loop the loop with a twist - university of icelandisa13/efni/einn/f.hl/edlisfraedi 1 v... · loop...

skiladæmi 8Due: 11:59pm on Wednesday, October 28, 2015

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

Loop the Loop with a Twist

In this problem you will consider the motion of a cylinder of radius that is rolled from a certain height so that it "loopsthe loop," that is, rolls around the track with a loop of radius shown in the figure without losing contact with the track.

Unless otherwise stated, assume that friction is sufficient that thecylinder rolls without slipping. The radius of the cylinder ismuch smaller than the radius of the loop.

Part A

Compared to an object that does not roll, but instead slides without friction, should a rolling object be released from thesame,a greater, or a lesser height in order just barely to complete the loop the loop?

Hint 1. How to approach the question

In order just to complete the loop without falling off, each object will need a certain critical speed at the topof the loop. The value of will be the same for all objects. Each object will need to be dropped from a certainheight in order to achieve a speed at the top of the loop. The most straightforward way to find is to useconservation of mechanical energy.

Hint 2. The difference between a sliding and a rolling object

The object that slides will have only translational kinetic energy. The rolling object, however, will have bothtranslational kinetic energy and rotational kinetic energy.

ANSWER:

Correct

Part B

Find the minimum height that will allow a solid cylinder of mass and radius to loop the loop of radius .

rcyl h

rloop

rcylrloop

vcritvcrit

h vcrit h

The rolling object should be released from a greater height.

The rolling object should be released from a lesser height.

The rolling object should be released from exactly the same height.

The answer depends on the moment of inertia of the rolling object.

h m rcyl rloopTypesetting math: 100%

Express in terms of the radius of the loop.

Hint 1. How to approach the problem

First, determine the minimum speed the cylinder needs to have at the top of the loop in order to stay in contactwith the track. Then, compute the total mechanical energy of the cylinder at the top of the loop (potential pluskinetic energy). Apply conservation of energy to find the height from which the object must be released.

Hint 2. Find the critical speed at the top of the loop

To stay in contact with the track, the cylinder must exceed some critical speed at the top of the loop. Findan expression for .

Express your answer in terms of the loop's radius and the acceleration due to gravity .


Use what you know about circular motion to determine the minimum accleration needed to keep theobject in contact with the track. Then identify all of the forces acting on the object. Apply Newton'ssecond law to relate these two quantities.

Hint 2. Find the acceleration at the top of the loop

In order to stay in contact with the track, the object must have an appropriate centripetal acceleration.Find , the acceleration of a cylinder in the vertical direction at the top of the loop, in terms of its speed .

Express in terms of the speed of the object at the top of the loop and the radius of theloop. Your answer should obey the convention that a positive (negative) value for indicates anupward (downward) acceleration.

ANSWER:

Hint 3. Find the sum of vertical forces

Assume that the object is traveling at the critical velocity, so that it just barely touches the track at thetop of the loop. Find the sum of the forces acting on the object at the top of the loop.

Express your answer in terms of given quantities and , the magnitude of the accleration due togravity. The variable should not appear in your answer.

ANSWER:

ANSWER:

Hint 3. Find the potential energy at top of loop

Find the potential energy of the cylinder at the top of the loop, taking the gravitational potential to be zeroat the bottom of the loop.

Express your answer in terms of , , and .

h rloop

h

vcritvcrit

rloop g

acv

ac v rloopac

= ac−( )v2

rloop

gac

= ∑ Fy −mg

= vcrit (g )rloop− −−−−−√

Utop

m g rloop

Typesetting math: 100%

ANSWER:

Hint 4. Find the kinetic energy at the top of the loop

Find the total kinetic energy of the cylinder at the top of the loop. Assume that it reaches the top of theloop with speed .

Express your answer in terms of and any other given quantities.

Hint 1. Find the rotational kinetic energy

Find the rotational kinetic energy of the cylinder at the top of the loop.

Express your answer in terms of the cylinder's mass and its translational speed .

Hint 1. Formula for rotational kinetic energy

If an object has moment of inertia with respect to an axis about which it is rotating with angularvelocity , its rotational kinetic energy is given by .

Hint 2. Formula for the moment of inertia of a cylinder

If the cylinder has mass and radius , the appropriate moment of inertia is .

Hint 3. Find the angular velocity of the cylinder

Find an expression for , the magnitude of the angular velocity of the cylinder.

Express your answer in terms of the cylinder's translational speed and its radius .

ANSWER:

ANSWER:

ANSWER:

ANSWER:

Correct

= Utop 2mgrloop

Ktopv

v

Kr

m v

I

ω I12 ω2

m r = mIcm12

r2

ω

v rcyl

= ωv

rcyl

= Kr m14

v2

= Ktop m34

v2

= h 114

rloop


Exercise 10.51

The rotor (flywheel) of a toy gyroscope has mass 0.130 . Its moment of inertia about its axis is .The mass of the frame is 2.50×10−2 . The gyroscope is supported on a single pivot with its center of mass a horizontaldistance of 4.00 cm from the pivot. The gyroscope is precessingin a horizontal plane at the rate of one revolution in 2.10 .

Part A

Find the upward force exerted by the pivot.

ANSWER:

Correct

Part B

Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min.

ANSWER:

Correct

PSS 10.1 Rotational Dynamics for Rigid Bodies

Learning Goal:

To practice ProblemSolving Strategy 10.1 Rotational Dynamics for Rigid Bodies.

While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into aroom and attempts to swing the heavy door shut before the dragon gets to her . The door is initially perpendicular to thewall, so it must be turned through 90 to close it. The door is 3.00 tall and 1.25 wide, and it weighs 750 . You canignore the friction at the hinges. If Exena applies a force of 220 at the edge of the door and perpendicular to it, how muchtime does it take her to close the door?

kg 1.20 × kg ⋅10−4 m2

kg

s

= 1.52 F N

= 1620 ω rev/min

∘ m m NN


ProblemSolving Strategy: Rotational dynamics for rigid bodies

IDENTIFY the relevant concepts:In some cases you may be able to use an energy approach. However, if the target variable is a force, a torque, anacceleration, an angular acceleration, or an elapsed time, using is almost always the best approach.

SET UP the problem using the following steps:

1. Sketch the situation and select the body or bodies to analyze.2. Draw a freebody diagram for each body and label unknown quantities with algebraic symbols. Show the shapeof the body accurately, with all dimensions and angles that you will need for calculations of torque.

3. Choose coordinate axes for each body, and indicate a positive sense of rotation for each rotating body.

EXECUTE the solution as follows:

1. Write an equation of motion for each body by applying , , or both to each body.2. There may be geometrical relationships between the motions of two or more bodies, as with a string thatunwinds from a pulley while turning it. Express these relationships in algebraic form.

3. Check that the number of equations matches the number of unknown quantities. Then, solve the equations tofind the target variable(s).

EVALUATE your answer:Check that the algebraic signs of your results make sense. Whenever possible, check the results for special cases orextreme values of quantities. Ask yourself: “Does this result make sense?”

IDENTIFY the relevant concepts

You are asked to find the time it takes Exena to close the door (a rigid body that rotates about the hinges). Energyconsiderations do not easily provide information about time. Instead, applying to describe the rotational motionof the door will be the best approach.

SET UP the problem using the following steps

Part A

In the following freebody diagrams, let be the force exerted by Exena on the door, the weight of the door, and the magnitude of the normal force exerted by the hinges. The diagrams show a view of the door from above, with theedge of the door in contact with the hinges at the top of each diagram. Any force directed out of the plane of the figureis represented by a dot, and any force directed into the plane of the figure is represented by a cross. Which freebodydiagram correctly describes the system under investigation?

Σ = Iτz αz

Σ = mF a Σ = Iτz αz

Σ = Iτz αz

F F g n


ANSWER:

CorrectIf we ignore the friction at the hinges, there are only three forces acting on the door: its weight, the forces exertedby Exena, and the normal force exerted by the hinges.

Part B

Which set of axes shown in the figure represents the best orientation for the coordinate axes? As in Part A, thediagrams show a view of the door from above, with the edge of the door in contact with the hinges at the top of eachdiagram.

ANSWER:

diagram A

diagram B

diagram C

diagram D

set A

set B

set C

set D


Correct

It is most convenient to take a point on the axis of rotation as the origin of the coordinate axes. Here, the axis ofrotation is parallel to the door and passes through the hinges, so the z axis is directed out of the plane of the figurethrough the edge of the door at the top of the diagram. It also helpful to choose the positive x axis in the directionof the force that causes the rotation. Keep in mind that counterclockwise torque is positive.

EXECUTE the solution as follows

Part C

If Exena applies a force of 220 at the edge of the door and perpendicular to it, how much time does it take her toclose the door?

Express your answer in seconds to three significant figures.


To find the time it takes Exena to shut the door, you need to first find the angular acceleration of the door, whichcan be found using the equation for rotational motion. You can then use the equations associated with rotationalkinematics to solve for time.

Hint 2. Find the net torque acting on the door

Find the net torque around the z axis (i.e., the direction coming out of the plane of the diagram shown inPart B at the origin) acting on the door.

Express your answer in newtonmeters to four significant figures.

Hint 1. The definition of torque

The tendency of a force to cause or change rotational motion around a chosen axis is measured by thetorque, which is the product of the magnitude of the force and the perpendicular distance betweenthe axis of rotation and the line of action of the force, or

.

The distance is also called the moment arm of the force .

Hint 2. Determine which forces exert a torque

As shown in the freebody diagram in Part A, there are three forces acting on the door: the weight, thenormal force, and the force applied by Exena. Which one of these forces exerts a nonzero torque on thedoor around the z axis (i.e., the direction coming out of the plane of the diagram at the origin)?

Check all that apply.

ANSWER:

Hint 3. Find the moment arm of the force applied by Exena

What is the moment arm of the force applied by Exena? Recall that Exena applies a force perpendicularto the door and the door is 3.00 high and 1.25 wide.

Express your answer in meters to four significant figures.

N t

τnet

F l

τ = Fl

l F

the normal force

the weight

the force applied by Exena

lm m


ANSWER:

ANSWER:

Hint 3. Find the moment of inertia

What is the moment of inertia of the door with respect to an axis along its vertical edge?

Express your answer in kilogrammeters squared to four significant figures.

Hint 1. The moment of inertia of a rectangular plate

The moment of inertia of a thin rectangular plate of width and height with respect to an axis alongthe edge of length is

,

where is the mass of the plate.

Hint 2. Find the mass of the door

You are given the weight of the door. What is the mass of the door?

Express your answer in kilograms to four significant figures.

ANSWER:

ANSWER:

Hint 4. Find the angular acceleration of the door

What is , the angular acceleration of the door?

Express your answer in radians per second per second to three significant figures.

Hint 1. The equation for rotational motion

As mentioned in the IDENTIFY step, to study the rotational motion of the door, it is most convenient touse the equation , which states that the net torque acting on a body is equal to the product ofits angular acceleration and its moment of inertia .

ANSWER:

= 1.250 l m

= 275.0 τnet N ⋅ m

I

I a hh

I = M13

a2

M

m

= 76.45 m kg

= 39.82 I kg ⋅ m2

α

Σ = Iτz αz

α I

= 6.91 α rad/s2


Hint 5. Identify the rotational kinematics equation needed to find the time

As established in Part A, the door undergoes only rotational motion. Assume that Exena turns the door an angle in a time interval . Let be the width of the door; the gravitational acceleration; and , respectively, thespeed and acceleration of the center of mass of the door; and its angular acceleration. Then, which of thefollowing expressions will be most useful in solving this problem?

ANSWER:

ANSWER:

Correct

EVALUATE your answer

Part D

How long would it take Exena to close the door if she doubled the magnitude of the force applied to the door?

Express your answer in seconds to three significant figures.

ANSWER:

Correct

This makes sense. If she pushes twice as hard, it will take less time to shut the door.

Rotating Spheres

Two massive spheres are mounted on a light rod that can be rotated by a string wrapped around a central cylinder, forminga winch as shown in the figure. A force of magnitude is applied to the string to turn the system. With respect to thevariables given in the figure, the equation for the magnitude of the angular acceleration is

.

θ t l g v aα

θ = α12

t2

θ = g12

t2

l = a12

t2

a = αl

= 0.674 t s

= 0.477 t s

Fα

α = rF

3mR2


Assume that the spheres are small enough that they may be considered point masses and that the masses of the rod andcylinder can be neglected.

Part A

If the sphere on the left is moved closer to the central cylinder and placed at a distance from the axis of rotation,what is the magnitude of the angular acceleration of the modified system? Assume that the rest of the systemdoesn't change.


The rotational analogue of Newton's second law states that the angular acceleration of the system is equal tothe net torque exerted on the system divided by the system's moment of inertia. By moving one of the spherescloser to the central cylinder, will either of these two quantities change?

Hint 2. Find the net torque

Write an expression for the magnitude of the net torque exerted on the system.

Express your answer in terms of some or all of the variables , , , and .

Hint 1. Torque

The magnitude of the torque of a force of magnitude with respect to a point is defined as theproduct of times the perpendicular distance between the line of action of the force and the point .That is,

.

ANSWER:


When the sphere of mass on the right is at a distance from the axis of rotation, and the sphere of mass on the left is at a distance from the axis of rotation,what is the moment of inertia of the system with

respect to the axis of rotation?


R/2α

τnet

F m r R

τ F OF l O

τ = Fl

= τnet rF

m R2m R/2 I

F m r R


Hint 1. Moment of inertia

The moment of inertia of a system of objects with respect to a given axis is defined as the sum of theproduct of the mass of each object in the system and the square of its perpendicular distance fromthe axis, or

,

where is the distance of the object of mass from the axis of rotation, and so on.

ANSWER:

ANSWER:

Correct

If one of the spheres is moved closer to the axis of rotation, the angular acceleration of the system increasesbecause now the system has a smaller moment of inertia. Specifically, the moment of inertia of the modifiedsystem is half the moment of inertia of the original system, which means that the angular acceleration of themodified system is twice that of the original system.

Consider again the original system. Instead of applying the force to the string, a force with the same magnitude is appliedto the rod at a point from the sphere of mass and in a direction perpendicular to the rod, as shown in the figure.

Part B

What is the magnitude of the angular acceleration of the system?

Imi ri

I = + + + ⋯m1 r21 m2 r2

2 m3 r23

r1 m1

= I m32 R

2

α = rF

3mR2

α = 3rF

2mR2

α = 2rF

3mR2

α = rF

2mR2

FR/2 2m

αTypesetting math: 100%


As you did in the previous part, use the rotational analogue of Newton's second law to find the angularacceleration of the system. Application of the force of magnitude directly to the rod still causes the rod torotate around an axis that passes through its midpoint, as in the original system. However, now theperpendicular distance between the line of action of and the axis of rotation is different. Will this affect eitherthe net torque or the moment of inertia of the system?

Hint 2. Find the net torque

When the force of magnitude is applied to the rod at from the sphere of mass and in a directionperpendicular to the rod, what is the magnitude of the net torque exerted on the system?


ANSWER:


What is the moment of inertia of the system with respect to the axis of rotation, when the force is applied tothe rod at from the sphere of mass and in a direction perpendicular to the rod?


ANSWER:

ANSWER:

CorrectIn conclusion, the angular acceleration of a rotating system will change if either the moment of inertia of thesystem or the net torque on it changes.

Exercise 10.12

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shownin . The pulley is a uniform disk with mass 11.0 and radius 39.0 and turns on frictionless bearings. You measurethat the stone travels a distance 12.9 during a time interval of 2.00 starting from rest.

F

F

F R/2 2mτnet

F m r R

= τnetRF2

I FR/2 m

F m r R

= I 3mR2

α = rF

3mR2

α = F6mR

α = F6m

α = F

3mR2

kg cmm s


Part A

Find the mass of the stone.

Express your answer with the appropriate units.

ANSWER:

Correct

Part B

Find the tension in the wire.

Express your answer with the appropriate units.

ANSWER:

Correct

Alternative Exercise 10.123

A horizontal plywood disk with mass 6.60 and diameter 1.06 pivots on frictionless bearings about a vertical axisthrough its center. You attach a circular modelrailroad track of negligible mass and average diameter 0.960 to the disk.A 1.10 , batterydriven model train rests on the tracks. To demonstrate conservation of angular momentum, you switchon the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.600 relative to thetracks.

Part A

Find the magnitude and direction of the angular velocity of the disk relative to the earth.

ANSWER:

= 10.6 m kg

= 35.5 T N

kg mm

−kgm/s

= 0.268 ωdisk rad/sTypesetting math: 100%

Correct

Part B

ANSWER:

Correct

Exercise 10.42

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotationaxis of . She then tucks into a small ball, decreasing this moment of inertia to . While tucked, shemakes two complete revolutions in 1.1 .

Part A

If she hadn't tucked at all, how many revolutions would she have made in the 1.6 from board to water?

Express your answer using two significant figures.

ANSWER:

Correct

Score Summary:Your score on this assignment is 99.2%.You received 6.94 out of a possible total of 7 points.

counterclockwise

clockwise

18kg ⋅ m2 3.6kg ⋅ m2

s

s

0.58 rev


loop the loop with a twist - university of icelandisa13/efni/einn/f.hl/edlisfraedi 1 v... · loop...

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