Mastering Physics Ch 12 HW College Physics I Brian Uzpen LCCC

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Ch 12 HW

Ch 12 HW Due: 9:30am on Monday, May 2, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy

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 symbols are defined as follows: is the angular angular position of the particle at time . is the initial angular position of the particle. is the angular velocity of the particle particle at time . is the initial angular velocity of the particle. is the angular acceleration of the particle. is the time that has elapsed since the particle was located at its initial position. In answerin answering g the following questions, assume that the angula angularr acceleration is constant and nonzero nonzero::

.

Part A True or false: The The quantity quantity represented represented by

is a function of time (i.e., is not constant).

ANSWER: true false

Correct

Part B True or false: The quantity repre represented sented by

is a function of of time (i.e., is not constant).

ANSWER: true false

Correct Keep in mind that

represents represents an initial initial value, not a variable. It refers refers to the angular position of an object at some initial moment.

Part C True or false: The quantity repre represented sented by

is a function of time (i.e., is not constant).

ANSWER: true false Typesetting math: 56%

https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Part D True or false: The quantity represented represented by

is a function of time (i.e., (i.e., is not constant).

ANSWER: true false

Correct The angular angular velocity

always varies with with time when when the angular angular acceleration is nonzero. nonzero.

Part E Which of the following equations equations is not an explicit function of time ? Keep in mind that an equation that is an explicit function of time involves variable.

as a

ANSWER:

Correct An equation that is not an explicit function of time is useful when you do not know or do not need the time.

Part F 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 angula angularr velocity equals

until the angula angularr velocity equals

Correct

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to . At time

, particle B, which also under undergoes goes constant angular acceleration, has twice the angular acceleration, half

the angular velocity, and the same angular position that particle A had at time

.

Part G Which of the following equations describes the angular position of particle B?

Hint 1. How to approach the problem The general equation for the rotation of B can be written as , https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW where wher e B at time

and are the position and angular speed, respectively, of particle B at time . Note the time factors of and not . This chan change ge is because because nothing is known known about particle B at time . Instead you have information about particle .

Express the quantities on the right-ha right-hand nd side of this equation for

in terms of particle A's variables and constants of of motion.

ANSWER:

Correct Note that particle B has a smaller initial angular velocity but greater angular acceleration. Also, it has been in motion for less time than particle A.

Part H How long after the time

does the angula angularr velocity of particle B equal that of particle A?

Hint 1. How to approach the problem Write expressions for the angular velocity of A and B as functions of time, either by comparison of the above equations with the general kinematic equations or by differentiating the above equation equations. s. Then equate the 2 expressions and solve for .

ANSWER:

The two particles never have the same angular velocity.

Correct

An Exhausted Bicyclist An exhausted bic yclis t pedals s omewh omewhat at erra erratically tically when exercising on a static bicycl e. The angular v elocity of the whee wheels ls follows the equation , where wher e

represents repr esents time (measure (measured d in seconds),

= 0.500

,

= 0.250

and

= 2.00

.

Part A There is a spot of paint on the front whee wheell of the bicycle. Take Take the position of the spot at time to be at angle radians radian s with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular angular displacement displacement has the spot of paint paint undergon undergone e between between time 0 and 2 seconds? Express your answer in radians using three significant figures.


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Hint 1. How to approach the problem The spot of paint rotates together with the wheel. Therefore, the equation of the angular velocity of the wheel can be used to describe the rotation of the spot of paint. Since the angular velocity is the rate of change of the angular displacement with time, you need to integrate the function over the time interval from 0 to 2 seconds.

Hint 2. Find the angular displacement Given the equation of the angular velocity

, find an expression for the angula angularr displacement

as a function of time.

Express your answer as a function in terms of time (t).

Hint 1. Integration The angular displacement

of the spot of paint is given by .

Note the initial conditions

and

.

ANSWER: =

ANSWER: =

0.793 0.7 93

Correct

Part B Express the angula angularr displacement under undergone gone by the spot of paint at should you need it.

seconds in degre degrees. es. Remember Remember to use use the unro unrounded unded value from from Part A,

Express your answer in degrees using three significant figures.

Hint 1. Radians and degrees Recall that the value

of an angle in degr degrees ees is

, where

is the value of the angle measured in radians.

ANSWER: =

45.5 45 .5

Correct

Part C What distance

has the spot of paint paint moved in 2 seconds if the radius of of the wheel wheel is 50 centimeters?

Express your answer in centimeters using three significant figures.

Hint 1. Distance traveled by a particle that rotates about a certain axis https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW Consider a particle that rotates about the z axis in the xy the xy plane. plane. The angular diplacement

traveled by the particle is given by

, where wher e

is the distance traveled by the particle in the the xy xy plane plane and and

is the distance of the particle particle from the z axis.

ANSWER: =

39.7 39 .7

Correct

Part D Which one of the following statements describes the motion of the spot of paint at

seconds?

Hint 1. How to approach the problem As the bicycl ist pedals, the whee wheels ls rotate in the positive direction as we have defined it. This makes the angular velocity of the wheel always nonnegative, and consequently the angular velocity of the spot of paint is always nonnegative. In addition, the angular velocity of the spot of paint increases if its angular acceleration is positive, and it decreases if its angular acceleration is negative.

Hint 2. Find the angular acceleration at What is the angula angularr acceleration

seconds

of the spot of paint at

seconds?

Express your answer in radians per second per second.

Hint 1. Find the angular acceleration of the spot of paint Write an expression for the angula angularr acceleration

of the spot of paint as a function of time.

Express your answer as a function in terms of time (t).

Hint 1. Angular acceleration The angular acceleration

of a particle that rotates with angular velocity

is the rate of change of

with time; that is,

.

ANSWER: =

ANSWER: =

0.827 0.8 27

ANSWER: The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is constant and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is positive and the magnitude of the angular speed is increasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is decreasing. The angular acceleration of the spot of paint is negative and the magnitude of the angular speed is increasing.


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Linear and Rotational Kinematics Ranking Task The pulley in represents different pulleys that are attached with outer radius and inner radius indicated in the table. The horizontal rope is pulled to the right at a constant linear speed that is the same in each case, and none of the two separate ropes slips in its contact with the pulley.

Part A Rank these scenarios on the basis of the linear speed of the block. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Relating the two rope speeds By pulling the horizontal rope at constant linear speed, the pulley is given a constant angular velocity. This angular velocity in turn causes the rope attached to the block to wind up at a constant linear speed. Since both ropes are attached to the same pulley, each of their speeds must satisfy the relationship , where wher e

is the angular velocity of the pulley and

is the radius for the rope.

ANSWER:


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Analysis of a single pulley Several points on the pulley are indicated in . Each letter designates a point on either the pulley or one of the two ropes and point D is at the center of the axle. The horizontal rope is pulled to the right at a constant linear speed, and neither rope slips in its contact with the pulley.

Part B Rank the designated points on the basis of their linear or tangential speed. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Determining speed Since each point is located on the same pulley (or is located on a rope attached to the same pulley), each point’s linear or tangential speed is determined by its radial distance from the rotation axis via .

ANSWER:

Correct

Part C Rank the designated points in on the basis of the magnitude of their linear or radial acceleration. Rank from largest to smallest. To rank items as equivalent, overlap them.


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Hint 1. Constant speed The horizontal rope is pulled at constant linear speed, giving the pulley a constant angular speed. Any point on the pulley, however, is undergoing circular motion and must have a nonzero radial acceleration (except for the point on the axis of the pulley). Any point not on the pulley does not have a radial acceleration, because they move in a straight line at constant speed.

ANSWER:

Correct

Problem 12.1 A skater holds her arms outstretched as she spins at 100

.

Part A What is the speed of her hands if they are 140

apart? apar t?

Express your answer with the appropriate units. ANSWER: 7.33

Correct

Problem 12.6 https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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The three masses shown shown in the figure are connected by massless, rigid rods. Assume that that 220 and = 350 .

=

Part A What is the

-coordinate -coordin ate of of the center of mass?

Express your answer to two significant figures and include the appropriate units. ANSWER: =

5.2

Correct

Part B What is the -coord -coordinate inate of the center of mass? Express your answer to two significant figures and include the appropriate units. ANSWER: =

3.3

Correct

Problem 12.15 The three masses shown in the figure are connected by massless, rigid rods.

Part A Find the coordinates of the center of of mass. Find the

-coordinate. -coord inate.


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Express your answer to two significant figures and include the appropriate units. ANSWER: =

6.0

Correct

Part B Find the -coordina -coordinate. te. Express your answer to two significant figures and include the appropriate units. ANSWER: = 4.0

Correct

Part C Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page. Express your answer to two significant figures and include the appropriate units. ANSWER: = 2.0×10−3

Correct

Part D Find the moment of inertia about an axis that passes through masses B and C. Express your answer to two significant figures and include the appropriate units. ANSWER: = 1.3×10−3

Correct

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 an object object with a pivot pivot point p at the origin origin of of the coordinate coordinate system shown . The force force vector acts on the object at a point in the xy plane. The vector

lies in the xy the xy plane, plane, and this force of magnitude

is the position vector relative relative to the pivot point point p to the point wher where e

is applied. applied.

The torque on the object due to the force is equal to the cross produ product ct . When, as in this proble problem, m, the force vector and lever arm both lie in the xy the xy plane of the paper or computer screen, only the z component component of torque is nonzero. When the torque vector is parallel to the z axis ( ), it is easie easiest st to find the mag magnitud nitude e and sign of the torq torque, ue, , in term terms s of the ang angle le betw between een the position and force vectors using one of two simple methods: the Tangential Component of the Force Force method or the Moment Arm of the Force method. Force method. Note that in this problem, the positive z directi on is perpendicular to the comput computer er screen and points toward you (given by the right-hand rule so a positive torque would cause counterclockwise rotation about the z axis.


),

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Tangential component of the force

Part A

Decompose the force vector into radial (i.e., parallel parallel to ) and tangential (per (perpendicular pendicular to ) components as shown. Find the magnitude of the radial and tangential components, and . You may ass assume ume that is between zero and 90 degrees. Enter your answer as an ordered pair. Express

and

in terms of

and .

Hint 1. Magnitude of Use the given angle between the force vector

and its radial componen componentt

to compute the magnitude

.

ANSWER: =

Correct

Part B Is the following statement true or false? The torque about point p is proportiona proportionall to the length

of the position vector

.

ANSWER: true false

Correct


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Part C Is the following statement true or false? Both the radial and tangential components of

generate torque about point p.

ANSWER: true false

Correct

Part D Is the following statement true or false? In this problem, the tangential force vector would tend to turn an object clockwise around pivot point p. ANSWER: true false

Correct

Part E Find the torque

about the pivot pivot point p due due to force force

Express your answer in terms of

and

. Your Your answer should correctly express express both the magnitude magnitude and sign of .

or in terms of

, , and .

ANSWER: =

Correct

Moment arm of the force In the figure, the dashed dashed line extending extending from the force force vector is called the line of action of line of action is called the moment arm of the force.

. The perp perpendicular endicular distance

from the pivot pivot point p to the

Part F

What is the length,

, of the moment arm of the force


about point p?

and .


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ANSWER: =

Correct

Part G Find the torque

about p due to

. Your answer should correct correctly ly express both the magnitude and sign 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 Torque is defined defined as the cross product product between between the position and and force vectors. When both torque is nonzero, and the cross product simplifies to:

and

lie in the xy the xy plane, only the z component component of

. Note that a positive value for

indicates a counterclockwise direction direction about about the z axis.

2. Torque Torque is generated generated by the component of

that is tangential to the position vector

(the tangential component of force): force): .

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

,

.

Problem 12.18

Part A In , what is the net torque about the axle? Express your answer using two significant figures and include the appropriate units.

ANSWER: = -0.20

Correct


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± PSS 12.1 Rotational Dynamics Problems Learning Goal: To practice Problem-Solving Strategy 12.1 for rotational dynamics problems. Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 10.0 degree degrees s from the vertical? vertical? A typical pencil pencil has an average length of 15.0 15.0 and an average mass of 10.0 . Assume the tip of the pencil does not slip as it falls. PROBLEM-SOLVING STRATEGY 12.1 Rotational dynamics problems MODEL: Model the object as a simple shape. VISUALIZE: Draw a pictorial representation to clarify the situation, define coordinates and symbols, and list known information.

Identify the axis about which the object rotates. Identify forces, and determine their distance from the axis. For most problems, it will be useful to draw a free-body diagram. Identify any torques caused by the forces and the signs of the torques. SOLVE: The mathematical representation is based on Newton’s second law for rotational motion:

or

.

Find the moment of inertia in the table of common shapes of objects, or if needed, calculate it as an integral or by using the parallel-axis theorem. Use rotational kinematics to find angles and angular velocities. ASSESS: Check that your result has the correct units, is reasonable, and answers the question.

Model There are two reasonable approximations to consider for the pencil in this problem: a cylinder and a thin rod. However, in this problem we will treat the pencil as a uniform thin rod of length 15.0 and mass 10.0 .

Visualize

Part A The pencil rotates about an axis perpendicular to the plane of the figure. Which of the labeled points is the point that the axis of rotation passes through?

ANSWER: A B C D E

Correct https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Part B Identify and draw the forces that act on the pencil. Be certain to draw each force at the correct location (the point at which the force acts on the pencil). The black dot represents the center of mass of the pencil. Draw the vectors starting at the points of application of the corresponding forces. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. ANSWER:

Correct

Solve

Part C What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical? Express your answer numerically in radians per second squared to three significant figures.

Hint 1. Find the moment of inertia In Part A, you determined the point about which the pencil rotates. To determine the angular acceleration of the pencil, you will need to know its moment of inertia inertia about about that point. What What is the moment of inertia inertia of the pencil pencil about an end, end, given our our approximation approximation that it is a uniform uniform thin rod of length 15.0 and mass 10.0 ? Express your answer numerically in kilogram-meters squared to four significant figures.

Hint 1. The moment of inertia of a thin rod The formula for moment of inertia of a thin rod about an end is

, wher where e

is the mass and

is the length. You have the mass

and length in grams and centimeters, respectively. Be sure to convert these quantities to the appropriate units before entering them into the formula.

ANSWER: = 7.500×10 −5

Hint 2. Find the net torque


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Ch 12 HW What is the net net torque torque you identified in Part A)?

acting on the pencil pencil when torque is calculated about about the end that is in contact with the groun ground d (i.e., (i.e., the point point that

Express your answer numerically in newton-meters to four significant figures.

Hint 1. How to approach the problem Recall that the magnitude of a torque is defined by , wher where e is the distance from the pivot point to the point of application for the force, force, is the magnitude of the force that that exerts the torque, torque, and is the angle betwee between n the force and and the line connecting the pivot to the point of application.

Hint 2. Find the distance from the axis for each force acting on the pencil What is the distance

between betwee n the point of application of

application of and the axis? What is the distance rotation in this problem is horizontal.

and the axis of rotation? What is the distance

between betwee n the point of application of

between betwee n the point of

and the axis? Recall that the axis of

Express your answers numerically in meters to three significant figures separated by a comma. ANSWER: ,

,

=

0,7.50×10−2,0

,

Hint 3. Find the weight of the pencil What is the magnitude

of the weight of the pencil?

Express your answer numerically in newtons to four significant figures. ANSWER: = 9.800×10−2

ANSWER: = −1.276×10−3

ANSWER: =

-17.0 -1 7.0

Correct

Assess

Part D Calculate the time

for the pencil pencil to hit the ground, ground, assuming assuming that it falls from standing perfectly perfectly vertical and maintains this angular angular acceleration.

Express your answer numerically in seconds to three significant figures.

Hint 1. How to approach the problem Recall that angles measured clockwise are negative. In falling from vertical to striking the ground, the pencil travels clockwise and goes through a change in angle of degrees, or . Use the angular acc acceleration eleration that you found in Part C along with one of the angular kinematics equations to find the time for the pencil to traverse .

Hint 2. Determine which equation to use Which of the following equations for motion under constant angular acceleration \textti p{\alpha }{alpha} }{alpha} should you use to find \texttip{t}{t} \texttip{t}{t}? ? Let \texttip{\omega _{\rm 0}}{omega_0} be 0}}{omega_0} be the initial angular speed, \textti p{\omeg p{\omega a }{ome }{omega} ga} the angular speed at time \texttip{t}{t} \texttip{t}{t},, and \texttip{\Delta \theta }{Deltatheta} }{Deltatheta} the change in angle during time \texttip{t}{t} \texttip{t}{t}.. https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW ANSWER: \omega = \omega_0 +\alpha t \large{\Delta \theta = \omega_0 t + \frac {_1}{^2} \alpha t^2} \omega^2 = \omega_0^2 + 2 \alpha \Delta \theta

ANSWER: \texttip{t}{t} =

0.430 0.4 30

\rm s

Correct In this part, you assumed that the angular acceleration from partway through the fall is the constant angular acceleration through the entire fall. This is similar to the calculation using the average angular acceleration. Therefore, you should get a value similar to (i.e., with the same order of magnitude as) what you find in real life. You calculated a time of around half a second. This is certainly reasonable, based on your experience with pencils. If you had found an answer such as 50 seconds or 1/100 second, then you would need to rethink your answer and check over your work.

Acceleration of a Pulley A string is wra wrapped pped around a uniform s olid cylinder of radius \texttip{r}{r} \texttip{r}{r},, as shown in the figure . 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 \texttip{m}{m} \texttip{m}{m}.. Note that the positive y direction is downward and counterclockwise torques are positive.

Part A Find the magnitude \textti p{\alpha }{alpha} }{alpha} of the angular acceleration of the cylinder as the block descends. Express your answer in terms of the cylinder's radius \textt radius \texttip{r}{r} ip{r}{r} and and the magnitude of the acceleration due to gravity \textt gravity \texttip{g}{g} ip{g}{g}..

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: F= ma. ma. 2. The pulley rotates. To analyze its motion, you should use Newton's second law in its angular form: \tau= I \alpha. \alpha. 3. Using the geometry of the situation, you need to find the relationship between \texttip{a}{a} \texttip{a}{a} and \textti p{\alpha }{alpha} }{alpha}. 4. Finally, solve the system of three equations to obtain an expression for \textti p{\alpha }{alpha} }{alpha}.

Hint 2. Find the net force on the block The block has two forces acting on it: the tension of the string and its own weight. What is the net force \texttip{F}{F} \texttip{F}{F} acting on the block? Use the coordinate system shown in the figure. Express your answer in terms of \textt of \texttip{m}{m} ip{m}{m},, \textt \texttip{g}{g} ip{g}{g} (the (the magnitude of the acceleration due to gravity), and \textt and \texttip{T}{T} ip{T}{T} (the tension in the string). ANSWER: F = ma= = m g-T


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Hint 3. Find the net torque on the pulley The tension \texttip{T}{T} \texttip{T}{T} in 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 \textt radius \texttip{r}{r} ip{r}{r} and and the tension \textt tension \texttip{T}{T} ip{T}{T} in the string.

Hint 1. Formula for torque Recall that \tau=rF\sin(\phi) \tau=rF\sin(\phi),, where \texttip{F}{F} \texttip{F}{F} is the force causing the torque, \texttip{r}{r} \texttip{r}{r} is the distance from the pivot to the point at which the force acts, and \texttip{\phi }{phi} is }{phi} is the angle between the position vector of the point mentioned above and the force vector.

ANSWER: I\alpha= = -T r

Hint 4. Relate linear and angular acceleration The string does not stretch. Therefore, there is a geometric constraint between the linear acceleration \texttip{a}{a} \texttip{a}{a} and the angular acceleration \texttip{\alpha }{alp }{alpha} ha}.. What is the cylinder's angular acceleration \textt ip{\alpha }{alpha} }{alpha} in terms of the linear acceleration \texttip{a}{a} \texttip{a}{a} of the block? Express your answer in terms of \textt of \texttip{a}{a} ip{a}{a} and \textt and \texttip{r}{r} ip{r}{r}.. Be careful with your signs. ANSWER: \textti p{\alpha }{alpha} }{alpha} = \large{\frac{-a}{r}}

Hint 5. Putting it together Solve the system of equations to eliminate \texttip{T}{T} \texttip{T}{T} and and obtain an expression for \textti p{\alpha }{alpha} }{alpha}.

ANSWER: \textti p{\alpha }{alpha} }{alpha} = \large{\frac{g}{1.5r}}

Correct Note that the magnitude of the linear acceleration of the block is \large{\frac{2}{3}g} \large{\frac{2}{3}g},, which does not depend on the value of \texttip{r}{r} \texttip{r}{r}..

PSS 12.2 Static Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 12.2 for static equilibrium problems. Tweedledum and Tweedledee are carrying a uniform wooden board that is \texttip{L}{L} \texttip{L}{L} = 3.00 {\rm m} m} long and has a mass \texttip{M}{M} \texttip{M}{M} = 15.0 {\rm kg} . If Tweedledum applies an upward force of magnitude \textti p{F_{ p{F_{\rm \rm 1}}{ 1}}{F_1} F_1} = 60.0 {\rm N} N} at the left end of the board, at what point and with what magnitude \textti p{F_{ p{F_{\rm \rm 2}}{ 2}}{F_2} F_2} of force does Tweedledee have to lift for the board to be carried? PROBLEM-SOLVING STRATEGY 12.2 Static equilibrium problems MODEL: Model the object as a simple shape. VISUALIZE: Draw a pictorial representation showing all forces and distances. List known information.

Pick any point you wish as a pivot point. The net torque about this point is zero. Determine the moment arms of all forces about this pivot point. Determine the sign of each torque about this pivot point. SOLVE: The mathematical representation is based on the fact that an object in total equilibrium has no net force and no net torque:

\vec{F}_{\rm net}=\vec{0} net}=\vec{0} and \tau_{\rm net}=0. net}=0. Write equations for \sum F_x =0, =0, \sum F_y=0, F_y=0, and \sum \tau =0. =0. Solve the three simultaneous equations. ASSESS: Check that your result is reasonable and answers the question.


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Model Model the board as a uniform rod, with its center of mass at the center. Note that in order for Tweedledum and Tweedledee to carry the board safely, the board must be in static equilibrium. If the board were not in equilibrium, the net force acting on the board would exert a torque on it, causing the board to rotate, likely hitting one of the carriers.

Visualize

Part A Which of the following diagrams represents the correct free-body diagram for this situation?

ANSWER: A B C D

Correct

Part B Take the left end of the board as a pivot point. Sort the forces acting on the board according to whether they exert a positive, negative, or zero torque about this point. Drag the appropriate items to their respective bins.

Hint 1. The definition of the positive sense of rotation By convention, the positive sense of rotation is counterclockwise.

ANSWER:


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Correct Now that you have identified the correct free-body diagram, complete your pictorial representation of the problem by defining symbols representing the important variables, such as distances and magnitudes of forces. Make sure you identify what the problem is trying to find. Here, we will call \texttip{d}{d} the \texttip{d}{d} the distance between the left end of the board and the point at which Tweedledee exerts a force of magnitude \textt ip{F_{\r ip{F_{\rm m 2}}{ 2}}{F_2} F_2}.. Therefore, your target variables are \texttip{d}{d} \texttip{d}{d} and \textti p{F_{ p{F_{\rm \rm 2}}{ 2}}{F_2} F_2}.. A complete pictorial representation representation f or this problem problem should look like the sketch shown below. Keep in mind t hat by convention the positive sense of rotation is counterclockwise.

Solve

Part C If Tweedledum applies a force of \textti p{F_{ p{F_{\rm \rm 1}}{ 1}}{F_1} F_1} at the left end of the board, at what distance \texttip{d}{d} \texttip{d}{d} from the left end and with what magnitude of force \textt ip{F_{\r ip{F_{\rm m 2}}{ 2}}{F_2} F_2} does Tweedledee have to lift for the board to be carried? Express your answers numerically in newtons and meters to three significant figures separated by commas.

Hint 1. Set up the equilibrium condition for the net force For the board to be in equilibrium, there should be no translational motion, i.e., the net force acting on the board must be zero. This equilibrium condition becomes \Sigma F_y = 0 in 0 in this problem because all the forces acting on the board are vertical. Write an expression for \Sigma F_y, F_y , the sum of the vertical components of all forces acting on the board. Assume that the vertical axis is https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW positive in the upwards direction. Express your answer in terms of some or all of the variables \textt variables \texttip{F_{\rm ip{F_{\rm 1}}{F_1} 1}}{F_1},, \textt \texttip{F_{\rm ip{F_{\rm 2}}{F_2} 2}}{F_2},, \textt \texttip{L}{L} ip{L}{L},, and \texttip{M}{M}.. Use \textt \texttip{M}{M} Use \texttip{g}{g} ip{g}{g} for the acceleration due to gravity. ANSWER: \Sigma F_y=0= = F_{1}+F_{2}-M g

Hint 2. Set up the equilibrium condition for the net torque Because the board is not rotating, the net torque \tau_{\rm net}=\Sigma \tau \tau about any point must be zero. Below is a list of equations for \Sigma \tau \tau for different pivot points. In each case, the x the x coordinate of the pivot point is given assuming that the x the x axis is parallel to the board with the origin located at the left end of the board. Which of the following is the correct equation for \Sigma \tau for \tau for the given pivot point? Torques that would cause a counterclockwise rotation are positive. There may be more than one correct answer. Pivot point A x= x=0 B

x =L =L

\Sigma \tau \ large{\frac {1}{2}Mg L + F_2 d = 0} \ la large{\ fr frac {1 {1}{2}Mg L - F_2 (L - d) - F_1 L = 0}

C x= x =0

\ large{- \ frac {1}{2}Mg L + F_2 d = 0}

D x=d

\la \l arge{- F_ F_1 1 d - F_ F_2 2 \l \le eft ft((d - \f \frrac{ c{1 1}{2}L\r \rig igh ht) = 0}

E

\large{x=\frac{1}{2}L \large{x=\frac{1} {2}L}}

\large{- \frac{1}{2 \frac{1}{2}} F_1 L + F_2 \left(d - \frac{1}{2 \frac{1}{2}L\righ }L\right)=0} t)=0}

F

\large{x=\frac{1}{2}L \large{x=\frac{1} {2}L}}

\large{\frac{1}{2 \large{\fra c{1}{2}} F_1 L + F_2 \left(d - \frac{1}{2 \frac{1}{2}L\right) }L\right)=0} =0}

Enter the letter(s) of the correct equation(s) in alphabetical order. For instance, if you thought that choices A, B, and D were correct, you would enter ABD.

Hint 1. Finding the torque The torque \texttip{\tau }{tau} due }{tau} due to a force is the product of the magnitude \texttip{F}{F} \texttip{F}{F} of the force and the moment arm \texttip{l}{l} of the force: \tau=F\cdot l. l. As shown in the f igure, the moment arm of the force is the perpendicular perpendicular distance betwee between n the axis of rotation and the line of action of the force. If the line of action of the force goes through the axis of rotation that you have chosen, the moment arm of the force is equal to zero. Hence, this force will not cause a torque about the chosen axis.

ANSWER: BCE

ANSWER: \textti p{F_{ p{F_{\rm \rm 2}}{ 2}}{F_2} F_2},, \texttip{d}{d} = 87.0 87.0,2.53 ,2.53

{\rm N}, N}, {\rm m}


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Correct

Assess

Part D Now, assume that Tweedledum applies an upward force at the far left end of the board ( \texttip{F_{\rm left}}{F_left}) left}}{F_left} ) and that Tweedledee applies an upward force at the far right end of the board (\textti ( \textti p{F_{ p{F_{\rm \rm right}} right}}{F_righ {F_right} t}). ). With what force does each person lift? Recall that the mass of the board is \texttip{M}{M} \texttip{M}{M} and \texttip{g}{g} \texttip{g}{g} is is the acceleration due to gravity. ANSWER: F_{\rm left}=2Mg; left}=2Mg; F_{\rm right}=2Mg F_{\rm left}=0; left}=0; F_{\rm right}=Mg \large{F_{\rm left}=\frac{1}{2}Mg}; left}=\frac{1}{2}Mg} ; \large{F_{\rm right}=\frac{1}{2}Mg} \large{F_{\rm left}=\frac{1}{2}Mg}; left}=\frac{1}{2}Mg} ; \large{F_{\rm right}=\frac{3}{2}Mg} F_{\rm left}=Mg; left}=Mg; F_{\rm right}=0

Correct When two people hold the board at equal distances from the board's center of mass, they must each exert the same force on the board, that is, F_{\rm left} = F_{\rm right}. right}. If F_{\rm left} \neq F_{\rm right}, right} , the board would rotate. Since the net force on the board must be zero, F_{\rm left} + F_{\rm right} = Mg. Mg. Hence, each person must exert a force on the board equal to Mg/2 Mg/2.. Consider the original situation, in which the person on the right (Tweedledee) held the board closer to the center of mass than the person holding the left end of the board (Tweedledum). If you consider a rotation axis through the center of mass, the net torque about that axis must be zero. Since torque is directly proportional to both force and moment arm, when one person holds the board closer to the center of mass (smaller moment arm), that person must exert more force to create a torque with the same magnitude as the person holding the board farther from the center of mass. Your calculations confirmed this fact.

A Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass \textti p{m_{ p{m_{\rm \rm 1}}{ 1}}{m_1} m_1} and length \texttip{L}{L} \texttip{L}{L} is supported by two identical massless strings. Both strings are vertical. String A is attached at a distance d < L/2 L/2 from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass \textti p{m_{ p{m_{\rm \rm 2}}{ 2}}{m_2} m_2} is supported against gravity by the bar at a distance \texttip{x}{x} from the left end of the bar, as shown in the figure. \texttip{x}{x} Throughout this problem positive torque is that which spins an object counterclockwise. Use \texttip{g}{g} for \texttip{g}{g} for the magnitude of the acceleration due to gravity.

Part A Find \texttip{T_{\mit A}}{T_A}, A}}{T_A}, the tension in string A. Express the tension in string A in terms of \textt of \texttip{g}{g} ip{g}{g},, \textt \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{L}{L} ip{L}{L},, \textt \texttip{d}{d} ip{d}{d},, \textt \texttip{m_{\rm ip{m_{\rm 2}}{m_2} 2}}{m_2},, and \texttip{x}{x}.. \texttip{x}{x}

Hint 1. Choosing an axis Choose a rotation axis p axis p,, about which to apply the requirement \sum\tau_p = 0. 0. Since the system is in static equilibrium, the choice of rotation axis is arbitrary; however, there is a convenient choice of p A}}{T_A} by eliminating the torque from an unknown force. p to find \texttip{T_{\mit A}}{T_A} by


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Hint 2. Find the torque around the best axis It is convenient to choose the rotation axis to be through the point where string B is attached to the bar. This eliminates any torque from the tension in string B. Find the total torque about this point. Answer in terms of \textt of \texttip{T_{\mit ip{T_{\mit A}}{T_A A}}{T_A}}, \textt \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{m_{\rm ip{m_{\rm 2}}{m_2} 2}}{m_2},, \textt \texttip{L}{L} ip{L}{L},, \textt \texttip{x}{x} ip{x}{x},, \textt \texttip{d} ip{d} {d},, and \textt {d} and \texttip{g}{g} ip{g}{g}.. ANSWER: \sum\tau_B = \large{T_{A} d - \frac{m_{1} g L}{2} - m_{2} g x}

Hint 3. Summing the torques \sum\tau_p = 0 0 for a static system. Solve for \texttip{T_{\mit A}}{T_A}. A}}{T_A}.

ANSWER: \texttip{T_{\mit A}}{T_A} = \large{{\frac{g}{d}}\left({\frac{m_{1} L}{2}}+m_{2} x\right)}

Correct

Part B Find \texttip{T_{\mit B}}{T_B}, B}}{T_B}, the magnitude of the tension in string B. Express the magnitude of the tension in string B in terms of \textt of \texttip{T_{\mit ip{T_{\mit A}}{T_A A}}{T_A}}, \textt \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{m_{\rm ip{m_{\rm 2}}{m_2} 2}}{m_2},, and \texttip{g}{g}.. \texttip{g}{g}

Hint 1. Two different methods to find \texttip{T_{\mit B}}{T_B} There are two equivalent ways to find \texttip{T_{\mit B}}{T_B}. B}}{T_B}. One way is to balance the torques as was done in the calculation of \texttip{T_{\mit A}}{T_A}, A}}{T_A} , except using a different rotation axis. In this case, a convenient axis is through the point where string A is attached to the bar. The second, and easier, method is to use the second equation for static equilibrium, \sum\vec{F}=0 \sum\vec{F}=0..

Hint 2. Direction of forces Since both strings are vertical, all forces on the bar--the tension forces and the weights of the bar and block--act vertically. Thus, only vertical components of forces need be considered.

ANSWER: \texttip{T_{\mit B}}{T_B} = T_{A}-\left(m_{1}+m_{2}\right)g

Correct

Part C If the bar and block are too heavy the strings may break. Which of the two identical strings will break first? ANSWER: string A string B

Correct

Part D If the mass of the block is too large and the the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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What is the smallest possible value of \texttip{x}{x} \texttip{x}{x} such that the bar remains stable (call it \texttip{x_{\rm critical}}{x_critical})? critical}}{x_critical} )? Express your answer for \textt for \texttip{x_{\rm ip{x_{\rm critical}}{x_c critical}}{x_critical} ritical} in terms of \textt of \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{m_{\rm ip{m_{\rm 2}}{m_2} 2}}{m_2},, \textt \texttip{d}{d} ip{d}{d},, and \texttip{L}{L}.. \texttip{L}{L}

Hint 1. Nature of the unstable motion When the bar becomes unstable there are only two points about which the bar can rotate: the points where the strings attach to the bar. About which point will the bar rotate when x < x_{\rm critical}? critical} ? ANSWER: The point where string A is attached to the bar The point where string B is attached to the bar

Hint 2. Tension in string B at the critical point The tension in string B counteracts the clockwise rotation of the bar about the point where string A is attached to the bar. As \texttip{x}{x} is decreased, \texttip{T_{\mit B}}{T_B} B}}{T_B} is likewise decreased because the clockwise torque about this point decreases. The critical value \texttip{x_{\rm critical}}{x_critical} critical}}{x_critical} corresponds to when T_B = 0. 0. If \texttip{x}{x} \texttip{x}{x} is decreased further, \texttip{T_{\mit B}}{T_B} B}}{T_B} will continue to be zero and the counterclockwise torque due to the weight of the block will be greater than the clockwise torque due to the weight of the bar, causing the system to rotate.

Hint 3. Calculate the torques Add up the total torque about t he point in which s tring A attaches to the bar when the mass \textti p{m_{ p{m_{\rm \rm 2}}{ 2}}{m_2} m_2} is at \texttip{x_{\rm critical}} {x_critical}.. Remember that \texttip{T_{\mit B}}{T_B} {x_critical} B}}{T_B} has a special value at this point and that, owing to the choice of origin, the torque due to string A is 0. Remember to pay attention to the direction of the torques. Answer in terms of \textt of \texttip{m_{\rm ip{m_{\rm 2}}{m_2} 2}}{m_2},, \textt \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{d}{d} ip{d}{d},, \textt \texttip{L}{L} ip{L}{L},, \textt \texttip{g}{g} ip{g}{g},, and \textt and \texttip{x_{\rm ip{x_{\rm critical}}{x_critical}.. critical}}{x_critical}

Hint 1. Find the distance of the center of mass of the bar from string A What is the distance \texttip{d_{\rm 1}}{d_1} 1}}{d_1} of the center of mass of the bar from string A? Answer in terms of the given variables. ANSWER: \texttip{d_{\rm 1}}{d_1} = \large{\frac{L}{2}-d}

Hint 2. Find the distance of \texttip{m_{\rm 2}}{m_2} from 2}}{m_2} from the string A What is the distance \texttip{d_{\rm 2}}{d_2} of \textt ip{m_{ ip{m_{\rm \rm 2}}{ 2}}{m_2} m_2} from the string A? Answer in terms of the given variables. ANSWER: \texttip{d_{\rm 2}}{d_2} = d-x_{\rm{critical}}

ANSWER: \sum\tau_A = 0 = \large{m_{2} g \left(d-x_{\rm{critical}}\right)-m_{1} g \left(\frac{L}{2}-d\right)} \sum\tau_A

ANSWER: \texttip{x_{\rm critical}}{x_critical} = \large{d-\frac{m_{1}}{m_{2}}\left({\frac{L}{2}}-d\right)}

Correct


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Part E Note that \texttip{x_{\rm critical}}{x_critical}, critical}}{x_critical} , as computed in the previous part, is not necessarily positive. If x_{\rm critical} < 0, 0 , the bar will be stable no matter where the block of mass \textti p{m_{ p{m_{\rm \rm 2}}{ 2}}{m_2} m_2} is placed on it. Assuming that \textt ip{m_{ ip{m_{\rm \rm 1}}{ 1}}{m_1} m_1},, \texttip{d}{d} \texttip{d}{d},, and \texttip{L}{L} \texttip{L}{L} are held fixed, what is the maximum block mass \texttip{m_{\rm max}}{m_max} for which the bar will always always be stable? In other words, what is the maximum block mass such that x_{\rm critical} \leq 0? 0? Answer in terms of \textt of \texttip{m_{\rm ip{m_{\rm 1}}{m_1} 1}}{m_1},, \textt \texttip{d}{d} ip{d}{d},, and \textt and \texttip{L}{L} ip{L}{L}..

Hint 1. Requirement of stability If \texttip{x}{x} \texttip{x}{x} is calculated to be less than zero, the solution is unphysical. (The bar does not extend there to support it!) The minimum value that \texttip{x}{x} \texttip{x}{x} can have is obviously zero. If \texttip{m}{m} \texttip{m}{m} is less than the mass that would give x_{\rm critical}=0 then critical}=0 then the bar will be stable for any physical value of \texttip{x}{x} \texttip{x}{x}..

ANSWER: \texttip{m_{\rm max}}{m_max} = \large{\frac{m_{1}\frac{L}{2}-m_{1} d}{d}}

Correct

An Unfair Race This applet applet shows the results of releasing a frictionless block and a rolling disk with equal masses from the top of identical inclined planes.

Part A Which of the following is the best explanation of the results shown in the applet? ANSWER: The disk loses energy to friction as it rolls, but the box is frictionless and so it speeds up more quickly and gets to the bottom first. The potential energy of the disk is converted into translational and rotational kinetic energy, so the translational speed grows more slowly than that of the box, which has no rotational energy. The net forces on the two objects are equal, but the force on the disk gets partially used up in creating the torque necessary to make it roll. The net forces on the two objects are equal, but the force on the disk is not directed parallel to the ramp, and so does not create as great an acceleration down the ramp.

Correct

This applet applet shows shows the same situation, but it also shows, through bar graphs that change with time, the way that the energy is transformed as the box and the disk go down the inclined plane. Assume that the box and disk each have mas s \texttip{m}{m} \texttip{m}{m},, the top of the incline is at height \texttip{h}{h} \texttip{h}{h},, and the angle between the incline and the ground is \texttip{\theta }{theta} }{theta} (i.e., the incline is at an angle \texttip{\theta }{theta} }{theta} above the horizontal). Also, let the radius of the disk be \texttip{R}{R} \texttip{R}{R}..

Part B How much sooner does the box reach the bottom of the incline than the disk? Express your answer in terms of some or all of the variables \textt variables \texttip{m}{m} ip{m}{m},, \textt \texttip{h}{h} ip{h}{h},, \textt \texttip{\theta ip{\theta }{theta}, }{theta}, and \textt and \texttip{R}{R} ip{R}{R},, as well as the acceleration due to gravity \textt gravity \texttip{g}{g} ip{g}{g}..

Hint 1. How to approach the problem You can use conservation of energy to determine the final speed at the bottom of the ramp for each object. From that, you can determine the average speed and then use kinematics to determine the time to get to the bottom. The difference in times for the two objects to get to the bottom is what you are trying to find. If you take the direction down the inclined plane as the positive x axis, then the velocities are always positive and one-dimensional in the x the x direction, so the velocity and the speed have the same sign.


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Hint 2. Find the final speed of the box Determine the speed \texttip{v_{\rm box}}{v_box} box}}{v_box} of the box at the bottom of the ramp. Express your answer in terms of some or all of the variables \textt variables \texttip{m}{m} ip{m}{m},, \textt \texttip{h}{h} ip{h}{h},, \textt \texttip{\theta ip{\theta }{theta}, }{theta}, and \textt and \texttip{R}{R} ip{R}{R},, as well as the acceleration due to gravity \textt gravity \texttip{g}{g} ip{g}{g}..

Hint 1. Using conservation of energy Initially, the box has only gravitational potential energy mgh mgh.. At the bottom of the incline, all its potential energy has been converted to kinetic energy \large{\frac12 mv_{\rm box}^2}. box}^2}. Thus, \large{mgh=\frac 1 2 mv_{\rm box}^2}. box}^2} . You can solve this equation to find \texttip{v_{\rm box}}{v_box} box}}{v_box} in terms of the given variables.

ANSWER: \texttip{v_{\rm box}}{v_box} = \sqrt{2 g h}

Hint 3. Find the final speed of the disk Determine the speed \texttip{v_{\rm disk}}{v_disk} disk}}{v_disk} of the center of mass of the disk at the bottom of the ramp. Be certain to account for translational and rotational kinetic energy. Express your answer in terms of some or all of the variables \textt variables \texttip{m}{m} ip{m}{m},, \textt \texttip{h}{h} ip{h}{h},, \textt \texttip{\theta ip{\theta }{theta}, }{theta}, and \textt and \texttip{R}{R} ip{R}{R},, as well as the acceleration due to gravity \textt gravity \texttip{g}{g} ip{g}{g}..

Hint 1. Using conservation of energy Initially, the disk has only gravitational potential energy mgh mgh.. At the bottom of the incline, the potential energy has been entirely converted to kinetic energy. However, the kinetic energy is not simply \large{\frac 12mv_{\rm disk}^2}, disk}^2} , because the disk has rotational kinetic energy as well. The total kinetic energy is \large{\frac12 mv_{\rm disk}^2+\frac12 I\omega^2} , where \texttip{I}{I} \texttip{I}{I} is the moment of inertia and \texttip{\omega }{ome }{omega} ga} is is the angular speed. Thus, \large{mgh=\frac 12 mv_{\rm disk}^2+\frac 12 I\omega^2} . You need to find the values of \texttip{I}{I} \texttip{I}{I} and \textti p{\omeg p{\omega a }{ome }{omega} ga} in terms of the given variables before you solve for \texttip{v_{\rm disk}}{v_disk}. disk}}{v_disk} .

Hint 2. Moment of inertia for a disk The moment of inertia \texttip{I}{I} \texttip{I}{I} for a uniform disk of radius \texttip{R}{R} \texttip{R}{R} and mass \texttip{m}{m} is \large{I=\frac 1 2 m R^2}. R^2}.

Hint 3. Relating angular speed and the speed of the center of mass Recall that you are assuming that the disk is rolling without slipping. When a disk (or other object with circular cross section) rolls without slipping, it follows the relation v=\omega R, R, where \texttip{v}{v} \texttip{v}{v} is the speed of the center of mass, \texttip{R}{R} \texttip{R}{R} is the radius, and \textt ip{\omega }{omega} }{omega} is the angular speed.

ANSWER: \texttip{v_{\rm disk}}{v_disk} = \large{\sqrt{\frac{4 g h}{3}}}

Hint 4. Finding the average speed Since the forces remain constant throughout the motion, this must be motion with constant acceleration. Therefore, the velocity must change linearly. Recall that in motion with constant acceleration, the average velocity \texttip{v_{\rm avg}}{v_avg} avg}}{v_avg} is given by the formula v_{\rm avg}= (v_{\rm initial}+v_{\rm final})/2, final})/2 , where \texttip{v_{\rm initial}}{v_initial} initial}}{v_initial} and \texttip{v_{\rm final}}{v_final} final}}{v_final} are the initial and final velocities.

Hint 5. Finding the time from the average speed Recall that average velocity \texttip{v_{\rm avg}}{v_avg} avg}}{v_avg} is defined by \large{v_{\rm \large{ v_{\rm avg}=\frac{x_{\rm final}-x_{\rm i nitial}}{t_{ nitial}}{t_{\rm \rm final}-t_{ final}-t_{\rm \rm initial}}} initial}}},, where \texttip{x_{\rm final}}{x_final} final}}{x_final} and \texttip{x_{\rm initial}}{x_initial} initial}}{x_initial} are the final and initial positions, and \texttip{t_{\rm final}}{t_final} and https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW \texttip{t_{\rm initial}}{t_initial} initial}}{t_initial} are the final and initial times. The change in time is what you want to find, and the change in position is just the length of the inclined plane.

Hint 6. Find the length of the incline What is the length \texttip{L}{L} \texttip{L}{L} of of the inclined plane? The figure may be helpful. Express your answer in terms of one or both of the variables \textt variables \texttip{h}{h} ip{h}{h} and \texttip{\theta }{theta}. \texttip{\theta }{theta}.

ANSWER: \texttip{L}{L} = \large{\fra \large{\frac{h}{{ c{h}{{\sin}\left({\theta}\right)} \sin}\left({\theta}\right)}}}

ANSWER: \large{\frac{\sqrt{\frac{3h}{g}}-\sqrt{\frac{2h}{g}}}{{\sin}{\theta}}}

Correct You should look at your answer and consider limiting cases. A simple one is that the time difference should tend to zero as the length of the board shrinks to zero. Simply express the height of the board in terms of the length of the incline and you'll see that your answer indeed behaves this way. Your answer also predicts that the difference in time grows longer as \texttip{\theta }{theta} shrinks }{theta} shrinks toward zero while the height remains fixed (i.e., the difference in time grows longer as the length of the board grows longer). It might not be immediately obvious to you that this should happen, but it is not inconceivable, and you can do some simple experiments to see that it is actually true. As \texttip{\theta }{theta} }{theta} grows toward \pi/2\;{\rm rad}=90^\circ , you might expect the difference in time to go to zero, because if you drop a disk and a box they fall at the same rate. However, recall that your derivation included the assumption that the disk rolls without slipping, which is definitely not the case if the disk is simply dropped vertically. Therefore, this formula shouldn't apply to the case of simply dropping the disk and box. Can you think of a situation with a vertical drop in which the disk would obey obey v=\omega R? R?

Problem 12.46

Part A What is the magnitude of the angular momentum of the 2.70 {\rm kg} , kg} , 5.70-cm-diameter rotating disk in the figure ?


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ANSWER: 6.89×10−2

\rm{kg\,m^2/s}

Correct

Part B What is its direction? ANSWER: x direction - x x direction y direction -y direction z direction -z direction

Correct

Problem 12.70 Blocks of mass m_1 m_1 and m_2 m_2 are connected by a massless string that passes over the pulley in the figure . The pulley turns on frictionless bearings, and mass m_1 m_1 slides on a horizontal, frictionless surface. Mass m_2 m_2 is released while the blocks are at rest.

Part A Assume the pulley is massless. Find the acceleration of m_1 m_1.. Express your answer in terms of the given quantities. ANSWER: a_1 = \large{\frac{m_{2} g}{m_{1}+m_{2}}}

Correct

Part B Find the tension in the string. Express your answer in terms of the given quantities. ANSWER: https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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T = \large{\frac{m_{1} m_{2} g}{m_{1}+m_{2}}}

Correct

Part C Suppose the pulley has mass m_p m_p and radius R. Find the acceleration of m_1 m_1.. Verify that your answers agree with part A if you set m_p=0 m_p=0.. Express your answer in terms of the given quantities. ANSWER: a_1 = \large{\frac{2m_{2} g}{2m_{1}+2m_{2}+m_{p}}}

Correct

Part D Find the tension in the upper portion of the string. Verify that your answers agree with part B if you set m_p=0 m_p=0.. Express your answer in terms of the given quantities. ANSWER: T_{upper} = \large{\frac{2m_{1} m_{2} g}{2m_{1}+2m_{2}+m_{p}}}

Correct

Part E Find the tension in the lower portions of the string. Verify that your answers agree with part B if you set m_p=0 m_p=0.. Express your answer in terms of the given quantities. ANSWER: T_{lower} = \large{\frac{2m_{1} m_{2} g+m_{p} m_{2} g}{2m_{1}+2m_{2}+m_{p}}}

Correct

Problem 12.79 A 10 \rm g bullet g bullet traveling at 430 {\rm m/s} m/s} strikes a 11 {\rm kg} kg} , 0.90-\rm 0.90-\rm m-wide m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

Part A What is the angular velocity of the door just after impact? Express your answer to two significant figures and include the appropriate units. ANSWER: w=

1.3 \large{{\rm \frac{rad}{s}}}

Correct

Rotational Kinematics Ranking Task https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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The figure shows angular position versus time graphs for six different objects.

Part A Rank these graphs on the basis of the angular velocity of each object. Rank positive angular velocities as larger than negative angular velocities. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Determining angular velocity from an angular position versus time graph The slope of an angular position versus time graph is the "rise" (change in angular position) over the "run" (change in time). In physics, the ratio of change in angular position to change in time is defined as the angular velocity. Thus, the slope of an angular position versus time graph gives the angular velocity of the object being graphed.

ANSWER:

Correct

Part B Rank these graphs on the basis of the angular acceleration of the object. Rank positive angular accelerations as larger than negative angular accelerations. Rank from largest to smallest. To rank items as equivalent, overlap them. https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Hint 1. Determining angular acceleration Angular acceleration is prop proportional ortional to the change in an object's angular velocity. Since angula angularr velocity can be determined f rom the slope of an angular position versus time graph, angular acceleration can be determined from changes in slope.

ANSWER:

Correct

Balancing Torques Ranking Task A sign is to be hung from t he end of a thin pole, and the pole s uppor upported ted by a single cable. You Yourr design f irm brainstorms the six scenarios shown below. In scenarios A, B, and D, the cable is attached halfway between the midpoint and end of the pole. In C, the cable is attached to the mid-point of the pole. In E and F, the cable is attached to the end of the pole.

Part A Rank the design scenarios (A through F) on the basis of the tension in the supporting cable. Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. How to approach the problem In all cases, the pole and sign are held in equilibrium by the torque supplied by the cable. Therefore, by setting the torque due to the cable equal to the torque due to the sign and the pole, you can determine the relative tension in the cable. Note that the pole and sign are the same in each case, so the torque due to them is the same for each case.

Hint 2. The mathematical relationship The torque due to the cable is given by \tau = rF\sin\theta, rF\sin\theta, where \texttip{r}{r} \texttip{r}{r} is the distance from the hinge to the force and \texttip{\theta }{theta} }{theta} is the angle of the force relative to the pole. Since the pole is in equilibrium, the torque due to the cable is equal in magnitude to the net torque due to the sign and the weight of the pole. Thus, \tau _{\rm cable} = \tau _{\rm sign} + \tau_{\rm pole} , or rF\sin\theta = \tau _{\rm sign} + \tau_{\rm pole} , and therefore https://sessi on.master i ngphysi cs.com /myct/assi gnmentPr i ntVi ew?assi gnmentID= 4057478

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Ch 12 HW \large{F = \frac{\tau _{\rm sign} + \tau_{\rm pole}}{r\sin\theta}} . Substituting different values of \texttip{r}{r} \texttip{r}{r} and \texttip{\theta }{theta} }{theta} into this relationship can give you the relative sizes of the required cable forces.

ANSWER:

Correct

Video Tutor: Balancing a Meter Stick First, launch the video below. 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 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.


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ANSWER: 50 cm (in the middle) 25 cm 10 cm 75 cm (25 cm from the weight) 90 cm (10 cm from the weight)

Correct

Video Tutor: Walking the plank First, launch the video below. 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 \rm m m needed to keep the board from tipping over. To rank items as equivalent, overlap them.

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:


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Correct Score Summary: Your score on this assignment is 118%. You received 103.37 out of a possible total of 105 points, plus 20.56 points of extra credit.


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Mastering Physics Ch 12 HW College Physics I Brian Uzpen LCCC

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