Chapter 10 Homework

4/24/2014 Chapter 10 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2845893 1/32

Chapter 10 HomeworkDue: 10:00pm on Wednesday, April 16, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Torque about the z Axis

Learning Goal:

To understand two different techniques for computing the torque on an object due to an applied force.

Imagine an object with a pivot point p at the origin of the coordinate system shown . The force vector lies in the xyplane, and this force of magnitude acts on the object at apoint in the xy plane. The vector is the position vectorrelative to the pivot point p to the point where is applied.

The torque on the object due to the force is equal to thecross product . When, as in this problem, theforce vector and lever arm both lie in the xy plane of the paperor computer screen, only the z component of torque isnonzero.

When the torque vector is parallel to the z axis ( ), it iseasiest to find the magnitude and sign of the torque, , interms of the angle between the position and force vectorsusing one of two simple methods: the Tangential Componentof the Force method or the Moment Arm of the Force method.Note that in this problem, the positive z direction isperpendicular to the computer screen and points toward you(given by the right-hand rule ), so a positive torque would cause counterclockwise rotation about the z axis.

Tangential component of the force

Part A

Decompose the force vector into radial (i.e., parallel to ) and tangential (perpendicular to ) components as

shown. Find the magnitude of the radial and tangentialcomponents, and . You may assume that is

between zero and 90 degrees.

Enter your answer as an ordered pair. Express and in terms of and .

Hint 1. Magnitude of

F F

r F

F

= ×τ r F

= ττ kτ

θ

× =i j k

F

r r

Fr Ft θ

Ft

Fr F θ

F r

r



Use the given angle between the force vector and its radial component to compute the magnitude .

ANSWER:

Correct

Part B

Is the following statement true or false?

The torque about point p is proportional to the length of the position vector .

ANSWER:

Correct

Part C


Both the radial and tangential components of generate torque about point p.

ANSWER:

Correct

Part D


In this problem, the tangential force vector would tend to turn an object clockwise around pivot point p.

ANSWER:

F F r Fr

= ( , )Fr Ft Fcos(θ),Fsin(θ)

r r

true

false

F

true

false



Correct

Part E

Find the torque about the pivot point p due to force . Your answer should correctly express both the magnitudeand sign of .

Express your answer in terms of and or in terms of , , and .

ANSWER:

Correct

Moment arm of the force

In the figure, the dashed line extending from the force vector is called the line of action of . The perpendicular distance from the pivot point p to the line of action is called the moment arm of the force.

Part F

What is the length, , of the moment arm of the force

about point p?

Express your answer in terms of and .

ANSWER:

true

false

τ F

τ

Ft r F θ r

= τ −rFsin(θ)

F rm

rm

F

r θ

= rm rsin(θ)



Correct

Part G

Find the torque about p due to . Your answer should correctly express both the magnitude and sign of .

Express your answer in terms of and or in terms of , , and .

ANSWER:

Correct

Three equivalent expressions for expressing torque about the z axis have been discussed in this problem:

1. Torque is defined as the cross product between the position and force vectors. When both and lie in thexy plane, only the z component of torque is nonzero, and the cross product simplifies to:

.

Note that a positive value for indicates a counterclockwise direction about the z axis.

2. Torque is generated by the component of that is tangential to the position vector (the tangentialcomponent of force):

.

3. The magnitude of torque is the product of the force and the perpendicular distance between the z axis andthe line of action of a force, , called the moment arm of the force:

.

Exercise 10.3

A square metal plate 0.180 on each side is pivoted about an axis through point at its center and perpendicular tothe plate .

τ F τ

rm F r θ F

= τ −rFsin(θ)

F r

= × = r ∗ F ∗ sin(θ) = ττ r F k k

τ

F r

τ = r ∗ = r ∗ F sin(θ)Ft

rm

τ = ∗ F = r ∗ sin(θ) ∗ Frm

m O



Part A

Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forcesare = 21.0 , = 16.1 , and = 17.0 . The plate and all forces are in the plane of the page. Take

positive torques to be counterclockwise.

ANSWER:

Correct

Exercise 10.6

A metal bar is in the xy-plane with one end of the bar at the origin. A force 6.24 -3.25 is applied tothe bar at the point 2.46 , 3.76 .

Part A

What is the position vector for the point where the force is applied?

Enter the x and y components of the radius vector separated by a comma.

ANSWER:

Correct

Part B

F1 N F2 N F3 N

= 1.72 τ N ⋅ m

= (F N ) + (i N )jx = m y = m

r

, = 2.46,3.76 rx ry m



What are the magnitude of the torque with respect to the origin produced by ?

Express your answer with the appropriate units.

ANSWER:

Correct

Part C

What are direction of the torque with respect to the origin produced by ?

ANSWER:

Correct

Pivoted Rod with Unequal Masses

The figure shows a simple model of a seesaw. These consistof a plank/rod of mass and length allowed to pivotfreely about its center (or central axis), as shown in thediagram. A small sphere of mass is attached to the leftend of the rod, and a small sphere of mass is attached tothe right end. The spheres are small enough that they can beconsidered point particles. The gravitational force actsdownward. The magnitude of the acceleration due to gravity isequal to .

Part A

F

= 31.5 τ N⋅m

F

-direction

-direction

-direction

-direction

-direction

-direction

+x

+y

+z

−x

−y

−z

mr 2x

m1m2

g

I



What is the moment of inertia of this assembly about the axis through which it is pivoted?

Express the moment of inertia in terms of , , , and . Keep in mind that the length of the rod is , not .

Hint 1. How to approach the problem

The moment of inertia of the assembly about the pivot is equal to the sum of the moments of inertia of eachof the components of the assembly about the pivot point. That is, the total moment of inertia is equal to themoment of inertia of the rod plus the moment of inertia of the particle of mass plus the moment of inertia

of the particle of mass , all measured with respect to the pivot point.

Hint 2. Find the moment of inertia due to the sphere of mass

What is the moment of inertia of the particle of mass measured about the pivot point?

Express your answer in terms of given quantities.

Hint 1. Formula for moment of inertia

Consider an object consisting of particles with masses . Let be the distance of the th particle

from the axis of rotation. Then the moment of inertia of the object about the axis of rotation is givenby

.

ANSWER:

Hint 3. Find the moment of inertia due to the sphere of mass

What is the moment of inertia of the particle of mass measured about the pivot point?

Express your answer in terms of given quantities.

ANSWER:

Hint 4. Find the moment of inertia of the rod

What is the moment of inertia of the rod about the pivot point?

Express in terms of and .

Hint 1. General formula for the moment of inertia of a rod

Consider a rod of total length and mass , pivoted about its center. (In this problem, equals

I

mr m1 m2 x

2x x

m1

m2

m1

m1

mi ri i

I

I = ∑i mir2i

= I1 m1x2

m2

m2

= I2 m2x2

Ir mr x

L mr L

2x



.) What is the moment of inertia of the rod about its pivot point?

ANSWER:

ANSWER:

ANSWER:

Correct

Part B

Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, youranswer may include the symbol , the moment of inertia of the assembly, whether or not you have answered thefirst part correctly.)

What is the angular acceleration of the rod immediately after it is released?

Take the counterclockwise direction to be positive. Express in terms of some or all of the variables ,, , , , and .


The forces acting on the system (spheres and rod) are the weights of the spheres and the rod, and thereaction force from the pivot. Find the torque due to each of these forces about the pivot point and add themwith the correct signs. Finally, use Newton's second law for rotational motion: .

Hint 2. Find the torque due to the sphere of mass

Find the torque about the pivot due to the sphere of mass .

Express your answer in terms of given quantities. Keep in mind that the positive direction iscounterclockwise.

2x

mrL2

(1/3)mrL2

(1/4)mrL2

(1/12)mrL2

= Ir13

mrx2

= I ( + + )x2 m1 m2mr

3

I

α

α mr

m1 m2 x I g

τ = Iα

m1

m1



Hint 1. Formula for torque

The torque about the pivot point due to a force is

,

where is the vector from the pivot point to the point where the force is applied. The other

symbols have their usual meanings. If you are using any of the latter two expressions, you mustremember that if the force tends to cause a clockwise rotation, you need to include a negative sign inyour expression since the torque due to such a force is taken to be negative (by convention).

ANSWER:

Hint 3. Find the torque due to the sphere of mass

Find the torque about the pivot due to the particle of mass .

Express your answer in terms of given quantities. Keep in mind that the positive direction iscounterclockwise.

ANSWER:

Hint 4. Torque due to forces acting on the rod

Besides the two masses, there are two more forces to consider: the normal force acting at the pivot and thegravitational force acting on the rod. The normal force acts at the pivot point, so its distance from the pivotpoint is zero, and thus this force contributes zero torque. The gravitational force acts at the rod's center ofmass, which is also at the pivot point. Therefore, the torque due to the gravitational force about the pivotpoint is also zero for the rod.

Hint 5. Relating the angular acceleration to the net torque

Let the net torque acting on the system about the pivot point be denoted by . Find an expression for

.

Express your answer in terms of the system's moment of inertia and its resulting angularacceleration . (Use in your answer, not the expression for you found in Part A.)

ANSWER:

ANSWER:

F

τ = F sin ϕ = = (moment arm)Frpivot rpivotF⊥

r pivot

= τ1 gxm1

m2

m2

= τ2 − gxm2

τpivot

τpivot

I

α I I

= τpivot Iα



Correct

Substituting for , the value obtained in Part A yields

.

A large angular acceleration is often desirable. This can be accomplished by making the connecting rod lightand short (since both and appear in the denominator of the expression for ). For a seesaw, on the

other hand, and are usually chosen to be as large as possible, while making sure that the "rod" does not

get too heavy and unwieldy. This ensures that the angular acceleration is quite low.

Exercise 10.15

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential forceequal to 73.0 is applied to the rim of the wheel. The wheel has radius 0.100 . Starting from rest, the wheel has anangular speed of 15.0 after 2.21 .

Part A

What is the moment of inertia of the wheel?

Express your answer with the appropriate units.

ANSWER:

Correct

± Hoop on a Ramp

A circular hoop of mass , radius , and infinitesimal thickness rolls without slipping down a ramp inclined at an angle with the horizontal.

= α( − )gxm1 m2

I

I

α = ( − )gm1 m2

( + + )xmr

3m1 m2

mr x α

mr x

N mrev/s s

= 0.171 I kg⋅m2

m rθ



Part A

What is the acceleration of the center of the hoop?

Express the acceleration in terms of physical constants and all or some of the quantities , , and .


Draw a diagram showing the forces and torques on the hoop. Write the corresponding force and torqueequations. Consider the condition for no slipping. Finally, solve this system of equations for .

Hint 2. Find the torque about the center of mass

Write an expression for the total torque on the hoop about its center of mass. (By convention, a positivetorque produces a counterclockwise rotation, and a negative torque produces a clockwise rotation.)

Express the torque in terms of given quantities and the force of friction .

Hint 1. A formula for the magnitude of torque

A force whose line of action is a perpendicular distance (called the lever arm) from the chosencenter of rotation creates a torque with magnitude .

Hint 2. Existence and direction of the frictional force

If there is no slipping, a frictional force on the hoop must exist. In what direction does the frictionalforce act?

Hint 1. Finding the direction of the frictional force

To figure out the direction of friction, consider the process by which friction is brought intoplay. When the hoop is placed on the ramp, it will start sliding down the ramp due to (acomponent of) its weight. Friction "kicks in," stopping this sliding and causing the hoop to rollinstead.

a

m r θ

a

τ

Ffric

F L

τ = FL



ANSWER:

ANSWER:

Hint 3. Find an expression for the torque

Complete the general equation of rotational dynamics relating an object's moment of inertia and angularacceleration to the total torque acting on the object.

Express the torque in terms of the moment of inertia and angular acceleration .

ANSWER:

Hint 4. What is the moment of inertia of the hoop?

What is the moment of inertia of the circular hoop?


ANSWER:

Hint 5. Find the frictional force

Using Newton's second law to relate the forces acting on the hoop to the hoop's acceleration, find anexpression for the force of friction .

Express your answer in terms of , , , and .

Hint 1. Existence and direction of the frictional force

If there is no slipping, a frictional force on the hoop must exist. In what direction does the frictionalforce act?

Hint 1. Finding the direction of the frictional force

To figure out the direction of friction, consider the process by which friction is brought into

down the ramp (along the x axis)

up the ramp (along the x axis)

to the right but not along the x axis

to the left but not along the x axis

+−

+−

= τ − rFfric

τ

τ I α

= τ Iα

I

m r

= I mr2

Ffric

m θ a g



play. When the hoop is placed on the ramp, it will start sliding down the ramp due to (acomponent of) its weight. Friction "kicks in," stopping this sliding and causing the hoop to rollinstead.

ANSWER:

Hint 2. Find the total force on the hoop

Use Newton's laws to write an expression for , the mass of the hoop times the acceleration of

the center of mass of the hoop in the x direction.

Express your answer in terms of , the force of friction, and some or all of the othervariables supplied in the problem introduction.

ANSWER:

ANSWER:

Hint 6. Find the linear acceleration

If you've answered the previous parts, the only missing link is the relationship between linear accelerationand angular acceleration. Find the linear acceleration in terms of the angular acceleration . Use acoordinate system in which the positive x axis points down the ramp, and keep in mind thatcounterclockwise angular acceleration is positive.


ANSWER:

Hint 7. Putting it all together

If you've used the previous hints, you will have five equations involving five unknowns: , , , , and .

Solve this system of equations to eliminate the first four of these unknowns to obtain an expression for involving only the magnitude of the acceleration due to gravity and all or some of the given variables , ,

and .

down the ramp (along the x axis)

up the ramp (along the x axis)

to the right but not along the x axis

to the left but not along the x axis

+−

+−

max

Ffric

= max mgsin(θ) − Ffric

= Ffric −ma + mgsin(θ)

a α

r α

= a −rα

Ffric I α τ a

a

g m r

θ



ANSWER:

Correct

So the acceleration is independent of the hoop characteristics, that is, the mass and size (radius) of the hoop.This is quite generally true for objects freely rolling down a ramp; the acceleration depends only on thedistribution of mass, for example, whether the object is a disk or a sphere, but within each class theacceleration is the same. For example, all spheres will accelerate at the same rate, though this rate is differentfrom the rate for (all) disks.

Part B

What is the minimum coefficient of (static) friction needed for the hoop to roll without slipping? Note that it is

static and not kinetic friction that is relevant here, since the bottom point on the wheel is not moving relative to theground (this is the meaning of no slipping).

Express the minimum coefficient of friction in terms of all or some of the given quantities , , and .


In Part A of this problem, you found the linear acceleration of the hoop, assuming that it rolls withoutslipping. In one of the hints for the previous part, Newton's second law is used to derive an equation for

in terms of the force of gravity and the linear acceleration of the hoop. The force of friction must be largeenough to satisfy this equation if the hoop is to roll without slipping. However, for a given coefficient of friction

, there is a maximum limit on the magnitude of the frictional force. That means that there is some

minimum value of below which the hoop cannot roll without slipping. To find take the expression for

the maximum force of friction for a given value of , and then substitute this expression for into

Newton's second law.

Hint 2. Find the maximum value of the frictional force

If the coefficient of static friction is , what is , the maximum possible magnitude of the force of

static friction?

Express your answer in terms of the normal force and the coefficient of friction .

ANSWER:

Hint 3. What is the normal force?

Using Newton's laws, write down an expression for the normal force .

Express your answer in terms of , , and/or and any physical constants.

ANSWER:

= a gsin(θ)2

μmin

m r θ

a

Ffric

μ

μ μmin

μ Ffric

μ Ffric.max

N μ

= Ffric.max μN

N

m r θ




Let's make things a little clearer. Solve the following relationships that you've found for :

,

,

,

.

ANSWER:

Correct

Part C

Imagine that the above hoop is a tire. The coefficient of static friction between rubber and concrete is typically atleast 0.9. What is the maximum angle you could ride down without worrying about skidding?

Express your answer numerically, in degrees, to two significant figures.

ANSWER:

Correct

When roads are wet or icy though, the coefficient of friction between rubber and concrete drops to about 0.3

(or less), making skidding likely at much smaller angles.

Exercise 10.24

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is adistance above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll withoutslipping, but the right half has no friction because it is coated with oil.

Part A

How far up the smooth side will the marble go, measured vertically from the bottom?

Express your answer in terms of .

= N mgcos(θ)

μ

= μNFfric

N = mg cos(θ)− + mg sin(θ) = maFfric

a = (g/2)sin(θ)

= μmintan(θ)

2

θmax

= 61 θmax∘

μ

h

h



ANSWER:

Correct

Part B

How high would the marble go if both sides were as rough as the left side?

Express your answer in terms of .

ANSWER:

Correct

Exercise 10.20

A string is wrapped several times around the rim of a small hoop with radius 8.00 and mass 0.180 . The free endof the string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 65.0 ,calculate

Part A

the angular speed of the rotating hoop and

ANSWER:

= h′ h57

h

= h′′ h

cm kgcm

= 31.5 ω rad/s



Correct

Part B

the speed of its center.

ANSWER:

Correct

Pulling a String Adds Energy to a Wheel

A bicycle wheel is mounted on a fixed, frictionless axle, witha light string wound around its rim. The wheel has moment ofinertia , where is its mass, is its radius, and

is a dimensionless constant between zero and one. Thewheel is rotating counterclockwise with angular speed ,when at time someone starts pulling the string with aforce of magnitude . Assume that the string does not slip onthe wheel.

Part A

Suppose that after a certain time , the string has been pulled through a distance . What is the final rotational

speed of the wheel?

Express your answer in terms of , , , and .

Hint 1. Which physical principle to use

Apply the work-energy theorem: .

Hint 2. Find the final kinetic energy of the wheel

What is the kinetic energy of the wheel at time ?

= 2.52 v m/s

I = kmr2 m rk

ω0t = 0

F

tL L

ωfinal

L F I ω0

= + WKfinal Kinitial

Kfinal tL

L F I



Express the final kinetic energy in terms of some or all of the variables , , , and .

Hint 1. Find the initial kinetic energy of rotation

What is the kinetic energy of the wheel at time ?


Hint 1. Formula for kinetic energy

The formula for the kinetic energy of an object with moment of inertia , rotating about anaxis with angular speed , is

.

ANSWER:

Hint 2. Find the work done by the applied force

What is the work done on the wheel by the force acting through the string between and ?

Express your answer in terms of some or all of the variables defined in the problemintroduction.


The work done can be found either as the product of the torque and the angle turned by thewheel, or the product of the force and the distance moved by the point of application of theforce.

ANSWER:

ANSWER:

ANSWER:

L F I ω0

Kinitial t = 0

I ω0

K I

ω

K = I12

ω2

= KinitialIω0

2

2

W t = 0t = tL

= W FL

= Kfinal FL+ Iω02

2



Correct

Part B

What is the instantaneous power delivered to the wheel via the force at time ?

Express the power in terms of some or all of the variables given in the problem introduction.

Hint 1. How to compute power

In general, power is defined as an amount of work done per unit time: . In a situation such as

this, where the work is done by a torque , one has , so , where is the

angular speed with which the object is rotating. This expression is also equal to , where is the forceapplied by the string and is the speed at which the point of application of the force is moving. (Check this ifyou like.)

As a first step in computing the power with the first formula above, find , the torque due to the force .

Express your answer in terms of some or all of the variables in the problem introduction.


The formula for the magnitude of the torque due to a force applied at a point whose positionvector is with respect to some origin is

,

where is the smaller of the angles between and . (Also, .)

ANSWER:

ANSWER:

= ωfinal +2FLI

ω02

− −−−−−−−−√

P F t = 0

P = dW/dt

τ dW = τ dθ P = τ = τωdθ

dtω

Fv F

v

τ F

τ F

r

τ = rF sin ϕ

ϕ F r F = | |,r = | |F r

= τ Fr

= P F rω0



Correct

Note that your answer can be factored either as , where is the torque (force times

moment arm) applied to the wheel, or as where is the initial velocity of the point of

application of the force, which equals the initial velocity of a point on the rim of the wheel, since the string doesnot slip). So both ways of thinking about power here are completely equivalent.

Twirling a Baton

A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod ofmass 0.120 and length 80.0 .

Part A

Initially, the baton is spinning about a line through its center at angular velocity 3.00 . What is its angular

momentum?

Express your answer in kilogram meters squaredper second.

Hint 1. Angular momentum for a rigid body rotating about an axis of symmetry

The angular momentum of a rigid body that rotates about an axis of symmetry at angular velocity is

,

where is the moment of inertia of the object about the rotational axis.

Hint 2. Moment of inertia

For a uniform rod of length with mass , the moment of intertia about an axis passing through its centerperpendicular to the rod is

.

ANSWER:

(Fr)( ) = τωω0 Fr

F(r ) = Fvω0 r = vω0

kg cm

rad/s

L ω

= IL ω

I

l M

M112

l2



Correct

Part B

With a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axispassing through its end at the same angular velocity 3.00 as before. What is the new angular momentum of

the rod?

Express your answer in kilogram meters squaredper second.


If you know the moment of inertia of a rod about one of its ends, you can solve this part in the same way thatyou solved Part A.

Hint 2. Moment of inertia

For a rod of length with mass , the moment of intertia about an axis passing through either end is

.

ANSWER:

1.92×10−2 kg ⋅ /sm2

rad/s

l M

M13

l2

7.68×10−2 kg ⋅ /sm2



Correct

Here is another way to solve this problem. There is a theorem that relates the angular momentum of anobject about an arbitrary axis to the angular momentum of the object about the axis passing through its centerof mass :

,

where is the mass of the object, is the length of the position vector of the center of mass with respect

to the point chosen, and is the velocity of the center of mass with respect to the point chosen. Substituting

for the values on the right-hand side would yield the same angular momentum that you calculated.

Exercise 10.37

A 3.40 rock has a horizontal velocity of magnitude 12.0 when it is at point in the figure .

Part A

At this instant, what is the magnitude of its angular momentum relative to point ?

ANSWER:

Correct

Part B

What is the direction of the angular momentum in part (A)?

ANSWER:

L

Lcm

L = M +rcmvcm Lcm

M rcm

vcm

kg m/s P

O

= 196 L kg ⋅ /sm2



Correct

Part C

If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angularmomentum at this instant?

ANSWER:

Correct

Part D

What is the direction of the rate in part (C)?

ANSWER:

Correct

Record and Turntable

Learning Goal:

To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies.

Consider a turntable to be a circular disk of moment of inertia rotating at a constant angular velocity around anaxis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis ofthe disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is noexternal torque being applied to the axis.Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia ofthe record is . The initial angular velocity of the second disk is zero.

There is friction between the two disks.

After this "rotational collision," the disks will eventually rotate with the same angular velocity.

into the page

out of the page

= 213 ∣∣

dL

dt

∣∣ kg ⋅ /m2 s2

into the page

out of the page

It ωi

Ir



Part A

What is the final angular velocity, , of the two disks?

Express in terms of , , and .


Because there is friction between the record and turntable, you can't use energy conservation. However,since there are no net external torques acting on the system, angular momentum is conserved.

Hint 2. Initial angular momentum

Find the magnitude, , of the (combined) initial angular momenta of the two disks.

Express in terms of , , and/or .

ANSWER:

Hint 3. Final angular momentum

Find the magnitude, , of the (combined) final angular momenta of the two disks.

Express in terms of , , and/or .

ANSWER:

ANSWER:

ωf

ωf It Ir ωi

Li

Li ωi It Ir

= Li Itωi

Lf

Lf ωf It Ir

= Lf ( + )It Ir ωf



Correct

Part B

Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What isthe final rotational kinetic energy, , of the two spinning disks?

Express the final kinetic energy in terms of , , and the initial kinetic energy of the two-disksystem. No angular velocities should appear in your answer.

Hint 1. Initial rotational kinetic energy

What is the initial rotational kinetic energy of the two-disk system, .


Hint 1. Formula for rotational kinetic energy

The formula for the rotational kinetic energy of a rigid body with moment of inertia , spinning withan angular velocity is

ANSWER:

Hint 2. Final rotational kinetic energy

What is the final rotational kinetic energy of the two-disk system.

Express your answer in terms of , , and .

ANSWER:


Use the relationship between and (from the answer to the first part of this problem) to express in

terms of . Then use the equation relating and to express in terms of .

= ωfItωi

+I t Ir

Kf

It Ir Ki

Ki

It ωi

K I

ω

K = I12

ω2

= Ki12 Itωi

2

It Ir ωf

= Kf( + )It Ir ωf

2

2

ωi ωf Kf

ωi Ki ωi Kf Ki



ANSWER:

Correct

Some of the energy was converted into heat and sound as the frictional force, torque acted, stopping relativemotion.

Part C

Assume that the turntable deccelerated during time before reaching the final angular velocity ( is the timeinterval between the moment when the top disk is dropped and the time that the disks begin to spin at the sameangular velocity). What was the average torque, , acting on the bottom disk due to friction with the record?

Express the torque in terms of , , , and .

Hint 1. Average angular acceleration

What is the average angular acceleration, , of the bottom disk?

Express in terms of , , and .

Hint 1. Definition of average angular acceleration

The angular acceleration is the rate of change of angular velocity. The average angular acceleration isthe net change in angular velocity (final angular velocity minus initial angular velocity) divided by theelapsed time.

ANSWER:


The torque is given by.

ANSWER:

= KfItKi

+I t Ir

Δt Δt

⟨τ⟩

It ωi ωf Δt

⟨α⟩

⟨α⟩ ωi ωf Δt

= ⟨α⟩−ωf ωi

Δt

τ = I dωdt

= ⟨τ⟩ ( − )ωf ωi It

Δt



Correct

Exercise 10.44

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotationaxis of . She then tucks into a small ball, decreasing this moment of inertia to . While tucked,she makes two complete revolutions in 1.0 .

Part A

If she hadn't tucked at all, how many revolutions would she have made in the 1.7 from board to water?

Express your answer using two significant figures.

ANSWER:

Correct

Exercise 10.47

A small 14.0- bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The otherend of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 70.0 andis 120 in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 25.0relative to the table.

Part A

What is the angular speed of the bar just after the frisky insect leaps?

ANSWER:

Correct

Torque Magnitude Ranking Task

The wrench in the figure has six forces of equal magnitude acting on it.

18kg ⋅ m2 3.6kg ⋅ m2

s

s

0.68 rev

gg

cm cm/s

= 0.125 ω rad/s



Part A

Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measuredabout an axis centered on the bolt.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Definition of torque

Torque is a measure of the "twist" that an applied force exerts on an object. Mathematically, torque isdefined as

,where is the magnitude of the displacement vector from the rotation axis to the point of application of theforce of magnitude , and is the angle between this displacement and the applied force, as shown in thefigure.

The direction of a torque can be either counterclockwise (as above) or clockwise. This is determined by thedirection the object will rotate under the action of the force.

τ = rF sin θ

r

F θ



Hint 2. Maximum torque

Based on the mathematical definition of torque, torque is maximized when the force is large in magnitude,located a large distance from the axis of interest, and oriented perpendicular to the displacement , which isoften referred to as the lever arm of the force.

ANSWER:

Correct

Exercise 10.11

A machine part has the shape of a solid uniform sphere of mass 235 and diameter 4.50 . It is spinning about africtionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction forceof 0.0200 at that point.

Part A

Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.

ANSWER:

r

g cm

N



Part B

How long will it take to decrease its rotational speed by 28.0 ?

ANSWER:

Exercise 10.23

A solid ball is released from rest and slides down a hillside that slopes downward at an angle 66.0 from the horizontal.

Part A

What minimum value must the coefficient of static friction between the hill and ball surfaces have for no slipping tooccur?

ANSWER:

Exercise 10.31

A playground merry-go-round has radius 2.90 and moment of inertia 3000 about a vertical axle through itscenter, and it turns with negligible friction.

Part A

A child applies an 19.0 force tangentially to the edge of the merry-go-round for 25.0 . If the merry-go-round isinitially at rest, what is its angular speed after this 25.0 interval?

ANSWER:

= α rad/s2

rad/s

= t s

∘

= kmin

m kg ⋅ m2

N ss



Part B

How much work did the child do on the merry-go-round?

ANSWER:

Part C

What is the average power supplied by the child?

ANSWER:

Exercise 10.39

Part A

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of theclock face. The clock hand has a length of 15.0 and a mass of 6.00 . Take the second hand to be a slender

rod rotating with constant angular velocity about one end.

ANSWER:

Exercise 10.41

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called aneutron star. The density of a neutron star is roughly times as great as that of ordinary solid matter. Suppose werepresent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was

= ω rad/s

= W J

= P W

cm g

= L kg ⋅ /sm2

1014



8.0×105 (comparable to our sun); its final radius is 18 .

Part A

If the original star rotated once in 35 days, find the angular speed of the neutron star.

Express your answer using two significant figures.

ANSWER:

Score Summary:Your score on this assignment is 110%.You received 14.31 out of a possible total of 14 points, plus 1.04 points of extra credit.

km km

= ω2 rad/s

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Chapter 10 Homework

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