chap6

Chapter 6 Rotation and Angular Momentum.

Chapter 6 Rotation and Angular Momentum

Basic Requirements：

1. Understand the concept of translation and rotation;

2. Master the kinematic equations for constant angular acceleration;

3. Master the relationship between the linear and angular variables;

4. Master kinetic energy of rotation;

5. Master the calculation of the rotational inertia;

6. Master the parallel-axis theorem;

7. Learn to apply Newton's second law for rotation;

8. Master the work-kinetic energy theorem for rotation.9. Understand the concept of rolling;

10. Master the kinetic energy of rolling;

11. Master the forces of rolling;

12. Master angular momentum;

13. Learn to apply Newton's second law in angular form;

14. Understand the angular momentum of a system of particles;

15. Understand the angular momentum of a rigid body rotating about a fixed axis;

16. Master the conservation law of angular momentum.

Review and Summary

Static Equilibrium A rigid body at rest is said to be in static equilibrium. For such a

body , the vector sum of the external forces acting on it is zero:

(balance of forces ) (6-3)

If all the forces lie in the plane, this vector equation is equivalent to two

component equations:

and (balance of forces) (6-7,6-8)

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Static equilibrium also implies that the vector sum of the external torques acting on the

body about any point is zero, or

(balance of torques) (6-5)

If the forces lie in the plane, all torque vectors are parallel to the axis, and Eq.6-

5 is equivalent to the single component equation

(balance of torques). (6-9)

Angular Position To describe the rotation of a rigid body about a fixed axis, called

the rotation axis, we assume a reference line is fixed in the body, perpendicular to that axis and rotating with the body. We measure the angular position of this line

relative to a fixed direction. When is measured in radians,

(radian measure), (6-10)

Where is the arc length, of a circular path of radius and angle . Radian measure

is related to angle measure in revolutions and degrees by

(6-11)

Angular Displacement A body that rotates about a rotation axis , changing its angular

position from to , undergoes an angular displacement

(6-13)

Where is positive for counterclockwise rotation and negative for clockwise

rotation.Angular Velocity and Speed If a body rotates through an angular displacement in

a time interval , its average angular velocity is

. (6-14)

The (instantaneous) angular velocity of the body is

(6-15)

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Both and are vectors, with directions given by the right hand rule. they are

positive for counterclockwise rotation and negative for clockwise rotation.

The magnitude of the body’s angular velocity is the angular speed.

Angular Acceleration If the angular velocity of a body changes from to in a

time interval , the average angular acceleration of the body is

(6-16)

The (instantaneous) angular acceleration of a body is

(6-17)

Both and are vectors.

The Kinematic Equations for Constant Angular Acceleration Constant angular acceleration ( =constant) is an important special case of rotational motion .The

appropriate kinematic equations are

(6-18)

(6-19)

(6-20)

(6-21)

(6-22)

Linear and Angular Variables Related A point in a rigid rotating body , at a

perpendicular distance from the rotation axis , moves in a circle with radius . If

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the body rotates through an angle , the point moves along an arc with length given

by


Where is in radians .

The liner velocity of the point is tangent to the circle, the point’s liner speed is

given by (radian measure), (6-24)

Where is the angular speed (in radians pre second) of the body .

The liner acceleration of the point has both tangential and radial components. The

tangential component is


Where is the magnitude of the angular acceleration (in radians per second-squared )

of the body . The radial component of is


If the point moves in uniform circular motion. the period of the motion for the

point and the body is

(radian measure) (6-25,6-26)

Rotational Kinetic Energy and Rotational Inertia The kinetic energy of a rigid

body rotating about a fixed axis is given by

(radian measure) (6-40)

In which is the rotational inertia of the body, defined as

(6-39)

for a system of discrete particles and as

(6-43)

for a body with continuously distributed mass. The and in these expressions

represent the perpendicular distance from the axis of rotation to each mass element in

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the body. The Parallel-Axis Theorem The parallel-axis theorem relates the rotational inertia

of a body about any axis to that of the same body about a parallel axis through the

center of mass:

(6-42)

Here is the perpendicular distance between the two axes.

Torque Torque is a turning or twisting action on a body about a rotation axis due to a

force . If is exerted at a point given by the position vector relative to the axis,

then the magnitude of the torque is

Where is the component of perpendicular to , and is the angle between

and . The quantity is the perpendicular distance between the rotation axis and an

extended line running through the vector. This line is called the of action of .

Similarly, is the moment arm of .

The SI unit of torque is the Newton-meter ( ). A torque is positive if it tends to

rotate a body at rest counterclockwise and negative if it tends to rotate the body in the

clockwise direction.

Newton’s Second Law in Angular Form The rotational analog of Newton’s second

law is

, (6-45)

Where is the net torque acting on a particle or rigid body, is the rotational

inertia of the particle or body about that rotation axis. and is the resulting angular

acceleration about that axis.

Work and Rotational Kinetic Energy The equations used for calculating work and

power in rotational motion correspond to equations used for translational motion and

are

(6-56)

When is constant, Eq. 6-56 reduces to

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(6-57)

And (6-58)

The form of the work-kinetic energy theorem used for rotating bodies is

(6-60)

Rolling Bodies For a wheel of radius that is rolling smoothly (no sliding),

(6-67)

Where is the linear speed of the wheel’s center and is the angular speed of the

wheel about its center. The wheel may also be viewed as rotating instantaneously about the point of the “road” that is in contact with the wheel. The angular speed of

the wheel about this point is the same as the angular speed of the wheel about its

center. The rolling wheel has kinetic energy

(6-70)

Where is the rotational moment of the wheel about its center and is the mass

of the wheel. If the wheel is being accelerated but is still rolling smoothly, the

acceleration of the center of mass is related to the angular acceleration about

the center with

(6-71)

If the wheel rolls smoothly down a ramp of angle , its acceleration along an axis

extending up the ramp is

(6-75)

Torque as a Vector In three dimensions, torque is a vector quantity defined relative

to a fixed point (usually an origin); it is

, (6-76)

Where is a force applied to a particle and is a position vector locating the particle

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relative to the fixed point (or origin). The magnitude of is given by

(6-77,6-78,6-79)

Where is the angle between and , is the component of perpendicular to

, and is the moment arm of . The direction of is given by the right-hand rule

for cross products.

Angular Momentum of a particle The angular momentum of a particle with linear

momentum , mass , and linear velocity is a vector quantity defined relative to a

fixed point (usually origin); it is

(6-80)

The magnitude of is given by

(6-81)

(6-82)

(6-83)

Where is the angle between and , and are the components of and

perpendicular to , and is the perpendicular distance between the fixed point and

the extension of . The direction of is given by the right-hand rule for cross

products.

Newton’s Second Law in Angular Form Newton’s second law for a particle can be

written in angular form as

, (6-85)

Where is the net torque acting on the particle, and is the angular momentum of

the particle.

Angular Momentum of a System of Particles The angular momentum of a

system of particles is the vector sum of the angular momentum of the individual

particles:

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(6-88)

The time rate of change of this angular momentum is equal to the net external torque

on the system (the vector sum of the torques due to interactions of the particles of the

system with particles external to the system):

(system of particles) (6-91)

Angular Momentum of a Rigid Body For a rigid body rotating about a fixed axis is (rigid body, fixed axis). (6-93)

Conservation of Angular Momentum The angular momentum of a system

remains constant if the net external torque acting on the system is zero: (isolated system) (6-94)

Or (isolated system) . (6-95)

This is the law of conservation of angular momentum. It is one of the fundamental

conservation laws of nature, having been verified even in situations (involving high-

speed particles or subatomic dimensions) in which Newton’s laws are not applicable.

Examples

Example 1 .Show that the moment of inertia of a uniform hollow cylinder of inner

radius ,outer radius ,and mass , is ,as stated in the

figure, if the rotation axis is through the center along the axis of symmetry.

Solution：We know that the moment of inertia of a thin ring of radius is m So we divide the cylinder into thin concentric cylindrical rings or hoops of thickness , one of which is indicated in

Fig.6-1. If the density (mass per unit volume ) is ,then

,

Where is the volume of the thin ring of radius ,

thickness , and height h . Since

Fig. 6-1 Example 1

We have

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Then the moment of inertia is obtained by integrating (summing) over all these hoops :

，

Where we are given that the cylinder has uniform density , constant .(If this were

not so ,we would have to know as a function of before the integration could be

carried out) The volume of this hollow cylinder is ,so its mass

is

Since ,we have

(Answer)

As stated in figure. As a check ,note that for a solid cylinder , and we obtain,

with :

,

Which is that given in Fig.10-21c for a solid cylinder of mass and radius .

Example 2 what will be the speed of a solid sphere of mass and radius when it

reaches the bottom of an incline if it starts from rest at a vertical height and rolls

without slipping ? see Fig.6-2. Ignore losses due to dissipative forces ,and compare

your result to that for an object sliding down a frictionless incline.

Solution ： We use the law of conservation of energy, and we must now include rotational

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kinetic energy. The total energy at any point a vertical distance above the base of

the

incline is

,

Where is the speed of the . We equate Fig. 6-2 Example 2

the total energy at the top and to the total energy at the bottom (

);

.

The moment of inertia of a solid sphere about an axis through its is

.

Since the sphere rolls without slipping , the speed , , of the center of mass with

respect to the point of contact (which is momentarily at rest at any instant) is equal to the speed of a point on the edge relative to the center. We therefore have .

Hence

.

Dividing out the s and s, we obtain

So

. (Answer)

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Note first that is independent of both the mass and the radius of the sphere .

Also , we can compare this result for the speed of a rolling sphere to that for an object

sliding down a plane without rotating and without friction ( ), in which

case , which is greater. An object sliding without friction transforms its

initial potential energy into translational kinetic energy

(none into rotational kinetic energy ), so its speed is

greater.

Example 3 Suppose a 60-kg person stands at the

edge of a 6.0-m-diameter circular platform, which is

mounted on frictionless bearings and has a moment of

inertia of 1800kg. . The platform is at rest initially,

but when the person begins running at a speed of 4.2m/s

(with respect to the ground).around its edge the platform begins to rotate in the

opposite direction as in figure 6-3. Calculate the angular velocity of the platform.

Fig. 6-3 Example 3

Solution： The total angular momentum is zero initially .Since there is no nettorque, is conserved and will remain zero ,as in Fig.6-3.The person’s angular

momentum is and we take this as positive .the angular

momentum of the platform is .Thus

So

. (Answer)

The frequency of rotation is and the period

per revolution. (Answer)

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B An


Example 4 A uniform rod of length and mass is placed horizontally by

putting its left end on the edge of a table and holding its right end with your

hand . Then you release the end suddenly . find , at the instant end is

released (a) the acceleration of the rod’s center of mass

(b) the force exerted on the rod at end

Solution: Draw the free-body diagram of the rod .set the coordinate system as in the figure. At the moment the end is released there are two forces acting on the

rod: the gravitational force exerted at

the center of the rod , downward. The force

exerted at end from the table. Apply

the Newton’s second law for the center of mass

of the rod. Fig. 6-4 Example 4

In direction: (1)

In direction: (2)

According to the theorem of rotation we write:

(3)

from the relationship between the linear and angular quantities we have

(4)

and (5)

At the moment end is just released .the speed of the rod’s center of mass

thus

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NF

SF

gF


Substituting into Eq. (2), yields

Then solving equations (3), (4) and (5) for yields

.

Substituting into Eq. (1) we obtain

(Answer)

Example 5 A uniform solid ball of radius and mass rolls down a circular track

from rest without sliding .The track, radius , is in the vertical plane. At the

beginning the ball is in the same height as the center of the circular track , Find

(a) the ball’s speed as it rolls down to the bottom of the track,

(b) the normal force on the track from the ball at the same moment as in (a).

Solution：(a) Taking the circular track as the referenceframe , the system of the problem includes the ball ,

the track and the Earth. Draw the free-body diagram

of the ball in the process of rolling down the track as

shown in figure 6-5. Three forces , the gravitational

force , the normal force and the static Fig. 6-5 Example 5

frictional force are acting on the ball, But only the gravitational force , being a

conservative force, does work in the rolling process .Thus the mechanic energy of the

system conserved. When the ball is at the bottom of the track, its center is chosen as

the reference of the zero potential energy. From the conservation law of mechanic

energy , we have

(1)

Where is the rotational inertia of the ball about a axis through its center, is the

speed of the center of mass (COM) of the ball , Then we have

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(2)

(3)

Solve these three equations for yield

(b) When the ball rolls down to the bottom of the circular track, two forces, and

.exerted on the ball , Write Newton’s second law in the normal direction

Therefore

This is the normal force on the ball from the track, Write Newton’s third law it is

known that the normal force on the track from the ball of the bottom

(Answer)

Example 6 A thin rod of mass and length is connected with a small ball of the

same mass at the rod’s one end, and the other end of the rod is pivoted on a

frictionless hinge as shown in the figure ,The rigid body is held at rest horizontally and

then released. What is (a) the rotational inertia of the rigid

body about the hinge ? (b) the angular speed of the rigid body as its rod forms an angle with the vertical line ? (c)

the angular acceleration of the rigid body at the same

instant as in(b)? (d) the normal acceleration of the center of

mass of the rigid body at the same instant as in (b)?

Solution: (a) the rotational inertia of the rigid body

about the hinge is Fig. 6-6 Example

6

(1) (Answer)

(b) During the rotation of the rigid body only the gravitational force does work so the

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process obeys the conservation law of mechanic energy . But first of all we should find

out the center of mass of the rigid body by

(2)

When the rod forms an angle with the vertical line its center of mass has

descent a distance

(3)

Applying the conservation law of mechanic energy. We write

Substituting (1), (2) and (3) into above equation and solve for , yields

(Answer)

(c) To find out the angular acceleration of the rigid body at the same instant as in (b) we use Newton’s second law for rotation

(Answer)

(d) The normal acceleration of the center of mass of the rigid body at the same instant

as in (b) is

(Answer)

Example 7 As show in the figure, the masses of wheels and are and and

the radii of them are and respectively, there is a thin rope rolling around the two

wheels and connecting the wheels as shown in the figure, where wheel rotates

about the fixed axis. (1) when wheel drops what is the acceleration of the center

of the wheel? (2) what is pulling force of the rope?

Solution: We consider the problem as the combined motion of the two wheels. That

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is the rotation of wheels and the plane motion of wheel . Then we can apply the

law of rotation (the Newton’s second law for rotation) and the Newton’s second law to

get the answer. As always we begin with a free-body diagram as shown in thefigure Wheel rotates about the fixed axis (perpendicular to the page). From the

law of rotation

Fig. 6-7

Example 7

(1)

Wheel rotates about the instantaneous axis (perpendicular to the page) while its

center of mass is moving down translationally. According to second law

for the motion of the center of mass of wheel

(2)

From the law of rotation for wheel

(3)

By the relationship between the linear quantities and the angular quantities we have

, (4)

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Where and are accelerations of a point at the edge of wheel and

respectively.

There are two more relations which help the problem solving

(5)

(6)

Solving above six equations for and , obtaining

(Answer)

(Answer)

Example 8 A thin rod of mass and length can rotate about a frictionless axis

freely. A small ball of mass is suspended at the end of a massless cord of length

on the same axis as shown in the figure . At the beginning the rod rests in vertical position and the ball is pulled with its cord makes an angle of with the rod. Then the

ball is released and swing down making a elastic collision with the rod. The rod deflects a maximum angle . Write down enough equations to determine the angle

.

Solution: We analyze the whole process by

dividing it into four successive parts as follows

and then use conservation law in each of them

(a) During the swing process of the ball, only the

gravitational force does work, so the mechanic

energy of the ball is conserved.

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(1) (Answer)

Where is the speed of the ball just before it Fig. 6-8 Example 8

collides with the rod.

(b) As the ball collides with the rod, there is no external torque about the reference point , thus the angular momentum of the system (the ball plus the rod) is conserved

(2) (Answer)

Where and are the speed of the ball just before and after the collision, is the

angular speed of the rod just after the collision.

(c) Since the collision between the ball and the rod is elastic, the mechanic energy of

the system is conserved just before and after the elastic collision.

(3) (Answer)

(d) In the consequential swing process of the rod the mechanic energy of the rod also

is conserved

(4) (Answer)

Solving above four equations for we can get the answer.

Note that the most common mistake in solving this problem is apply the

conservation law of linear momentum during the collision between the ball and the rod

instead of equation (2). This is wrong because during the collision the horizontal

component of the force on the rod from the axis can not be neglected. This does not

satisfy the condition of applying the conservation law of linear momentum.

Problem Solving1 (12) The wheel in Fig.6-9 has eight equally spaced

spokes and a radius of . It is mounted on a fixed axle

and is spinning at . You want to shoot a

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long arrow parallel to this axle and through the wheel without hitting any of the

spokes. Assume that the arrow and the spokes are very thin. (a) What minimum speed

must the arrow have? (b) Does it matter where between the axle and rim of the wheel

you aim? If so,

what is the best location? Fig. 6-9 Problem 1

Solution: The angular speed of the rotating wheel is

The angle between the maximum angular displacement when a arrow passes throug

the wheel, is

The time required for a spaced spoke turns around theangle is

During this time interval the arrow must pass though the wheel with

the required speed

(Answer)

(b) Since in equal time interval the wheel turns a equal angle it does not matter where

between the axle and rim of the wheel you aim. (Answer)

2 (16) An early method of measuring the speed of light makes use of a rotating slotted

wheel. A beam of light passes through one of the slots at the outside edge of the wheel,

as in Fig.6-10, travels to a distant mirror, and returns to the wheel just in time to pass

through the next slot in the wheel. One such slotted wheel has a radius of 5.0㎝ and 500 slots around its edge. Measurements taken when the mirror is L=500m from the

wheel indicate a speed of light of 3.0×105㎞／ s.(a)What is the (constant) angular speed of the wheel? (b)What is the linear speed

of a point on the edge of the wheel?

Solution: The distant for light beam traveling

through one slot to the reflecting mirror and

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backing through the next slot in the wheel is

The time for the light bean travel that distance is

During this time interval the wheel just

turns one slot, the angular displacement is

Fig. 6-10 Problem 2

So the angular speed of the wheel is

(Answer)

The linear speed of a point on the edge of the wheel is

(Answer)

3 (20) In Fig.6-11, a cylinder having a mass of 2.0㎏ can rotate about its central axis

through point O. Forces are applied as shown , ,

and .Also, and . Find the

(a)magnitude and(b)direction of the angular acceleration of the cylinder.(During the

rotation, the forces maintain their same angles relative to the cylinder.)

Solution: We apply the Newton’s second law for rotation to find out the angular

acceleration of the cylinder. Since passes

through the rotational axis it produce no torque

about the same axis therefore

Taking the counterclockwise direction

as positive for torques and the angular

Acceleration of the cylinder, we have Fig. 6-11 Problem 3

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(Answer)

The direction of the angular acceleration of the cylinder is counterclockwise.

4 (22) Figure 6-12 shows a rigid assembly of a thin hoop (of mass m and radius ) and a thin radial

rod (of mass m and length ). The assembly is

upright, but if we give it a slight nudge, it will rotate

around a horizontal axis in the plane of the rod and hoop,

through the lower end of the rod. Assuming that the

energy given to the assembly in such a nudge is

negligible, what would be the assembly’s angular speed Fig. 6-12 Problem 4

about the rotation axis when it passes through the upside-down (inverted) orientation?

Solution: In the rotating process only the gravitational force does work on the rigid

assembly, so we can use the conservation law of mechanical energy to find out the

answer. But first we must find out the position of the center of mass of the assembly

Next we should work out the rotational inertia of the rigid assembly about a horizontal

axis in the plane of the rod and hoop through the low end of the rod

Choose point as the reference level for .

From the conservation law of mechanical energy, we have

Solving for yields

(Answer)

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5 (23) A tall, cylindrical chimney falls over when its base is ruptures. Treat the

chimney as a thin rod of length . At the instant it makes an angle of with

the vertical as it falls, what are (a) its angular speed, (b) the radial acceleration of the

top, and (c) the tangential acceleration of the top.(Hint: Use energy considerations, not

a torque.) (d) At what angle is the tangential acceleration equal to ?

Solution: In the system under consideration ( the chimney plus the earth ),only the

gravitational force does work. So we can apply the conservation law to get the

answer. Supposing the mass of the chimney is M and taking the position of the center of mass of the chimney when it makes an angle of with the vertical as it falls

to be the zero potential energy level.(a) Form the principle of conservation of

mechanic energy we have

= （Answer）（b）The linear speed

of the top of the chimney is

So the radial acceleration of the top is

(Answer)

(c) The tangential acceleration of the top can be obtained by Newton’s second law for

rotation

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The angular acceleration of the rotationally falling chimney is when

So the tangential acceleration of the top is

(Answer)

(d) From equation when

we have ，Then

(Answer)

6 (31) Four particles, each of mass 0.20㎏, are placed at the vertices of a square with

sides of length 0.50 ｍ .The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through

one of the particles. The body is released from rest with rod AB horizontal, as shown

in Fig.6-13. (a)What is the rotational inertia of the body about axis A? (b)What is the

angular speed of the body about axis at the instant rod AB swings through the vertical

position?

Solution: (a) The rotation inertia of the body about axis A is

2

2 (answer)

Note that the particle at vertex A does not have rotational

inertia about axis A. Fig. 6-13 Problem

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6

(b) During the rotation of the body only the gravitational force does work, so the body

is mechanic energy is conserved.

The center of the mass C of the body is located at the center of the square, at the instant rod AB swings through the vertical position the center of mass C is a distance

below its initial position. Thus the potential energy of the body in the initial position is

. At the vertical position the kinetic energy of the body is .

From the conservation law of mechanic energy we have

Solving for yields

(Answer)

7 (40) Figure 6-14 shows a rigid structure consisting of a circular hoop of radius

and mass , and square made of four thin bars, each of length and mass . The

rigid structure rotates at a constant speed about a vertical axis, with a period of rotation

of . Assuming and ,calculate (a) The structure’s

rotational inertia about the axis of rotation and (b) its angular momentum about that

axis.

Solution: (a) The structure’s rotational inertia about the axis of rotation is

2 2 2 2 2

2 2 2 (Answer)

(b) The angular speed of the rotating rigid structure is

Thus the angular momentum of the rigid structure

about the rotational axis is

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(Answer) Fig. 6-14 Problem 7

8 (43) In Fig.6-15, a small block slides down a frictionless surface through

height and then sticks to a uniform rod of mass and length

.The rod pivots about point through angle before momentarily stopping. Find .

Solution : The whole process can be divided into three

parts :

(1), The small block slides down the frictionless surface through height , In this part only the gravitational force

, being a conservitive force ,does work, so the law of

conservation of mechanic energy holds

(1)

Where is the speed of the block before it collides with Fig. 6-15 Problem 8

the rod.

(2) The small block collides with the rod and sticks to it. During this interaction there

is no net torque acting on the block –rod system relative to the point , the angular

momentum of the system is conserved (Note: since there is a net force acting on the

rod at point by the pivot, the law of conservation of linear momentum does not

hold! )

(2)

Where is the angular speed of the system about point just after the collision. is

the rotational inertia of the block-rod system about point ,which is

(3)

(3) The block-rod system swings up until it momentarily stops , During this process

the mechanic energy of the system is conserved , we thus write

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(4)

Where is the height change of the center of mass of the block-rod system in

the swing up process. In the vertical position the center of mass of the system is

below point at

So (5)

Substituting all the known values into equations (1)~(5), we can then get the answer as

the following steps:

Form Eq.(1)

From Eq.(3)

From Eq.2 we have

From Eqs. (4) and (5), we have

Thus the angle we look for is

(Answer)

9 (44) A certain gyroscope consists of a uniform disk with a radius mounted at

the center of an axle that is long and of negligible mass. The axle is horizontal

and supported at one end. If the disk is spinning around the axle at ,

what is the precession rate?

Solution: According to formula of the procession rate of a Gyroscope

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Where the rotational inertia of the disk about the horizontal axle is

The angular speed of the spinning dist is

Therefore the precession rate of the gyroscope is

(Answer)

10 (45) Figure6-16 shows an overhead view of a ring that can rotate about its center

like a merry-go-round. Its outer radius is , its inner radius is ,

its mass M is , and the mass of the crossbars at

its center is negligible. It initially rotates at an angular

speed of with a cat of mass

on its outer edge, at radius . By how much does the

cat increase the kinetic energy of the cat-ring system if

the cat crawls to the inner edge, at radius ?

Solution: During the process as the cat crawls to the inner edge from the out edge of

the rotating ring there is not torque acting along the rotating axis, so the angular

Fig. 6-16 Problem 10

momentum of the cat-ring system is conserved. From the conservation law of angular

momentum of a rigid body rotating about a fixed axis we have

(1)

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Where and are the rotational inertia of the cat-ring system with the cat on the

ring s inner edge and on the out edge, respectively. Work out for them

(3)

From the problem, the initial angular speed of the cat-ring system

Substituting , and into equation (1) and solving for lead to

Next let us calculate the kinetic energy of the car-ring system both at the initial stage

and at the final stage . The initial kinetic energy of the system is

The final kinetic energy of the system is

Thus the amount of the increased kinetic energy of the cat-ring system as the cat

crawls to the inner edge from the outer edge of the ring is

(Answer)

11 (46) A uniform wheel of mass and radius is mounted rigidly on an

axle through its center (Fig.6-17). The radius of the axle is , and the rotational

inertia of the wheel-axle combination about its central axis is . The wheel

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is initially at rest at the top of a surface that is inclined at angle with the

horizontal; the axle rests on the surface while the wheel extends into a groove in the

surface without touching the surface. Once released, the axle rolls down the surface by

, what are (a) its rotational kinetic energy and (b) its translational kinetic

energy?

Solution: The key idea here is that as the wheel-

axle combination rolls down the inclined surface

only the gravitational force does work. So this

process obey the conservation law of mechanic

energy. First we should find out the mass of the axle m. Let the mass of the wheel then the

rotational inertia of the wheel-axle combination is


Where and are the radius of the wheel and the axle, respectively. Substituting the

known values and solving for . We get

The potential energy of the rigid body before releasing is

From the conservation law of mechanic energy we have

Substituting the known values and solving for , the angular speed of the

combination when it moves down the surface by , lead to

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Therefore, the rotational kinetic energy of the wheel-axle combination is

(Answer)

The translational kinetic energy of the wheel-axle combination is

(Answer)

12 (47) In Fig.6-18, a constant horizontal force of magnitude is applied to

a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the

cylinder is , and the cylinder rolls smoothly on the horizontal surface. (a) What

is the magnitude of the acceleration of the center of mass of the cylinder? （b）What is the magnitude of the angular acceleration of the cylinder about the center of mass?

(c) In unit-vector notation, what is the frictional force acting on the cylinder?

Solution: this is the plane notion of a rigid body.

There are two forces exerted on the cylinder: the

applyed force and the frictional force .

Consider first the translation motion of the center of

mass of the cylinder. From Newton’s second law, we

have

(1) Fig. 6-18 Problem 12

Where is the magnitude of the acceleration of the center of mass of the cylinder.

Next consider the rotational motion of the cylinder. From Newton’s second law for

rotation we write

(2)

Where is the radius of cylinder, is the rotational inertia of the cylinder.

(3)

There is a relation between the linear acceleration of the center of mass of the cylinder

164


and the angular acceleration of the rolling cylinder

(4)

Solving above four equations. We obtain

(Answer)

(Answer)

The frictional force

Choose the positive direction of axis to the right. In unit-vector vector notation the

frictional force acting on the cylinder can be written as (Answer)

13 (49) In Fig.6-19, a small block has a horizontal velocity of magnitude

when it slides off a table of height . Answer the following in unit-

vector notation for a coordinate system in which the origin is at the edge of the table (at point ), the positive direction is horizontally away from the table, and the

positive y direction is up. What are the angular momentum of the block about point

(a) just after the block leaves the table and (b) just before the block strikes the floor? What are the torques on the block about point (c)

just after the block leaves the table and (d) just

before strikes the floor?Solution: Taking point as the reference point,

and are the position vectors, when the block

just leaves the table and before it strikes the floor.

(a) The angular momentum of the block about

point just after block leaves the table is


Its magnitude is

( Answer )

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Its direction can be determined by the right hand rule, which is perpendicular to the

plane formed by and , entering into this page ( ).

(Answer )(b) The angular momentum of the block about point , just before the block strikes

the floor is

We should find out the velocity of the block in that moment in horizontal direction

.

In vertical direction the block is in free fall motion, thus

The horizontal displacement of the block is

So the magnitude of the velocity of block just before it strikes the floor is

The angle made by velocity with the axis is

Thus the magnitude of angular momentum of the block in (b) is

(Answer)

The direction of is perpendicular to the plane formed by and , entering into

this page ( ), also by applying the right hand rule. (Answer)

166


(c) The torque on the block about point just after the block leaves the table is

Note that, at this moment, and are in the same vertical line but opposite in

direction. so the torque is zero. (Answer)

(d) Just before the block strikes the floor the torque on the block about point is

Since now the magnitude of the torque acting on the block is

(Answer)

The direction of the torque is perpendicular to the plane formed by and which is

entering into this page ( ). (Answer)

14 (50) An impulsive force acts for a short time on a rotating rigid body of rotational inertial . Show that

Where is the torque due to the force, is the moment arm of the force, is the

average value of the force during the time it acts on the body, and and are the

angular velocities of the body just before and just after the forces acts (The quantity

is called the angular impulse, analogous to , the linear

impulse).

Solution: The angular impulse, the time accumulation effect of a torque, is defined as

167


(1)

For a variable force , the torque produced by it is a function of time

(2)

Substituting Eq.(1) and set and , we obtain

(3)

While according to the Newton’s Second law for rotation and the definition of angular

acceleration, we have

(4)

Recast above equation obtaining

(5)

Combining equation (2)、(3) and (5) we get

(Answer)

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