chapter 8-rotational motion - university of...

Chapter 8- Rotational Motion

AssignmentAssignment 99Textbook (Giancoli, 6 th edition):

Due on Thursday, November 26.

1. On page 131 of Giancoli, problem 18.



4. A cue stick strikes a cue ball horizontally at a point a distance d above the centre of the ball as in the figure below. Find the value of d for which the cue ball will roll without slipping from the beginning. Express your answer in terms of the radius R of the ball. (Hint one: you will find the moment of inertia of a ball or sphere on page 208 of Giancoli. Hint two: write the “no slip" condition as aball = Rα where aball is the linear acceleration of a point at the centre of the ball.).

Assignments and midterm exam

Solutions are available on the web

Chapter 8

• Angular Quantities

• Constant Angular Acceleration

• Rolling Motion (Without Slipping)

• Centripetal Forces

• Torque

• Rotational Dynamics; Torque and Rotational Inertia

• Rotational Kinetic Energy

• Angular Momentum and Its Conservation

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

Rotational Dynamics; TorqueRotational Dynamics; Torque

m i

ri

(8-22)

Note that F|| does not exert torque since its lever arm is zero .

Or , using

ττττi is known as torque , or moment of the force.

(8-23)

ri

ri


We can extend equation 8-23 to apply to the entire door. For this purpose, we have repeat our calculations for each section “i ” of the door and add the results to find the net torque exerted on the door:

m i

ri

(8-24)

But, according to eq. 8-19:

We can then write the net torque in terms of the moment of inertia and angular acceleration

ri

ri

(8-25)


Equation 8-25 is the rotational equivalent of Newton’s 2nd law for linear motion.

Here, the moment of inertia I plays the same role as the object’s mass m in F = ma. It tells us how difficult is to set an object in rotational motion.

(8-25)

TodayTodayTodayToday

Problem 8-37 (textbook): A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m.N . If the mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?

Problem 8-37:

The torque on the rotor will cause an angular acceleration given by

The torque and angular acceleration will have the opposite sign of the initial angular velocity because the rotor is being brought to rest.

The rotational inertia is that of a solid cylinder. Substitute the expressions for angular

Iα τ=

The rotational inertia is that of a solid cylinder. Substitute the expressions for angular acceleration and rotational inertia into the equation

and solve for the angular displacement.

2 2 2oω ω α θ= +

Problem 8-37:

( ) ( )

( )( )

2 2

2 2

212

22

2

0

2 2 42

rev 2 rad 1 min4.80 kg 0.0710 m 10,300

1 revmin 1 rev 60 s 5865 rad

o

MR

I MRο ο ο ο

ω ω αθ

ω ω ω ω ωθ

α τ ττ

π

2 2 2 2

= +

− − − −= = = =

−= =

The time can be found from

( )1 revmin 1 rev 60 s

5865 rad4 1.20 N m 2 rad

993 rev

π= =

−

=

�

( )12 o tθ ω ω= +

( )2 9 9 3 re v2 6 0 s1 0 .9 s

1 0 , 3 0 0 re v m in 1 m ino

tθ

ω ω= = =

+

Problem 8-40 (textbook): A helicopter rotor blade can be considered a long thin rod, as shown in Fig. 8–46. (a) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 160 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. (b) How much torque must the motor apply to bring the blades up to a speed of 5.0 rev/s in 8.0 s?

Problem 8-40:

(a) The moment of inertia of a thin rod, rotating about its end, is given in Figure 8-21(g). There are three blades to add.

(b) The torque required is the rotational inertia times the angular acceleration, assumed constant.

( ) ( )( )22 2 2 2 21total 3

3 160 kg 3.75 m 2250 kg m 2.3 10 kg mI ML ML= = = = ≈ ×� �2250 kg.m2 2.3 x 103 kg.m2

assumed constant.

( ) ( )( )2 30total total

5.0rev/sec 2 rad rev2250kg m 8.8 10 m N

8.0 sI I

t

πω ωτ α

−= = = = ×� �

Type Linear Rotational Relation

Displacement x θ x = r θ

Velocity v ω v = r ω

Acceleration a αααα a = r αααα

Dynamics F (force) ττττ (torque) ττττ = F d sin( θθθθ)

Inertia m (mass) I (moment of inertia)

Angular Momentum and Its ConservationAngular Momentum and Its ConservationWe have defined several angular quantities in analogy to linear motion.

Inertia m (mass) I (moment of inertia)

Kinetic Energy

Newton’s 2nd Law

Equations of motion

Angular Momentum and Its ConservationAngular Momentum and Its Conservation

To complete the analogy to linear motion, few quantities as well as an important law are missing from the previous table:

1. Angular momentum (equivalent to linear momentum in linear motion)

2. Work in rotational motion (equivalent to work done by a force in translational motion)

3. Rotational Impulse (equivalent to linear impulse done by a force in linear motion)motion)

4. Generalized form of Newton’s 2nd law (equivalent to the generalized form of Newton’s 2nd law in terms of momentum in linear motion)

5. Conservation of total angular momentum (equivalent to conservation of total linear momentum in linear motion)

Let’s now find each of these items.


1) Angular momentum

In linear motion, linear momentum is written as:

The equivalents for mass and velocity in rotational motion are moment of inertia, I, and angular velocity, ω.

We can then define the following equivalent quantity in rotational motion:We can then define the following equivalent quantity in rotational motion:

L is called angular momentum .

(8-26)


2) Work in angular momentum

In linear motion, the work done by a force is defined as:

(only the parallel component of the force to the direction of motion matters)

The equivalents for force and displacement in rotational motion are torqueexerted by F, ττττ = Fd, and angular displacement θ.

We can then define the following equivalent quantity in rotational motion:We can then define the following equivalent quantity in rotational motion:

Eq. 8-27 represents the work done by the torque ττττ when rotating an object through an angle ∆θ.

Using the above equation, we can obtain the power P (rate with it the work is done):

(8-27)

(8-28)


3) Impulse in angular momentum

In linear motion, the impulse done by a force is defined as:

Similarly, we can define impulse in rotational motion as:

(8-29)


4) Generalized form of Newton’s 2 nd Law

In linear motion, the general form of Newton’s 2nd law can be expressed as:

This form includes general cases where the mass of an object can change � remember the mass of a rocket changing as it burns fuel to accelerate.

Similarly, we can write the generalized form of the rotational equivalent of Newton’s 2nd law in terms of the angular momentum L:

Here, changes in the momentum of inertia ( L = Iωωωω ) are taken into account in the formalism of Newton’s 2nd law.

(8-30)


5) Conservation of Angular Momentum

In linear motion, the total linear momentum is conserved if there is no net external force acting on the system:

In this case, the total linear momentum does NOT change with time � it is conserved. conserved.

Similarly, if there is no net torque acting on a system, we have:

And the total angular momentum does NOT change with time:

� The total angular momentum is conserved if there is NO external net torque acting on the system.

(8-31)

Angular Momentum and Its ConservationAngular Momentum and Its ConservationThere are clear applications of angular momentum conservation in your day-to-day life.

Let’s recall the angular momentum definition (eq. 8-26):

Assume now that no external torque is exerted on the skater depicted below and ignore friction.

As she brings her arms closer to her body, her moment of inertia I will change (will decrease �the distance between the “particles” in her body will be closer to her axis of rotation).

Since NO external torque is applied on her, her

total angular momentum should be conserved:

Therefore, if her momentum of inertia decreases,

her angular velocity has to increase to maintain

the total angular momentum constant.

Problem 8-53 (textbook): A person stands, hands at his side, on a platform that is rotating at a rate of 1.30 rev/s. If he raises his arms to a horizontal position, Fig. 8-48, the speed of rotation decreases to 0.80 rev/s.

(a) Why?

(b) By what factor has his moment of inertia changed?

Problem 8-53 (textbook):

(a)Consider the person and platform a system for angular momentum analysis. Since the force and torque to raise and/or lower the arms is internal to the system, the raising or lowering of the arms will cause no change in the total angular momentum of the system. However, the rotational inertia increases when the arms are raised. Since angular momentum is conserved, an increase in rotational inertia must be accompanied by a decrease in angular velocity.

(b)

The rotational inertia has increased by a factor of 1.6

1.30 rev s 1.625 1.6

0.80 rev si

i f i i f f f i i i i

f

L L I I I I I I Iω

ω ωω

= → = → = = = ≈


(a) What is the angular momentum of a figure skater spinning at 3.5 rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 15 cm, and a mass of 55 kg?

(b) How much torque is required to slow her to a stop in 5.0 s, assuming she does notmove her arms?


(a)

(b)

If the rotational inertia does not change, then the change in angular momentum is

( )( )22 21 12 2

rev 2 rad55 kg 0.15 m 3.5 14 kg m s

s 1 revL I MR

πω ω= = = =

�

If the rotational inertia does not change, then the change in angular momentum is strictly due to a change in angular velocity.

The negative sign indicates that the torque is in the opposite direction as the initial angular momentum.

20 1 4 k g m s2 .7 m N

5 .0 s

L

tτ ∆ −

= = = −∆

��

Problem 8-62 (textbook): A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 Kg.m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now? What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?


The angular momentum of the merry-go-round and people combination will be conserved because there are no external torques on the combination. This situation is a totally inelastic collision, in which the final angular velocity is the same for both the merry-go-round and the people.

Subscript 1 represents before the collision, and subscript 2 represents after the collision. The people have no initial angular momentum.

1 2 1 1 2 2 L L I Iω ω= → = →

.

If the people jump off the merry-go-round radially, then they exert no torque on the merry-go-round, and thus cannot change the angular momentum of the merry-go-round. The merry-go-round would continue to rotate at .0.80 rad/s.

1 2 1 1 2 2

( )( )( )

m-g-r m-g-r12 1 1 1 2

2 m-g-r people m-g-r person

2

22

4

1760 kg m 0.80 rad s 0.48 rad s

1760 kg m 4 65 kg 2.1 m

I II

I I I I M Rω ω ω ω= = =

+ +

= =+

�

�

chapter 8-rotational motion - university of...

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