end= (1 / 3) m L

2 , is related to the moment of inertial about an axis passing

through the center of mass, I

cm= (1 / 12) m L

2 , according to Equation (11.3.11),

1

3m L

2=

1

12m L

2+

1

4m L

2 . (11.3.12)

In this case it’s easy and useful to check by direct calculation. Use Equation (11.2.21) but with the limits changed to !x = 0 and !x = L , where !x = x + L / 2 ;

Iend

= (r!

)2dm

body

" = # ( $x2 )

$x =0

$x = L

" d $x = #$x3

3$x =0

$x = L

=m

L

(L)3

3%

m

L

(0)3

3=

1

3m L

2 (11.3.13)

14

11.4 Simple Pendulum and Physical Pendulum Simple Pendulum A pendulum consists of an object hanging from the end of a string or rigid rod pivoted about the point S . The object is pulled to one side and allowed to oscillate. If the object has negligible size and the string or rod is massless, then the pendulum is called a simple pendulum. The force diagram for the simple pendulum is shown in Figure 11.9.

Figure 11.9 A simple pendulum.

The torque about the pivot point S is given by

!!

S=!r

S , m" m!g = l r̂ " m g(# sin($) $̂ + cos($) r̂) = #l m g sin($) k̂ . (11.4.1)

The angular acceleration is a vector pointing along the z -axis (the direction of

!k in the

figure),

!! =

d2"

dt2

k̂ . (11.4.2)

The z -component of the angular acceleration vector is d

2! / dt

2 and can be positive or negative. The moment of inertia of a point mass about the pivot point S is

I

S= ml

2 . (11.4.3) The rotational dynamical equation is

!!

S= I

S

!" . (11.4.4)

15

Therefore

!l m g sin(") k̂ = ml2 d 2

"

dt2k̂ . (11.4.5)

Thus we have the equation of motion for the simple pendulum,

!l m g sin(") = ml2 d 2

"

dt2. (11.4.6)

When the angle of oscillation is small, then we can use the small angle approximation sin(!) " ! ; (11.4.7) the rotational dynamical equation for the pendulum becomes

d 2!

dt2" #

g

l! . (11.4.8)

This equation is similar to the object-spring simple harmonic differential equation,

d2x

dt2= !

k

mx . (11.4.9)

that describes the oscillation of a mass about the equilibrium point of a spring. Recall that the angular frequency of oscillation was given by

! spring =k

m (11.4.10)

So by comparison, the frequency of oscillation for the pendulum is approximately

!pendulum "g

l, (11.4.11)

with period

T =2!

"pendulum

# 2!l

g (11.4.12)

16

Physical Pendulum A physical pendulum consists of a rigid body that undergoes fixed axis rotation about a fixed point S (Figure 11.10). The gravitational force acts at the center of mass of the physical pendulum (see Appendix Chapter 4.1). Suppose the center of mass is a distance

lcm

from the pivot point S .

Figure 11.10 Physical pendulum. The analysis is nearly identical to the simple pendulum. The torque about the pivot point is given by

!!

S=!r

S , cm" m!g = l

cmr̂ " m g(# sin($) $̂ + cos($) r̂) = #l

cmm g sin($) k̂ . (11.4.13)

The moment of inertia about the pivot point is

I

S. If the body is a uniform rod of mass m

and length l pivoted at one end, then the moment of inertial about the pivot point is

I

S= (1 / 3)ml

2 . The rotational dynamical equation is

!!

S= I

S

!" . (11.4.14)

Therefore

!l

cmm g sin(") k̂ = I

S

d 2"

dt2k̂ . (11.4.15)

If

l ! l

cm and

ml

2! I

S then Equation (11.4.6) is identical to Equation (11.4.15).

Therefore the period of the physical pendulum for small oscillations can be obtained from Equation (11.4.12),

17

Tphysical

=2!

"pendulum

# 2!I

S

lcm

m g. (11.4.16)

11.5 Torque and Rotational Work Introduction When a torque

!

S is applied to an object, and the object rotates through an angle !" ,

then the torque does an amount of work, !W = "

S!# , on the object. By extension of the

linear work-energy theorem, the amount of work done is equal to the change in the rotational kinetic energy of the object,

W

rot=

1

2I

cm!

f

2"

1

2I

cm!

0

2= #K

rot. (11.5.1)

The rate of doing this work is the rotational power exerted by the torque,

Prot!

dWrot

dt= "

S

d#

dt= "

S$ . (11.5.2)

Rotational work Consider a rigid body rotating about an axis. Each individual element

!m

i in the rigid

body is moving in a circle of radius (r

S ,!)

i about the axis of rotation passing through the

point S . The element, !m

i, undergoes a small angular displacement, !" , under the

action of a tangential force,

!F

tan, i= F

tan, i!̂ , where !̂ is a unit vector pointing in the

tangential direction (Figure 11.7). The element will then have an associated displacement vector for this motion,

!!r

S ,i= (r

S ,")

i!# #̂ . In the limit as !" # 0 , all small quantities

are replaced by differentials; the infinitesimal work done by the tangential force is then

!W

i=!F

tan ,i" !!r

S , i= F

tan, i#̂ " (r

S ,$)

i!# #̂ = F

tan, i(r

S ,$)

i!# . (11.5.3)

Newton’s Second Law applied to the element

!m

i in the tangential direction states that

F

tan, i= !m

ia

tan, i. (11.5.4)

Using our result for tangential acceleration we have that

F

tan, i= !m

i(r

S ,")

i# . (11.5.5)

18

Thus the infinitesimal rotational work is

!W

i= !m

i(r

S ,")

i

2# !$ . (11.5.6)

Summing the infinitesimal work for all the mass elements, we get

!W = !Wi

i

" = !mi(r

S ,# )i

2

i

"$%&'()* !+ . (11.5.7)

In the limit that the discrete mass elements become infinitesimal continuous mass elements,

!m

i" dm , the summation becomes an integral over the body:

!W = !mi(r

S ," )i

2

i

#$%&'()* !+ = dm(r

S ," )2

body

,$

%&

'

() * !+ . (11.5.8)

Since the integral in this expression is just the moment of inertia about a fixed axis passing through the point S , we have for the infinitesimal rotational work

!W = I

S" !# . (11.5.9)

Since the z-component of the torque (in the direction along the axis of rotation) about S is given by

(!

S)

z= I

S" , (11.5.10)

the rotational work is the product of the torque and the angular displacement,

!W = ("

S)

z!# . (11.5.11)

Recall Eq. (11.2.8) that the component of the torque (in the direction along the axis of rotation) about S due to the tangential force,

!F

tan, i, acting on the element

!m

i is

(!

S ,i)

z= F

tan, i(r

S ,")

i, (11.5.12)

and the total torque is just the sum

(!S)

z= (!

S ,i)

z

i

" = Ftan,i

(rS ,#

)i

i

" . (11.5.13)

So the total work done is

19

!W = !Wi

i

" = Ftan, i

(rS ,#

)i!$

i

" = (%S)

z!$ . (11.5.14)

In the limit as the angular displacement approaches zero, d! " 0 , the infinitesimal rotational work is

dW = (!

S)

zd" . (11.5.15)

We can integrate this amount of rotational work as the angle coordinate of the rigid body changes from some initial value

! = !

0 to some final value

! = !

f,

W = dW! =!

0

! =!f

" = (#S)

zd!

! =!0

! =!f

" . (11.5.16)

Rotational Kinetic Energy We have found that the general motion of a rigid body consists of a translation of the center of mass with velocity

!v

cm, and a rotation about the center of mass with angular

velocity, !

cm.

Having defined translational kinetic energy, we now define the rotational kinetic energy for a rigid body about the center of mass. Each individual element

!m

i undergoes

circular motion about the center of mass with angular frequency !

cm in a circle of radius

(r

cm,!)

i. Therefore the velocity of each element is given by

!v

cm, i= (r

cm,!)

i"

cm#̂ . The

rotational kinetic energy is then

K

cm, i=

1

2!m

iv

cm, i

2=

1

2!m

i(r

cm,")

i

2#

cm

2 . (11.5.17)

We now add up the kinetic energy for all the mass elements

Kcm

= Kcm, i

i

! =1

2"m

i(r

cm,# )i

2

i

!$%&

'()*

cm

2=

1

2dm(r

cm,# )2

body

+$

%&

'

()*cm

2=

1

2I

cm*

cm

2 . (11.5.18)

The total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy,

Ktotal

= Ktrans

+ Krot=

1

2mv

cm

2+

1

2I

cm!

cm

2 . (11.5.19)

20

Rotational Work-Kinetic Energy Theorem We will now show that the rotational work is equal to the change in rotational kinetic energy. We begin with our result for the infinitesimal rotational work,

dW

rot= I

S! d" . (11.5.20)

Recall that the rate of change of angular velocity is equal to the angular acceleration,

! " d# dt and that the angular velocity is

! " d# dt . Note that in the limit of small

displacements,

d!

dtd" = d!

d"

dt= d!! . (11.5.21)

Therefore the infinitesimal rotational work is

dWrot= I

S! d" = I

S

d#

dtd" = I

Sd#

d"

dt= I

Sd## . (11.5.22)

Now we can integrate this amount of rotational work as the angular velocity of the rigid body changes from some initial value

! =!

0 to some final value

! =!

f,

Wrot= dW

rot

! =!0

! =!f

" = IS

d! !! =!

0

! =!f

" =1

2I

S!

f

2 #1

2I

S!

0

2 . (11.5.23)

When a rigid body is rotating about a fixed axis passing through a point S in the body, there is both rotation and translation about the center of mass unless S is the center of mass. So if we choose the point S in the above equation for the rotational work to be the center of mass, then

W

rot=

1

2I

cm!

cm, f

2"

1

2I

cm!

cm,0

2= K

rot, f" K

rot,0# $K

rot. (11.5.24)

Recall the work-kinetic energy theorem stated that the total translational work done by all the forces is equal to the change in the translational kinetic energy of the center of mass,

W

trans=

1

2mv

cm, f

2!

1

2mv

cm,0

2" #K

trans. (11.5.25)

Since we did not include the effect of rotational work, we now add the two contributions to the total work and find that the total work done on a rigid body is equal to the total change of the kinetic energy

21

Wtotal

=Wtrans

+Wrot=

1

2mv

cm, f

2 !1

2mv

cm,0

2"#$

%&'+

1

2I

cm(

f

2 !1

2I

cm(

0

2"#$

%&'= )K

trans+ )K

rot. (11.5.26)

Rotational Power The rotational power is defined as the rate of doing rotational work,

Prot!

dWrot

dt. (11.5.27)

We can use our result for the infinitesimal work to find that the rotational power is the product of the applied torque with the angular velocity of the rigid body,

P

rot!

dWrot

dt= ("

S)

z

d#

dt= ("

S)

z$ . (11.5.28)

11.6 Rotational and Translation of a Rigid Body We now return to our translating and rotating rod (Figure 11.1) that we first considered when we began our discussion of rigid bodies. We shall now show that the motion of any rigid body consists of a translation of the center of mass and rotation about the center of mass. We shall demonstrate this for a rigid body by dividing up the rigid body into point-like constituents. Consider two point-like constituents with masses

m

1 and

m

2. Choose a

coordinate system with a choice of origin such that body 1 has position

!r

1 and body 2

has position

!r

2 (Figure 11.11).

Figure 11.11 Center of mass coordinate system.

22

Recall in Chapter 4.1 we defined the center of mass vector,

!R

cm, of the two-body system

is as

!R

cm=

m1

!r

1+ m

2

!r

2

m1+ m

2

(11.6.1)

The position vector of the object 1 with respect to the center of mass is given by

!r

1! =!r

1"!R

cm=!r

1"

m1

!r

1+ m

2

!r

2

m1+ m

2

=m

2

m1+ m

2

(!r

1"!r

2) =

µ

m1

!r (11.6.2)

where

µ =m

1m

2

m1+ m

2

(11.6.3)

is the reduced mass and

!r =!r

1!!r

2 (11.6.4)

is the relative position vector. The displacement of object about the center of mass is given by taking the derivative of Eq. (11.6.2),

d!r

1

! =µ

m1

d!r (11.6.5)

A similar calculation for the position of object 2 with respect to the center of mass yields for the position and displacement with respect tot eh center of mass

!r

2

! =!r

2"!R

cm= "

µ

m2

!r. (11.6.6)

and

d!r

2

! = "µ

m2

d!r. (11.6.7)

Let i =1,2. An arbitrary displacement of the ith object is given respectively by

d!r

i= d!r

i

! + d!R

cm (11.6.8)

23

which is the sum of a displacement about the center of mass

d!r

i

! and a displacement of the center of mass

d!R

cm.

The displacement of objects 1 and 2 are constrained by the condition that the distance between the objects must remain constant since the body is rigid. In particular, the distance between objects 1 and 2 is given by

r

2= (!r

1!!r

2) " (!r

1!!r

2) (11.6.9)

Since this distance is constant we can differentiate Eq. (11.6.9), yielding the rigid body condition that

0 = 2(

!r

1!!r

2) " (d!r

1! d!r

2) (11.6.10)

Translation of the Center of Mass: This condition (Eq. (11.6.10)) can be satisfied if the relative displacement vector between the two objects is zero,

d!r = d

!r1! d!r2=!0 (11.6.11)

This implies, using, Eq. (11.6.5) and Eq. (11.6.7), that the displacement with respect to the center of mass is zero,

d!r

1

! = d!r

2

! =!0 (11.6.12)

Thus by Eq. (11.6.8), the displacement of each object is equal to the displacement of the center of mass,

d!r

i= d!R

cm (11.6.13)

which means that the body is undergoing pure translation. Rotation about the Center of Mass: Now suppose that

d!r = d

!r1! d!r2"!0 . The rigid body condition can be expressed in terms

of the center of mass coordinates. Using Eq. (11.6.5), the rigid body condition (Eq. (11.6.10)) becomes

0 = 2(

!r

1!!r

2) " d!r

1# (11.6.14)

24

In addition, the relative position vector between the two objects is independent of the choice of reference frame,

!r =!r

1!!r

2= (!r

1" +!R

cm) ! (!r

2" +!R

cm) =!r

1" !!r

2" =!r" (11.6.15)

The rigid body condition Eq. (11.6.10) in the center of mass reference frame is then given by

0 = 2(

!r

1! "!r

2! ) # d!!r

1 (11.6.16)

This condition is satisfied if the relative displacement is perpendicular to the line passing through the center of mass,

(!r

1! "!r

2! ) # d

!!r

1 (11.6.17)

By a similar argument

(!r

1! "!r

2! ) # d

!!r2

. In order for these displacements to correspond to a rotation about the center of mass, the displacements must have the same angular displacement.

Figure 11.12: infinitesimal angular displacements in the center of mass reference frame In Figure 11.12, the infinitesimal angular displacement of each object is given by

d!1=

d!r

1

"

"r1

(11.6.18)

25

d!2=

d!r

2

"

"r2

(11.6.19)

From Eq. (11.6.5) and Eq. (11.6.7), we can rewrite Eq. (11.6.19) as

d!1=

µ

m1"r

1

d!r (11.6.20)

d!2=

µ

m2"r

2

d!r (11.6.21)

Recall that in the center of mass reference frame

m

1!r

1= m

2!r

2 (11.6.22)

and hence the angular displacements are equal,

d!

1= d!

2= d! (11.6.23)

Therefore the displacement of the ith object

d!r

i differs from the displacement of the

center of mass d!R

cm by a a vector that corresponds to an infinitesimal rotation in the

center of mass reference frame

d!r

i

! = d!r

i" d!R

cm (11.6.24)

So we have shown that the displacement of a rigid body is the the vector sum of the displacement of the center of mass (translation of the center of mass) and an infinitesimal rotation about the center of mass. 11.7 Kinetic Energy and Torque for Fixed Axis Rotation and Translation We have found that the general motion of a rigid body consists of a translation of the center of mass with velocity

!v

cm, and a rotation about the center of mass with angular

velocity, !

cm. The total kinetic energy is then the sum of the translational kinetic energy

and the rotational kinetic energy,

Ktotal

= Ktrans

+ Krot=

1

2mv

cm

2+

1

2I

cm!

cm

2 . (11.7.1)

26

When a rigid body undergoes translation and fixed axis rotation the z-component of the torque about a point S can be decomposed into the sum of two contributions, the z-component of the torque about a point S due to the total force acting at the center of mass, and the z-component of the torque about the center of mass. Consider the torque about the point S due to the total force

!F

i acting on the ith particle (Figure 11.13).

Figure 11.13 Torque about a point S due to total force

!F

i acting on the ith particle

The position vectors are related by

!r

S ,i=!!ri+!r

S ,cm (11.7.2)

From Eq. (11.2.2), the torque about S is given by

!!

S , i=!r

S , i"!F

i (11.7.3)

Substituting Eq. (11.7.2) into Eq. (11.7.3) yields

!!

S , i= (!"ri+!r

S ,cm) #!F

i=!"ri#!F

i+!r

S ,cm#!F

i (11.7.4)

The total torque about S is then the sum over all the particles

!!

S=

!"ri#!F

i

i=1

i=N

$ + (!r

S ,cm#!F

i

i=1

i=N

$ ) (11.7.5)

The vector

!r

S ,cm is the same constant for each term in the summation and hence can be

pulled outside of the summation yielding

!!

S=

!"ri#!F

i

i=1

i=N

$ +!r

S ,cm#

!F

i

i=1

i=N

$ (11.7.6)

27

Note that the total force is just the sum

!Fi

i=1

i=N

! =!Ftotal (11.7.7)

So the torque about the point S is then

!!

S=

!"ri#!F

i

i=1

i=N

$ +!r

S ,cm#!F

total (11.7.8)

The first term is the torque about the center of mass and is independent of the choice of the point S . The second term is the torque about the point S when all the external forces act at the center of mass. For fixed axis rotation, the z-component of the torque about the point S is then

(!!

S)

z= (!r

S , cm"!F

ext)

z+ ( #

!r

i"!F

i)

z

i=1

N

$ (11.7.9)

28

11.A.1 Appendix First Order Correction to the Period for Large Amplitudes for the Simple Pendulum In Section 11.4 we found that using the small angle approximation the period for a simple pendulum is

T =2!

"pendulum

# 2!l

g.

How good is this approximation? If the pendulum is pulled out to an initial angle

!

0 that

is not small (such that our first approximation, sin(!) " ! no longer holds) then our expression for the period is no longer valid. We then would like to calculate the first-order (or higher-order) correction to the period of the pendulum. Let’s first consider the mechanical energy, a conserved quantity in this system. Choose an initial state when the pendulum is released from rest at an angle

!

0; this need not be at

time t = 0 , and in fact later in this derivation we’ll see that it’s inconvenient to choose this position to be at t = 0 . Choose for the final state the position and velocity of the bob at an arbitrary time t . Choose the zero point for the potential energy to be at the bottom of the bob’s swing (Figure 11.10).

Figure 11.10 Energy states for a simple pendulum. The initial energy is

E

0= K

0+U

0= m g l (1! cos"

0) . (11.7.10)

The tangential velocity of the bob at an arbitrary time t is given by

vtan

= ld!

dt. (11.7.11)

So the kinetic energy at the final state is

29

Kf=

1

2mv

tan

2=

1

2m l

d!dt

"#$

%&'

2

. (11.7.12)

The final energy is then

Ef= K

f+U

f=

1

2m l

d!dt

"#$

%&'

2

+ m g l(1( cos!) . (11.7.13)

Since the tension in the string is always perpendicular to the displacement of the bob, the tension does no work and mechanical energy is conserved,

E

f= E

0. Thus

1

2m l

d!dt

"#$

%&'

2

+ m g l (1( cos!) = m g l (1( cos!0) . (11.7.14)

We can solve Equation (11.7.14) for the angular velocity as a function of ! ,

d!

dt=

2g

l(cos! " cos!

0) . (11.7.15)

Integrating with respect to ! , we get

d!

(cos! " cos!0)

# =2g

ldt# . (11.7.16)

Recall the trigonometric identity cos! = 1" 2sin2(! / 2) . (11.7.17) Then

cos! " cos!

0= 2 sin2(!

0/ 2) " sin2(! / 2)( ) , (11.7.18)

so our integral becomes

d!

2 sin2(!0

/ 2) " sin2(! / 2)( )# =

2g

ldt# . (11.7.19)

We make a change of variables

30

sin! = sin(" / 2) / sin("

0/ 2) . (11.7.20)

Let

k = sin(!

0/ 2) , so that sin(! / 2) = k sin" , and

cos(! / 2) = 1" sin2(! / 2)( )

1 2

= 1" k2 sin2#( )

1 2

. (11.7.21) Please be sure to note that here

k = sin(!

0/ 2) is a dimensionless parameter of the

system, not a spring constant. Our integral equation can be rewritten as

d!

k 2 1" sin2#( )

$ =2g

ldt$ . (11.7.22)

Taking differentials in Equation (11.7.22) and rearranging, we have that

k cos!( )d! = 1 / 2( )cos " / 2( )d" . (11.7.23)

Thus

d! = 2kcos"

cos(! / 2)d" = 2k

1# sin2"

(1# k2 sin2")1 2

d" . (11.7.24)

Substituting this into the left-hand side of the integral yields

2k

k 2(1! sin2")

1! sin2"

(1! k2 sin2")1 2

da# = 2d"

(1! k2 sin2")1 2# . (11.7.25)

Thus the integral equation becomes

d!

(1" k 2 sin2!)1 2# =g

ldt# . (11.7.26)

This integral is called an elliptic integral. We will encounter another elliptic integral when we solve the Kepler Problem, where the orbits of an object under the influence of an inverse square gravitational force are determined. We shall find a power series solution to this integral by expanding the function

31

(1! k

2 sin2")!1 2= 1+

1

2k

2 sin2" +3

8k

4 sin4" + # # # . (11.7.27)

Then the integral equation becomes

1+1

2k 2

sin2! +

3

8k 4

sin4! + " " "

#$%

&'(

d!) =g

ldt) . (11.7.28)

Now let’s integrate over one period. Set t = 0 , when ! = 0 , the lowest point of the swing, so that

sin! = 0 and

! = 0 . One period T has elapsed the second time the bob

returns to the lowest point, or when ! = 2" . Putting in the limits of the ! -integral, we

can integrate term by term, noting that

1

2k

2sin

2! da

!=0

!=2"

# =1

2k

21

21$ cos 2!( )( )d!

!=0

!=2"

#

=1

2k

2 1

2! "

sin(2!)

2

#$%

&'( !=0

!=2)=

1

2)k

2=

1

2) sin2(*

0/ 2) . (11.7.29)

Thus we have that

1+1

2k 2

sin2! +

3

8k 4

sin4! + " " "

#$%

&'(

d!!=0

!=2)

* =g

ldt

t=0

t=T

* , (11.7.30)

which becomes after integration

2! +

1

2! sin2("

0/ 2) + # # # =

g

lT . (11.7.31)

We can now solve for the period,

T = 2!l

g1+

1

4sin2("

0/ 2) + # # #

$%&

'()

. (11.7.32)

If the initial angle is still small compared to 1 (measured in radians),

!

0<< 1 , then

sin2(!

0/ 2) " !

0

2 / 4 and the period is approximately

T ! 2"l

g1+

1

16#

0

2$%&

'()= T

01+

1

16#

0

2$%&

'()

, (11.7.33)

32

where

T0= 2!

l

g (11.7.34)

is the period of the simple pendulum with the standard small angle approximation. The first order correction to the period of the pendulum is then

!T1"

1

16#

0

2T

0. (11.7.35)