Download HW #9: Chapter 7--Rotational Motion

3/10/2016
HW #9: Chapter 7--Rotational Motion
HW #9: Chapter 7­­Rotational Motion
Due: 8:59pm on Tuesday, March 22, 2016
To understand how points are awarded, read the Grading Policy for this assignment.
Flywheel Kinematics
A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular
acceleration . The flywheel is assumed to be at rest at time in Parts A and B of this problem.
Part A
Find the time it takes to accelerate the flywheel to Express your answer in terms of if the angular acceleration is .
and .
Hint 1. A linear analogy
The relationship between velocity and time for an object undergoing constant linear accleration is given by
,
where is initial velocity at time , is time, and is linear acceleration. An analogous formula for the angular
velocity applies when the angular acceleration is constant.
Hint 2. Angular velocity as a function of time
The relationship between angular velocity and time for an object undergoing constant angular accleration is
.
This formula applies for initial angular velocity at initial time .
ANSWER:
= Correct
Part B
Find the angle through which the flywheel will have turned during the time it takes for it to accelerate from rest up to
angular velocity .
Express your answer in terms of some or all of the following: , , and .
Hint 1. A linear analogy
The relationship between velocity is given by
and displacement An analogous formula for angular velocity for an object undergoing constant linear accleration
.
applies when the angular acceleration is constant.
Hint 2. Angular velocity as a function of displacement
For an object undergoing constant angular acceleration , the relationship between angular velocity angular displacement is
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and
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HW #9: Chapter 7--Rotational Motion
angular displacement where is
is the angular velocity at the initial time , and ,
the initial angular displacement.
ANSWER:
= Correct
Part C
Assume that the motor has accelerated the wheel up to an angular velocity with angular acceleration in time . At
this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of
. Find , the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero
angular velocity).
Express your answer in terms of some or all of the following: , , and .
Hint 1. How to approach the problem
Solve this part using the same technique that you used for Part A. Just keep in mind that the initial conditions have
changed (in this case, the wheel is initially spinning) and that the angular acceleration is different.
ANSWER:
= s Correct
Graphs of Linear and Rotational Quantities Conceptual Question
While working on her bike, Amanita turns it upside down and gives the front wheel a counterclockwise spin. It spins at
approximately constant speed for a few seconds. During this portion of the motion, she records the x and y positions and
velocities, as well as the angular position and angular velocity, for the point on the rim designated by the yellow­orange dot in
the figure. Let the origin of the coordinate system be at the center of the wheel, the positive x direction to the right, the positive
y position up, and the positive angular position counterclockwise. The graphs begin when the point is at the indicated position.
One graph may be the correct answer to more than one part.
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HW #9: Chapter 7--Rotational Motion
Part A
Which of the graphs corresponds to x position versus time?
Hint 1. Distinguishing between linear and angular quantities
As the bicycle wheel spins, every point on the wheel experiences the same angular velocity and angular
acceleration, corresponding to the rate at which the wheel is turning and the rate at which this speed is changing,
respectively. However, the linear velocity and linear acceleration of the various points on the wheel are not the
same. Additionally, since individual points on the wheel move in both the x and y directions, the velocity and
acceleration of these points have x and y components.
Hint 2. Describing the x position
As the wheel turns, the yellow­orange dot moves in a circle around the origin. As it moves along this circular path,
its position with respect to the y axis changes cyclically.
Hint 3. Determine the initial value of the x position versus time graph
At its initial location, is the x position of the yellow­orange dot positive, negative, or zero?
ANSWER:
Positive
Negative
Zero
ANSWER:
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HW #9: Chapter 7--Rotational Motion
Graph A
Graph B
Graph C
Graph D
Graph E
Graph F
Graph G
Graph H
Correct
Part B
Which of the graphs corresponds to angular position versus time?
Hint 1. Describing the angular position
The angular position of the wheel is the angle through which it has rotated since its initial orientation. Since the
wheel is spinning at a constant speed, this angle is continuously increasing at a constant rate.
ANSWER:
Graph A
Graph B
Graph C
Graph D
Graph E
Graph F
Graph G
Graph H
Correct
Part C
Which of the graphs corresponds to y velocity versus time?
Hint 1. Describing the y velocity
As the wheel turns, the yellow­orange dot moves in a circle around the origin. As it moves along this circular path,
its position with respect to the x­axis changes cyclically.
Hint 2. Determine the initial value of the y velocity versus time graph
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HW #9: Chapter 7--Rotational Motion
At its initial location, is the y velocity of the yellow­orange dot positive, negative, or zero?
ANSWER:
Positive
Negative
Zero
ANSWER:
Graph A
Graph B
Graph C
Graph D
Graph E
Graph F
Graph G
Graph H
Correct
Part D
Which of the following graphs corresponds to angular velocity versus time?
Hint 1. Describing the angular velocity
The angular velocity of the wheel is determined by its angular speed and the direction in which it spins. The
problem states that the wheel is spinning at a constant speed.
ANSWER:
Graph A
Graph B
Graph C
Graph D
Graph E
Graph F
Graph G
Graph H
Correct
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HW #9: Chapter 7--Rotational Motion
The End of the Song
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD
spins down with a constant angular acceleration.
Part A
If the CD rotates clockwise at 500 (revolutions per minute) while the last song is playing, and then spins down to
zero angular speed in 2.60 with constant angular acceleration, what is , the magnitude of the angular acceleration of
the CD, as it spins to a stop?
Express your answer in radians per second squared.
Hint 1. Angular acceleration
Since the CD spins down with a constant angular acceleration, the instantaneous angular acceleration is equal to
the average angular acceleration. Thus, the change in angular speed of the CD measured in a time interval divided by the length of the time interval yields the acceleration of the CD as it spins to a stop.
ANSWER:
= 20.1 Correct
Part B
How many revolutions does the CD make as it spins to a stop?
Express your answer using three significant figures.
Hint 1. Find the angular displacement
Consider a point P on the CD and take its angular coordinate to be zero when the CD reaches the end of the last
song. What is the angular displacement undergone by point P as the CD spins to a stop?
Express your answer in radians.
Hint 1. Angular displacement under constant acceleration
Consider a particle that rotates about an axis with constant angular acceleration . At time , the angular
displacement as a function of time is
,
where and are, respectively, the initial angle and the initial angular velocity of the particle.
ANSWER:
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HW #9: Chapter 7--Rotational Motion
ANSWER:
10.8 revolutions Correct
Angular Motion with Constant Acceleration
Learning Goal:
To understand the meaning of the variables that appear in the equations for rotational kinematics with constant angular
acceleration.
Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many
machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their
rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a
particle with constant nonzero angular acceleration. The kinematic equations for such motion can be written as and . Here, the meaning of the symbols is as follows:
is the angular position of the particle at time .
is the initial angular position of the particle.
is the angular velocity of the particle at time .
is the initial angular velocity of the particle.
is the angular acceleration of the particle.
In answering the following questions, assume that the angular acceleration is constant and nonzero: .
Part A
True or false: The quantity represented by is a function of time (i.e., is not constant).
ANSWER:
true
false
Correct
Part B
True or false: The quantity represented by is a function of time (i.e., is not constant).
ANSWER:
true
false
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Correct
Keep in mind that initial moment.
represents an initial value, not a variable. It refers to the angular position of an object at some
Part C
True or false: The quantity represented by is a function of time (i.e., is not constant).
ANSWER:
true
false
Correct
Part D
True or false: The quantity represented by is a function of time (i.e., is not constant).
ANSWER:
true
false
Correct
The angular velocity always varies with time when the angular acceleration is nonzero.
Part E
Which of the following equations is not an explicit function of time , that is, does not involve as a variable, and is
therefore useful when you do not know or do not need the time?
ANSWER:
Correct
Part F
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HW #9: Chapter 7--Rotational Motion
In the equation , what does the time variable represent?
Choose the answer that is always true. Several of the statements may be true in a particular problem, but only
one is always true.
ANSWER:
the moment in time at which the angular velocity equals the moment in time at which the angular velocity equals the time elapsed from when the angular velocity equals until the angular velocity equals Correct
A turntable is rotating at until it reaches . You then flip a switch, and the turntable speeds up, with constant angular acceleration,
.
Part G
Suppose you are asked to find the amount of time , in seconds, it takes for the turntable to reach its final rotational
speed. Which of the following equations could you use to directly solve for the numerical value of ?
ANSWER:
More information is needed before can be found.
Correct
Knowing and would not be enough information to allow you to solve for ; you would need additional
information (namely, the value of the angular acceleration ) before you could find a numerical value for .
Part H
You are now given an additional piece of information: It takes five complete revolutions for the turntable to speed up from to . Which of the following equations could you use to directly solve for the numerical value of the
angular acceleration ?
ANSWER:
More information is needed before can be found.
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HW #9: Chapter 7--Rotational Motion
Correct
Net Torque on a Pulley
The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A,
while the mass of the pulley is equal to the mass of block A. The
blocks are let free to move and the cord moves on the pulley
without slipping or stretching. There is no friction in the pulley axle,
and the cord's weight can be ignored.
Part A
Which of the following statements correctly describes the system shown in the figure?
Check all that apply.
Hint 1. Conditions for equilibrium
If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration
and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then
pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case
in the current situation where block B has twice the mass of block A?
Hint 2. Rotational analogue of Newton's second law
The net torque and is given by
of all the forces acting on a rigid body is proportional to the angular acceleration of the body
,
where is the moment of inertia of the body.
Hint 3. Relation between linear and angular acceleration
A particle that revolves with angular acceleration has tangential acceleration equal to
,
where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or
stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord
moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius
of the pulley.
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HW #9: Chapter 7--Rotational Motion
ANSWER:
The acceleration of the blocks is zero.
The net torque on the pulley is zero.
The angular acceleration of the pulley is nonzero.
Correct
Part B
What happens when block B moves downward?
Hint 1. How to approach the problem
To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical
expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley's
angular acceleration.
Hint 2. Find the net torque on the pulley
Let's assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley?
Take the positive sense of rotation to be counterclockwise.
Express your answer in terms of , , and .
Hint 1. Torque
The torque of a force with respect to a point is defined as the product of the magnitude times the
perpendicular distance between the line of action of and the point . In other words,
.
ANSWER:
= ANSWER:
The left cord pulls on the pulley with greater force than the right cord.
The left and right cord pull with equal force on the pulley.
The right cord pulls on the pulley with greater force than the left cord.
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HW #9: Chapter 7--Rotational Motion
Correct
Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in
both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the
present situation, the tensions must be different. If they were equal, the pulley could not have an angular
acceleration.
The Center of It All
Learning Goal:
To learn the definition of the center of mass for systems of particles and be able to locate it.
Imagine throwing a rock upward and away from you. With negligible air resistance, the rock will follow a parabolic path before
hitting the ground. Now imagine throwing a stick (or any other extended object). The stick will tend to rotate as it travels
through the air, and the motion of each point of the stick (taken individually) will be fairly complex. However, there will be one
point that will follow a simple parabolic path: the point about which the stick rotates. No matter how the stick is thrown, this
special point will always be located at the same position within the stick. The motion of the entire stick can then be described
as a combination of the translation of that single point (as if the entire mass of the stick were concentrated there) and the
rotation of the stick about that point. Such a point, it turns out, exists for every rigid object or system of massive particles. It is
called the center of mass.
To calculate the center of mass for a system of massive point particles that have coordinates following equations are used:
and masses , the
,
.
In this problem, you will practice locating the center of mass for various systems of point particles.
Part A
Two particles of masses system located?
and (
) are located 10 meters apart. Where is the center of mass of the
ANSWER:
less than 5 meters from the particle of mass exactly 5 meters from the particle of mass more than 5 meters but less than 10 meters from the particle of mass more than 10 meters from the particle of mass Correct
Part B
For the system of three particles shown, which have masses located?
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, , and as indicated, where is the center of mass
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HW #9: Chapter 7--Rotational Motion
ANSWER:
to the left of the particle of mass between the particle of mass between the particle of mass and the particle of mass and the particle of mass to the right of the particle of mass Correct
Perhaps you "feel" that the center of mass should be located to the right of the particle of mass , since the
particle to the right of it has greater mass than the particle to the left. A calculation, however, allows one to pinpoint
the exact location of the center of mass.
Part C
For the system of particles described in Part B, find the x coordinate of the center of mass. Assume that the particle
of mass is at the origin and the positive x axis is directed to the right.
Express your answer in terms of .
ANSWER:
= Correct
Part D
Let us now consider a two­dimensional case. The system includes three particles of equal mass located at the
vertices of an isosceles triangle as shown in the figure. Which arrow best shows the location of the center of mass of the
system? Do not calculate.
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HW #9: Chapter 7--Rotational Motion
ANSWER:
1
2
3
4
5
Correct
Part E
What is the x coordinate of the center of mass of the system described in Part D?
Express your answer in terms of .
ANSWER:
= Correct
From the symmetry of the situation, you can see that the x and y coordinates of the center of mass are the same.
Part F
A system of four buckets forms a square as shown in the figure. Initially, the buckets have different masses (it is not
known how these masses are related). A student begins to add water gradually to the bucket located at the origin. As a
result, what happens to the coordinates of the center of mass of the system of buckets?
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HW #9: Chapter 7--Rotational Motion
Hint 1. The movement of the center of mass
As the bucket at the origin gets heavier, will the center of mass move closer to the origin or away from it?
ANSWER:
The x coordinate stays the same; the y coordinate increases.
The x coordinate stays the same; the y coordinate decreases.
The x coordinate increases; the y coordinate stays the same.
The x coordinate decreases; the y coordinate stays the same.
The x coordinate increases; the y coordinate increases.
The x coordinate decreases; the y coordinate decreases.
The x coordinate stays the same; the y coordinate stays the same.
There is not enough information to answer the question.
Correct
Part G
Find the x coordinate of the center of mass of the system of particles shown in the figure.
Express your answer in meters to two significant figures.
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ANSWER:
= 0.56 Correct
Part H
Find the y coordinate of the center of mass of the system of particles described in the previous part.
Express your answer in meters to two significant figures.
ANSWER:
= 0.40 Correct
Torque Magnitude Ranking Task
The wrench in the figure has six forces of equal magnitude acting on it.
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HW #9: Chapter 7--Rotational Motion
Part A
Rank these forces (A through F) on the basis of the magnitude of the torque they apply to the wrench, measured about
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 is defined as
,
where is the magnitude of the displacement vector from the rotation axis to the point of application of the force of
magnitude , and is the angle between this displacement and the applied force, as shown in the figure.
The direction of a torque can be either counterclockwise (as above) or clockwise. This is determined by the
direction the object will rotate under the action of the force.
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 is often referred
to as the lever arm of the force.
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HW #9: Chapter 7--Rotational Motion
ANSWER:
Correct
Rotating Spheres
Two massive spheres are mounted on a light rod that can be rotated by a string wrapped around a central cylinder, forming a
winch as shown in the figure. A force of magnitude is applied to
the string to turn the system. With respect to the variables given in
the figure, the equation for the magnitude of the angular
acceleration is
.
Assume that the spheres are small enough that they may be considered point masses and that the masses of the rod and
cylinder can be neglected.
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HW #9: Chapter 7--Rotational Motion
Part A
If the sphere on the left is moved closer to the central cylinder and placed at a distance from the axis of rotation,
what is the magnitude of the angular acceleration of the modified system? Assume that the rest of the system doesn't
change.
Hint 1. How to approach the problem
The rotational analogue of Newton's second law states that the angular acceleration of the system is equal to the
net torque exerted on the system divided by the system's moment of inertia. By moving one of the spheres closer
to the central cylinder, will either of these two quantities change?
Hint 2. Find the net torque
Write an expression for the magnitude of the net torque exerted on the system.
Express your answer in terms of some or all of the variables , , , and .
Hint 1. Torque
The magnitude of the torque of a force of magnitude with respect to a point is defined as the
product of times the perpendicular distance between the line of action of the force and the point .
That is,
.
ANSWER:
= Hint 3. Find the moment of inertia
When the sphere of mass on the left is at a distance to the axis of rotation?
on the right is at a distance from the axis of rotation, and the sphere of mass from the axis of rotation,what is the moment of inertia of the system with respect
Express your answer in terms of some or all of the variables , , , and .
Hint 1. Moment of inertia
The moment of inertia of a system of objects with respect to a given axis is defined as the sum of the
product of the mass of each object in the system and the square of its perpendicular distance from
the axis, or
,
where is the distance of the object of mass from the axis of rotation, and so on.
ANSWER:
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HW #9: Chapter 7--Rotational Motion
ANSWER:
Correct
If one of the spheres is moved closer to the axis of rotation, the angular acceleration of the system increases
because now the system has a smaller moment of inertia. Specifically, the moment of inertia of the modified system
is half the moment of inertia of the original system, which means that the angular acceleration of the modified system
is twice that of the original system.
Consider again the original system. Instead of applying the force to the string, a force with the same magnitude is applied
to the rod at a point from the sphere of mass and in a direction perpendicular to the rod, as shown in the figure.
Part B
What is the magnitude of the angular acceleration of the system?
Hint 1. How to approach the problem
As you did in the previous part, use the rotational analogue of Newton's second law to find the angular
acceleration of the system. Application of the force of magnitude directly to the rod still causes the rod to rotate
around an axis that passes through its midpoint, as in the original system. However, now the perpendicular
distance between the line of action of and the axis of rotation is different. Will this affect either the net torque or
the moment of inertia of the system?
Hint 2. Find the net torque
When the force of magnitude is applied to the rod at from the sphere of mass and in a direction
perpendicular to the rod, what is the magnitude of the net torque exerted on the system?
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Express your answer in terms of some or all of the variables , , , and .
ANSWER:
= Hint 3. Find the moment of inertia
What is the moment of inertia of the system with respect to the axis of rotation, when the force is applied to
the rod at from the sphere of mass and in a direction perpendicular to the rod?
Express your answer in terms of some or all of the variables , , , and .
ANSWER:
= ANSWER:
Correct
In conclusion, the angular acceleration of a rotating system will change if either the moment of inertia of the system
or the net torque on it changes.
Video Tutor: Balancing a Meter Stick
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment.
Then, close the video window and answer the question at right. You can watch the video again at any point.
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Part A
Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place
the pivot to keep the meter stick in balance?
Hint 1. How to approach the problem.
For the meter stick to be in equilibrium, the net torque on it must be zero. Torques about the fulcrum may be
exerted by the mass hanging from the end of the stick and by the stick’s own weight.
Use the condition that the net torque must be equal to zero to obtain a relationship involving 1) the distance
between the left end of the stick and the fulcrum and 2) the distance between the center of mass of the stick and
the fulcrum. These two distances must add up to a constant. You should get two equations that you can solve for
the location of the fulcrum.
ANSWER:
10 cm
25 cm
90 cm (10 cm from the weight)
50 cm (in the middle)
75 cm (25 cm from the weight)
Correct
Video Tutor: Walking the plank
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment.
Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A
In the video, the torque due to the mass of the plank is used in the calculations. For this question, ignore the mass of the
board. Rank, from largest to smallest, the mass needed to keep the board from tipping over.
To rank items as equivalent, overlap them.
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HW #9: Chapter 7--Rotational Motion
Hint 1. How to approach the problem
In equilibrium, the sum of the clockwise torques about an axis must equal the sum of the counterclockwise
torques.
ANSWER:
Correct
Pivoted Rod with Unequal Masses
The figure shows a simple model of a seesaw. These consist of a plank/rod of mass and length allowed to pivot freely
about its center (or central axis), as shown in the diagram. A small sphere of mass is attached to the left end of the rod,
and a small sphere of mass is attached to the right end. The spheres are small enough that they can be considered point
particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to .
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HW #9: Chapter 7--Rotational Motion
Part A
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 each of the
components of the assembly about the pivot point. That is, the total moment of inertia is equal to the moment 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 given by
.
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:
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HW #9: Chapter 7--Rotational Motion
= 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 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, your
answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first 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 .
, , Hint 1. How to approach the problem
The forces acting on the system (spheres and rod) are the weights of the spheres and the rod, and the reaction
force from the pivot. Find the torque due to each of these forces about the pivot point and add them with the
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HW #9: Chapter 7--Rotational Motion
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 is
counterclockwise.
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 must remember that if
the force tends to cause a clockwise rotation, you need to include a negative sign in your 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 is
counterclockwise.
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 the
gravitational force acting on the rod. The normal force acts at the pivot point, so its distance from the pivot point is
zero, and thus this force contributes zero torque. The gravitational force acts at the rod's center of mass, which is
also at the pivot point. Therefore, the torque due to the gravitational force about the pivot point 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 angular acceleration . (Use in your answer, not the expression for you found in Part A.)
ANSWER:
= https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4141227
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HW #9: Chapter 7--Rotational Motion
ANSWER:
= 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 light and
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.
Acceleration of a Pulley
A string is wrapped around a uniform solid cylinder of radius , as shown in . The cylinder can rotate freely about its axis. The
loose end of the string is attached to a block. The block and
cylinder each have mass .
Part A
Find the magnitude of the angular acceleration of the cylinder as the block descends.
Express your answer in terms of the cylinder's radius and the magnitude of the acceleration due to gravity .
Hint 1. How to approach the problem
1. The block does not rotate. To analyze its motion, you should use Newton's second law in its linear
form: .
2. The pulley rotates. To analyze its motion, you should use Newton's second law in its angular form: .
3. Using the geometry of the situation, you need to find the relationship between and .
4. Finally, solve the system of three equations to obtain an expression for .
Hint 2. Find the net force on the block
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HW #9: Chapter 7--Rotational Motion
The block has two forces acting on it: the tension of the string and its own weight. What is the net force acting
on the block? Use the coordinate system shown in the figure.
Express your answer in terms of tension in the string).
, (the magnitude of the acceleration due to gravity), and (the
ANSWER:
= Hint 3. Find the net torque on the pulley
The tension in the string produces a torque that acts on the pulley. What is the torque?
Express your answer in terms of the cylinder's radius and the tension in the string.
Hint 1. Formula for torque
Recall that , where is the force causing the torque, is the distance from the pivot to the
point at which the force acts, and is the angle between the position vector of the point mentioned above
and the force vector.
ANSWER:
= Hint 4. Relate linear and angular acceleration
The string does not stretch. Therefore, there is a geometric constraint between the linear acceleration and the
angular acceleration . What is the cylinder's angular acceleration in terms of the linear acceleration of the
block?
Express your answer in terms of and . Be careful with your signs.
ANSWER:
= Hint 5. Putting it together
Solve the system of equations to eliminate and obtain an expression for .
ANSWER:
= Correct
Note that the magnitude of the linear acceleration of the block is https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4141227
, which does not depend on the value of .
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HW #9: Chapter 7--Rotational Motion
Problem 7.8
The tune­up specifications of a car call for the spark plugs to be tightened to a torque of 50 . You plan to tighten the plugs
by pulling on the end of a 21­cm­long wrench. Because of the cramped space under the hood, you'll need to pull at an angle
of 125 with respect to the wrench shaft.
Part A
With what force must you pull?
Express your answer using two significant figures.
ANSWER:
290 Correct
Problem 7.17
Hold your arm outstretched so that it is horizontal. Estimate the mass of your arm and the position of its center of gravity.
Part A
What is the gravitational torque on your arm in this position, computed around the shoulder joint?
ANSWER:
Correct
Problem 7.20
A 3.20­ ­long, 410 steel beam extends horizontally from the point where it has been bolted to the framework of a new
building under construction. A 67.0 construction worker stands at the far end of the beam.
Part A
What is the magnitude of the torque about the point where the beam is bolted into place?
ANSWER:
= 8530 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4141227
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HW #9: Chapter 7--Rotational Motion
Correct
Problem 7.50
A combination lock has a 1.0 ­diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you grip
it between your thumb and forefinger with a force of 0.59 as you twist your wrist. Suppose the coefficient of static friction
between the knob and your fingers is only 0.15 because some oil accidentally got onto the knob.
Part A
What is the most torque you can exert on the knob without having it slip between your fingers?
Express your answer using two significant figures.
ANSWER:
= 8.9×10−4 Correct
Problem 7.51
A 71 man's arm, including the hand, can be modeled as a 79 ­long uniform cylinder with a mass of 3.3 .
Part A
In raising both his arms, from hanging down to straight up, by how much does he raise his center of gravity?
Express your answer using two significant figures.
ANSWER:
= 7.3 Correct
Problem 7.67­70
The grand jeté is a classic ballet maneuver in which a dancer executes a horizontal leap while moving her arms and legs up
and then down. At the center of the leap, the arms and legs are gracefully extended . The goal of the leap is to create the
illusion of flight. The center of mass ­ and hence the center of gravity ­ of an extended object follows a parabolic trajectory
when undergoing projectile motion. But when you watch a dancer leap through the air, you don’t watch her center of gravity,
you watch her head. If the translational motion of her head is horizontal ­ not parabolic ­ this creates the illusion that she is
flying through the air, held up by unseen forces. The figure illustrates how the dancer creates this illusion. While in the air, she
changes the position of her center of gravity relative to her body by moving her arms and legs up, then down. Her center of
gravity moves in a parabolic path, but her head moves in a straight line. It’s not flight, but it will appear that way, at least for a
moment.
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HW #9: Chapter 7--Rotational Motion
Part A
To perform this maneuver, the dancer relies on the fact that the position of her center of gravity
ANSWER:
Is near the center of the torso.
Is determined by the positions of her arms and legs.
Moves in a horizontal path.
Is outside of her body.
Correct
Part B
Suppose you wish to make a vertical leap with the goal of getting your head as high as possible above the ground. At the
top of your leap, your arms should be
ANSWER:
Held at your sides.
Raised above your head.
Outstretched, away from your body.
Correct
Part C
When the dancer is in the air, is there a gravitational torque on her? Take the dancer’s rotation axis to be through her
center of gravity.
ANSWER:
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HW #9: Chapter 7--Rotational Motion
Yes, there is a gravitational torque.
No, there is not a gravitational torque.
It depends on the positions of her arms and legs.
Correct
Part D
In addition to changing her center of gravity, a dancer may change her moment of inertia. Consider her moment of inertia
about a vertical axis through the center of her body. When she raises her arms and legs, this
ANSWER:
Increases her moment of inertia.
Decreases her moment of inertia.
Does not change her moment of inertia.
Correct
Score Summary:
Your score on this assignment is 94.6%.
You received 17.03 out of a possible total of 18 points.
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