Mastering Physics Ch14

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Assignment 9

Due: 11:59pm on Sunday, March 14, 2010

Note: To understand how points are awarded, read your instructor's Grading Policy.

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Precessing Tilted Gyroscope

A gyroscope consists of a flywheel of mass , which has a moment of inertia for rotation about its axis. It is mounted on a rod of

negligible mass, which is supported at one end by a frictionless pivot attached to a vertical post, as shown in the diagram. The distancebetween the center of the wheel and the pivot is . The wheel rotates about its axis with angular velocity , where positive refers to

counterclockwise rotation as seen by an observer looking at the face of the wheel that is opposite the pivot. The rod is tilted upward, makingan angle with respect to the horizontal. Gravity acts downward with a force of magnitude .

Adopt a coordinate system with the z axis pointing upward and the x and y axes in the horizontal plane. The gyroscope is moving, but at, the rod is in the yz plane.

Part A

Assuming that the only significant contribution to the angular momentum comes from the spinning of the flywheel about its center, what is the angular momentum vector about the pivot at ?

Hint A.1 A formula for angular momentum

The formula for the angular momentum of an object with moment of inertia spinning with angular velocity is

.

Hint A.2 Find the direction of the angular velocity vector

Recall that the direction of the angular velocity vector is given by the right-hand rule. What angle does make with the positive y axis?

Express your answer in terms of some or all of the gi ven variables.

ANSWER:

Correct

Specify the components of with respect to the axes shown in the diagram. Write the components in order , , separated by commas.

ANSWER:, , =

Correct

Part B

At , what is the torque acting on the wheel about the pivot?

Hint B.1 What is the direction of the torque?

Hint not displayed

Hint B.2 Find the x component of the torque

Hint not displayed

Express your answer in terms of components , , , separated by commas.

ANSWER:, , =

Correct

Part C

The gyroscope is observed to precess about the vertical axis, with an angular velocity of precession , defined as positive for counterclockwise precession as seen from above. Find in terms of the

given quantities.

Hint C.1 Relevant laws for rotational dynamics

Hint not displayed

Hint C.2A relation between and

Hint not displayed

MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?

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ANSWER:

=

Correct

Thus the rate of precession is independent of ! The reason for this is that varying changes both and the torque due to gravity by the same factor, .

Introduction to Static Equilibrium

Learning Goal: To understand the conditions necessary for static equilibrium.

Look around you, and you see a world at rest. The monitor, desk, and chair—and the building that contains them—are in a state described as static equilibrium. Indeed, it is the fundamental objective of many branches of engineering to maintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the more unpredictable forces from wind and earthquakes.

The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angular acceleration. Hence static mechanical equilibrium (as opposed to thermal orelectrical equilibrium) requires that the forces acting on a body simultaneously satisfy two conditions:

and ;

that is, both external forces and torques sum to zero. You have the freedom to choose any point as the origin about which to take torques.

Each of these equations is a vector equation, so each represents three independent equations for a total of six. Thus to keep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tilt about either the x or y axis, nor can it rotate about its vertical axis.

Part A

Frequently, attention in an equilibrium situation is confined to a plane. An example would be a ladder leaning against a wall, which is in danger of slipping only in the plane perpendicular to theground and wall. By orienting a Cartesian coordinate system so that the x and y axes are in this plane, choose which of the following sets of quantities must be zero to maintain static equilibrium inthis plane.

Hint A.1 Simplifying the equations

The motion (or possible motion) is confined to a plane, the xy plane in this case, when there are no forces acting out of that plane (e.g., all or all z-component forces occur in pairs that are

applied at the same points). Recalling that torque is defined as a cross product, you can eliminate the need for two of the three equations for the components of torque since they will equal zero.

ANSWER:and and

and and

and and and

and and and and

Correct

Part B

As an example, consider the case of a board of length and negligible mass. Take the x axis to be the horizontal axis along the board and the y axis to be the vertical axis perpendicular to the board.

A mass of weight is strapped to the board a distance from the left-hand end. This is a static equilibrium problem, and a good first step

is to write down the equation for the sum of all the forces in the y direction since the only nonzero forces of that exist are in the

y direction.

What is ? Your equation for the net force in the y direction on the board should contain all the forces acting vertically on the board.

Express your answer in terms of the weight and the tensions in the two vertical ropes at the left and right ends and .

Recall that positive forces point upward.

ANSWER:

Correct

The only relevant component of the torques is the z component; however, you must choose your pivot point before writing the equations. This point could be anywhere; in fact, the pivot point doesnot even have to be at a point on the body. You should choose this point to your advantage. Generally, the best place to locate the pivot point is where some unknown force acts; this will eliminate thatforce from the resulting torque equation.

Part C

What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where acts)?

Express your answer in terms of the unknown quantities and and the known lengths and . Recall that counterclockwise torque is positive.

ANSWER:

Correct


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This gives us one equation involving two unknowns, and . We can use this result and to solve for and .

Part D

What is the equation that results from choosing the pivot point to be the left end of the plank (where acts)?

Express your answer in terms of , , , and the dimensions and . Not all of these variables may show up in the solution.

ANSWER:

Correct

Part E

What is the equation that results from choosing the pivot point to be the right end of the plank (where acts)?

Express your answer in terms of , , , and the dimensions and . Not all of these variables may show up in the solution.

ANSWER:

Correct

Part F

Solve for , the tension in the right rope.

Hint F.1 Choose the correct equation

Hint not displayed

Express your answer in terms of and the dimensions and . Not all of these variables may show up in the solution.

ANSWER:

=

Correct

Part G

Solve for , the tension in the left rope.

Hint G.1 Choose the correct equation

Hint not displayed


ANSWER:

=

Correct

Part H

Solve for the tension in the left rope, , in the special case that . Be sure the result checks with your intuition.


ANSWER:=

Correct

Only one set of forces, exactly balanced, produces static equilibrium. From this perspective it might seem puzzling that so much of the world is static. One must realize, however, that manyforces—like those of the tensions in the ropes here or those between the floor and an object resting on it—increase very quickly as the object moves. If there is a slight imbalance of the forces,the object accelerates so that its position changes until the object has adjusted itself to restore the force balance. It then oscillates about this point until friction or some other dissipativemechanism causes it to become stationary at the exact equilibrium point.

Tipping Crane

Learning Goal: To step through the application of to prevent a crane from tipping over.

A crane of weight has a length (wheelbase) , and its center of mass is midway between the wheels (i.e., the mass of the lifting arm is negligible). The arm extending from the front of the crane has

a length and makes an angle with the horizontal. The crane contacts the ground only at its front and rear tires.


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Part A

While watching the crane in operation, an observer mentions to you that for a given load there is a maximum angle that the crane arm can make with the horizontal without tipping the crane

over. Is this correct?

ANSWER: yes

no

Correct

Part B

Later that week, while watching the same crane in operation, a different observer mentions to you that there is a maximum load the crane can lift without tipping, and you can find that maximum load

by observing the minimum angle that the crane arm makes with the horizontal. Is this correct?

ANSWER: yes

no

Correct

Part C

Select the correct explanation for why you can determine the maximum load given that is the minimum angle the crane arm can make with the horizontal.

ANSWER: At this angle the load weighs the most.

At this angle the load exerts the most torque about the pivot point.

At this angle the load exerts the least torque about the pivot point.

At this point the load weighs the least.

Correct

Part D

You know that the torques must sum to zero about _________ if an object is in static equilibrium.

Pick the most general phrase that correctly completes the statement.

ANSWER: the center of mass

any point on the body

any point on or off the body

any point where any force acts on the body

Correct

Part E

This implies that you can pick the point about which to sum the torques to simplify the calculation. Often it is best to pick the point where an unknown force acts, so that the torque due to that force iszero. In this problem the simplest equations result if you take torques about __________.

ANSWER: the point where the rear tires touch the ground

the center of mass

the point where the front tires touch the ground

the point on the ground directly under the load

Correct

Part F

Given the angle , what is the maximum weight (or load) that this crane can lift without tipping forward? (Recall that weight has units of force.)

Hint F.1 Find the torque due to the weight of the crane

What is the magnitude of the torque exerted about the front wheels of the crane by the weight of the crane?

Answer in terms of , , and other given quantities. Take the sign of counterclockwise torques to be positive.

ANSWER:=

Correct

Hint F.2 Find the torque due to the weight of the load

What is the torque about the front wheels exerted by a load of weight when held at an angle from the horizontal?

Answer in terms of , , and , and take the sign of counterclockwise torques to be positive.

ANSWER:=

Correct

In a simple problem like this, it may seem easier to simply compute the magnitude of the two torques and set them equal to each other as a condition of equilibrium. Experience shows that


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students who explicitly account for the direction of the torque with the sign make fewer mistakes on problems with several torques, so we strongly recommend that you adopt this procedure.

Note that at the minimum angle the magnitude of the torque from the load is a maximum.

Hint F.3 Find the torque from the normal force on the rear wheels

What is the torque about the front wheels from the normal force on the rear wheels when the crane is just on the verge of tipping over?

Hint F.3.1 Normal force when tipping begins

Hint not displayed

Express your answer in terms of the weights and distances given in the introduction.

ANSWER:= 0

Correct

Hint F.4 Calculate the sum of torques

What must the total sum of the torques around the front wheels be if the crane is not moving?

Express your answer in newton-meters.

ANSWER:= 0

Correct

Use this information to find a relation between the two torques and therefore solve for the maximum weight of the load.

Express the maximum load in terms of and other quanitites given in the problem introduction.

ANSWER:

=

Correct

Part G

What is , the largest weight of the load that is safe to lift regardless of the angle of the crane's arm?

Express in terms of , , and .

ANSWER:

=

Correct

Part H

Notice that we have the weight of the crane exerting a torque about the front wheels of the same crane. To create a torque, a force must be present, so it would seem that somehow the weight of thecrane is exerting a force upon its front wheels. However, the crane is one object, and it follows from Newton's laws that an object cannot exert a net force upon itself. This crane seems to be defyingNewton's laws. What's going on here?

ANSWER: Newton's laws don't apply to torques.

The rear wheels exert a downward force on the front wheels.

The crane is not accelerating so forces don't matter.

The earth exerts forces on the crane and the load.

Correct

Part I

Assume you get a summer job as a crane operator. On the first day you are lifting a heavy piece of machinery. Even though you have the arm at above the horizontal, the crane begins to tip slowly

forward. Consider the following possible actions:

Release the brake on the lifting cable so that the l oad accelerates downward.1.Decrease so that the load accelerates downward.2.

Increase while simultaneously letting out the lifting cable so that the load accelerates downward.3.

Put the crane wheels in gear and accelerate the crane forward.4.

None of these solutions is ideal, but which will have the short-term effect of restoring contact of the crane's rear wheels with the ground?

ANSWER: all but 1

all but 2

all but 3

all but 4

all of them

Correct

In cases 1 and 2, the load will either free fall until it hits the ground or will be stopped before that, at which point the torque from the load will be greater than it was when the crane began to tip.In case 4, the crane must continue to accelerate indefinitely, or else it will begin to tip again.

The only action that potentially avoids disaster is option 3--increasing the angle of the lifting arm so that the load swings closer to the crane and therefore exerts less torque. You must


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simultaneously lower the load or else increasing will involve lifting the load, which increases the torque exerted by the load on the crane, causing the crane to tip more quickly. However, if you

lower the load too quickly, then reapplying the brake to the lifting cable could cause a greater torque on the crane than when it first began to tip. The best solution would be to let out the liftingcable at a rate such that the load stays at exactly the same height as you increase the angle of the crane. Obviously a delicate operation!

A Person Standing on a Leaning Ladder

A uniform ladder with mass and length rests against a smooth wall. A do-it-yourself enthusiast of mass stands on the ladder a

distance from the bottom (measured along the ladder). The ladder makes an angle with the ground. There is no friction between the wall

and the ladder, but there is a frictional force of magnitude between the floor and the ladder. is the magnitude of the normal force

exerted by the wall on the ladder, and is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem,consider counterclockwise torques to be positive. None of your answers should involve (i.e., simplify your trig functions).

Part A

What is the minimum coeffecient of static friction required between the ladder and the ground so that the ladder does not slip?

Hint A.1 Method for finding

Recall that, in general, . In this problem, is as small as it can be without allowing the ladder to slip, and therefore . Therefore, . Now, you need to

use the fact that the ladder is in translational and rotational equilibrium (i.e., the net force is zero, and the net torque is zero) to find expressions for and in terms of the desired quantities.

Hint A.2 Expression for

What is , the magnitude of the vertical force exerted by the ground on the ladder?

Express your answer in terms of , , and .

ANSWER:=

Correct

Hint A.3 Choice of origin of torque

The best choice of origin in which to add up the torques would be the point at which the ladder touches the wall. By summing torques around this point we can ignore the normal force (because

the moment arm has zero length) and find in terms of already given quantities.

Hint A.4 Sum the torques

Hint not displayed

Hint A.5 Putting it all together

Hint not displayed

Express in terms of , , , , and .

ANSWER:

=

Correct

Part B

Suppose that the actual coefficent of friction is one and a half times as large as the value of . That is, . Under these circumstances, what is the magnitude of the force of friction

that the floor applies to the ladder?

Hint B.1 Relation between frictional force and

Hint not displayed

Express your answer in terms of , , , , , and . Remember to pay attention to the relation of force and .

ANSWER:

=

Correct

A Bar Suspended by Two Vertical Strings

A rigid, uniform, horizontal bar of mass and length is supported by two identical massless strings. Both strings are vertical. String A is

attached at a distance 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 is supported against gravity by the bar at a distance from the left end of the bar, as

shown in the figure.


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Throughout this problem positive torque is that which spins an object counterclockwise. Use for the magnitude of the acceleration due to

gravity.

Part A

Find , the tension in string A.

Hint A.1 Choosing an axis

Choose a rotation axis p, about which to apply the requirement . Since the system is in static equilibrium, the choice of rotation axis is arbitrary; however, there is a convenient choice of

p to find by eliminating the torque from an unknown force.

Hint A.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 , , , , , , and .

ANSWER:

=

Answer Requested

Hint A.3 Summing the torques

Hint not displayed

Express the tension in string A in terms of , , , , , and .

ANSWER:

=

Correct

Part B

Find , the magnitude of the tension in string B.

Hint B.1 Two different methods to find

There are two equivalent ways to find . One way is to balance the torques as was done in the calculation of , 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, .

Hint B.2 Direction of forces

Hint not displayed

Express the magnitude of the tension in string B in terms of , , , and .

ANSWER:=

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 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).

What is the smallest possible value of such that the bar remains stable (call it )?

Hint D.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 ?

ANSWER: The point where string A is attached to the bar

The point where string B is attached to the bar

Correct


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Hint D.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 is decreased, is likewise decreased because the clockwise torque

about this point decreases. The critical value corresponds to when . If is decreased further, 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 D.3 Calculate the torques

Hint not displayed

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

ANSWER:

=

Correct

Part E

Note that since , as computed in the previous part, is not necessarily positive. If , the bar will be stable no matter where the block of mass is placed on it.

Assuming that , , and are held fixed, what is the maximum block mass for which the bar will always be stable? In other words, what is the maximum block mass such that ?

Hint E.1 Requirement of stability

Hint not displayed

Answer in terms of , , and .

ANSWER:

=

Correct

Sliding Dresser

Sam is trying to move a dresser of mass and dimensions of length and height by pushing it with a horizontal force applied at a height above the floor. The coefficient of kinetic friction

between the dresser and the floor is and is the magnitude of the acceleration due to gravity. The ground exerts upward normal forces of

magnitudes and at the two ends of the dresser. Note that this problem is two dimensional.

Part A

If the dresser is sliding with constant velocity, find , the magnitude of the force that Sam applies.

Hint A.1 Force required in terms of normal forces

Hint not displayed

Hint A.2 Normal force

Hint not displayed

Express the force in terms of , , and .

ANSWER:=

Correct

Part B

Find the magnitude of the normal force . Assume that the legs are separated by a distance , as shown in the figure.

Hint B.1 Torques

Hint not displayed

Hint B.2 Rotation

Hint not displayed

Express this normal force in terms of , , , , and .


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ANSWER:

=

Correct

Part C

Find the magnitude of the normal force . Assume that the legs are separated by a distance , as shown in the figure.

Hint C.1 Torques

Hint not displayed

Answer in terms of , , , , and .

ANSWER:

=

Correct

Part D

Find , the maximum height at which Sam can push the dresser without causing it to topple over.

Hint D.1 Torque dependence

Hint not displayed

Hint D.2 Limiting case

Hint not displayed

Express your answer for the maximum height in terms and .

ANSWER:

=

Correct

Young's Modulus

Learning Goal: To understand the meaning of Young's modulus, to perform some real-life calculations related to stretching steel, a common construction material, and to introduce the concept of breaking stress.

Hooke's law states that for springs and other "elastic" objects

,

where is the magnitude of the stretching force, is the corresponding elongation of the spring from equilibrium, and is a constant that depends on the geometry and the material of the spring. If

the deformations are small enough, most materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be useful to operate with an expressionthat is similar to Hooke's law but describes the properties of various materials, as opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length and

cross-sectional area stressed by a force of magnitude . As a result, the bar stretches by .

Let us define two new terms:

Tensile stress is the ratio of the stretching force to the cross-sectional area:

.

Tensile strain is the ratio of the elongation of the rod to the initial length of the bar:

.

It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too large. That constant, whichis an inherent property of a material, is called Young's modulus and is given by

Part A

What is the SI unit of Young's modulus?

Hint A.1 Look at the dimensions

Hint not displayed

ANSWER:

Correct

Part B

Consider a metal bar of initial length and cross-sectional area . The Young's modulus of the material of the bar is . Find the "spring constant" of such a bar for low values of tensile strain.


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Hint B.1 Use the definition of Young's modulus

Hint not displayed

Express your answer in terms of , , and .

ANSWER:

=

Correct

Part CTen identical steel wires have equal lengths and equal "spring constants" . The wires are connected end to end, so that the resultant wire has length . What is the "spring constant" of the

resulting wire?

Hint C.1 The spring constant

Hint not displayed

ANSWER:

Correct

Part D

Ten identical steel wires have equal lengths and equal "spring constants" . The wires are slightly twisted together, so that the resultant wire has length and its cross-sectional area is ten times

that of the individual wire. What is the "spring constant" of the resulting wire?

Hint D.1 The spring constant

Hint not displayed

ANSWER:

Correct

Part E

Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is . The wires are connected end to end, so that the resultant wire has length .

What is the Young's modulus of the resulting wire?

ANSWER:

Correct

Part F

Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is . The wires are slightly twisted together, so that the resultant wire has length

and is ten times as thick as the individual wire. What is the Young's modulus of the resulting wire?

ANSWER:

Correct

By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulus depends on the material, which remains unchanged. To change the Young'smodulus, one would have to change the properties of the material itself, for instance by heating or cooling it.

Part G

Consider a steel guitar string of initial length meter and cross-sectional area square millimeters. The Young's modulus of the steel is pascals. How far ( )

would such a string stretch under a tension of 1500 newtons?

Use two significant figures in your answer. Express your answer in millimeters.

ANSWER: = 15


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Correct

Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, for greater values of tensile strain, the material no longer behaves elastically. If thestrain and stress are large enough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the "stretching limit" of steel.

Part H

Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the

maximum stress the material can bear without deteriorating, is about pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of

the acceleration due to gravity remains constant at 9.8 meters per second per second.

Hint H.1 Why does the cable break?

Hint not displayed

Hint H.2 Find the stress in the cable

Hint not displayed

Use two significant figures in your answer, expressed in kilometers.

ANSWER: 26Correct

This is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is small enough to justify the assumption of virtually constant acceleration due to gravity.When making such assumptions, one should always check their validity after obtaining a result.

Problem 11.78

A bale is 0.25 wide, 0.50 high, and 0.80 long, with mass 30.0 . The center of gravity of each bale is at its geometrical center. It is dragged along a horizontal surface with constant speed by

a force (the figure ). The coefficient of kinetic friction is 0.35.

Part A

Find the magnitude of the force .

Express your answer using two significant figures.

ANSWER:= 100

Correct

Part B

Find the value of at which the bale just begins to tip.


ANSWER:= 0.36

Correct

Problem 11.66

One end of a uniform meter stick is placed against a vertical wall . The other end is held by a lightweight cord that makes an angle with the

stick. The coefficient of static friction between the end of the meter stick and the wall is 0.35.


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Part A

What is the maximum value the angle can have if the stick is to remain in equilibrium?


ANSWER:= 19

Correct

Part B

Let the angle between the cord and the stick is = 16 . A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance from the wall. What is the minimum value

of for which the stick will remain in equilibrium?


ANSWER:= 40

Correct

Part C

When = 16 , how large must the coefficient of static friction be so that the block can be attached 15 from the left end of the stick without causing it to slip?


ANSWER:= 0.60

Correct

Problem 11.85

A mass of 11.8 , fastened to the end of an aluminum wire with an unstretched length of 0.50 , is whirled in a vertical circle with a constant angular speed of 127 . The cross-sectional area

of the wire is 1.2!10!2

.

Part A

Calculate the elongation of the wire when the mass is at the lowest point of the path.


ANSWER:= 6.9!10

!3

Correct

Part B

Calculate the elongation of the wire when the mass is at the highest point of its path.


ANSWER:= 5.5!10

!3

Correct

Score Summary:

Your score on this assignment is 97.6%.

You received 97.55 out of a possible total of 100 points.


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Mastering Physics Ch14