Advanced Placement PHYSICS 1

Presenter 2014-2015

PHYSICS 1 Rotational Motion A

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AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material.

Copyright © 2013 National Math + Science Initiative®, Inc., Dallas, TX. All rights reserved.

Rotational Motion

What I Absolutely Have to Know to Survive the AP* Exam

The rotational kinematic equations are rotational relationships between the angular displacement, angular velocity, angular acceleration, and time that are only true when the angular acceleration is constant (i.e. when the angular acceleration is not a function of θ as one example). There exists an almost perfect parallel between translational and rotational motion. In a system in which there is both rotation and translation, you must include both rotational and translational kinetic energy in the same conservation of energy expressions. Linear and Angular analogs - variables

Linear Angular x Linear Distance (m) Rotational distance

(radians) θ

Δx Linear Displacement (m) Rotational displacement (radians)

Δθ

v Linear Velocity (m/s) Rotational velocity (radians/s)

ω

a Linear Acceleration (m/s2) Rotational acceleration (radians/s2)

α

m Mass (kg) Rotational inertia (kg.m2)

Ι

F Force (N) Torque (N.m) τ Rotational Inertia for a system of point masses = 2mrI Σ= Rotational Inertia for common objects (not necessary to memorize)

Solid Cylinder or Disc I = 1

2mr 2

Hoop about center axis 2mrI =

Solid Sphere 2

52mrI =

Torque Just as a non-zero net force causes a linear acceleration, a non-zero net torque will cause an angular acceleration. A torque can be thought of as a twist, just as a force is a push or pull. It is a torque that affects an object’s angular velocity. Torque is not energy, however and the units are mN or N.m, not Joules. Torque = τ = r⊥F = rF sinθ Where


Rotation I: Rotational Kinematics & Energy

F = force that is being applied to object (N) r = displacement from the point of rotation to the point of force application (m) θ = angle between the force vector and the displacement vector Relationships between the linear and angular variables when an object is rotating around a fixed axis or rolling without slipping.

Where r = the radius of the rotating object in meters aT = tangential acceleration in m/s2 aC = centripetal acceleration (also called radial acceleration) in m/s2

Key Formulas and Relationships

Linear and Angular analogs – kinematics and energy equations

Linear Angular Constant Motion

x = xo + vt θ = θo +ωt

Motion with Constant Acceleration

v = vo + at

x = 12

(vo + v)t

x = xo + vot +12

at2

vo2 = v2 + 2ax

ω =ω o +αt

θ = 12

(ω o +ω )t

θ = θo +ω ot +12αt2

ω o2 =ω 2 + 2αθ

2nd Law of Motion

a = Σ

F

m=Fnet

m

α = Σ

τI

=τ net

I

Kinetic Energy and Power Translational kinetic energy Rotational kinetic energy

2

21mvK = 2

21 ωIK =

Rolling object has both rotational and translational

rrvaa

rarvrx

RC

T

22

ω

αωθ

===

===



kinetic energies 22

21

21 mvIK += ω

Power Power

FvtxFP =Δ= τωθτ =Δ=

tP

Strategy on Rotation Problems

1. Draw a free-body diagram (FBD) for each object in the problem, showing the forces acting, the chosen coordinate system and the axes of rotation.

2. Write maF =Σ for each translating object and ατ I=Σ for each rotating object. 3. Relate the translational and rotational velocities and accelerations when applicable, like rolling

without slipping or a string on a pulley without slipping. 4. If mechanical energy is conserved, then conservation of energy methods provide a useful method of

calculating final linear and rotational velocities.

Multiple Choice

1. A bowling ball of mass M and radius R rolls without slipping down an inclined plane as shown above. The inclined plane forms an angle θ with the horizontal. The coefficient of static friction between the ball and the plane is µs. Which of the following diagrams correctly shows the forces acting on the object?



Resolve forces on an incline - FBD Direction of forces

c) correctly shows the forces. Weight always pulls an object down, a normal force is always perpendicular to the surface, and friction is always parallel to the surface. In this case, friction points up the ramp because it opposes the motion of the bottom of the ball, causing the ball to roll.

c

2. An automobile moves in a circle of radius 110 meters with a constant speed of 33 m/s. What is the angular velocity of the car about the center of the circle in radians per second?

a) 0.30 rad/s b) 0.30(π) rad/s c) 0.30(2 π) rad/s d) 110/33 rad/s

Relationship between the linear and angular variables when an object is rotating around a fixed axis.

Given and Unknownmrsmv

110/33

==

Equation rvrv

=

=

ω

ω

Solution 33 /100

0.30 rad/s

m sm

ω

ω

=

=

a



3. Which of the following statements are true about an object’s rotational inertia?

Select two answers. a. Rotational inertia is proportional to the object’s mass regardless of choice

of axis. b. Rotational inertia is inversely proportional to the object’s speed. c. Rotational inertia has the units of kg.m/s2. d. Rotational inertia depends on the choice of the axis of rotation.

Rotational Inertia for a solid (continuous mass distribution) object

I = Σmr 2 Rotational inertia depends on the choice of axis of rotation, r. Rotational inertia is proportional to the object’s mass regardless of choice of axis.

a,d

4. A wheel rotates through 10.0 radians in 2.5 seconds as it is brought to rest with a

constant angular acceleration. What was the initial angular velocity of the wheel before the braking began?

a) 0.25 rad/s b) 0.625 rad/s c) 2.0 rad/s d) 8.0 rad/s

Angular kinematic equations

sradsrad

t

t

o

o

o

/8

05.2

)10(2

2

)(21

0 =

−=

−=

+=

ω

ω

ωθω

ωωθ

d

5. A solid disc has a rotational inertia that is equal to I = ½ MR2, where M is the

disc’s mass and R is the disc’s radius. It is rolling along a horizontal surface with out slipping with a linear speed of v. How are the translational kinetic energy and the rotational kinetic energy of the disc related?

a) Rotational kinetic energy is equal to the translational. b) Translational kinetic energy is larger than rotational. c) Rotational kinetic energy is larger than translational. d) The answer depends on the density of the disc.



Calculation of mechanical energy problem

The ball will have both translational kinetic energy and rotational kinetic energy.

2

2

222

2

41_

21

21

21_

21_

MvKERotational

RvMRIKERotational

MvKEnalTranslatio

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛==

=

ω

b

6. The discs shown above are connected by a belt that rotates both discs at the same

time. The belt does not slip as it rotates the discs. The discs have different radii as shown, with R1 < R2. What is the relation between the angular velocities of the discs? a. 21 ωω = b. 2211 RR ωω =

c. 2

2

1

1

RRωω

=

d. 2

2

1

1

ωωRR

=

Relationships between the linear and angular variables when an object is rotating around a fixed axis

Both wheels have the same linear speed at their outside edges because both wheels are being rotated by the same belt that is moving without slipping.

2211

21

RRvv

ωω ==

b



7. A wrench has five equal forces V, W, X, Y, and Z applied to it. Which two forces produce the same clockwise torque?

a) V and W b) V and X c) V and Z d) W and X

Definition of torque

τ = r⊥F = rF sinθ Both V and X produce a clockwise torque that is equal in magnitude. V and Z produce a torque that is equal in magnitude, but Z produces a counterclockwise torque.

b

8. A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown above. Where should a third force of magnitude 5 newtons be applied so that the piece of plywood is in equilibrium?

a.

b.



c.

d.

Torque and static equilibrium

The net force must equal zero, so the possible answers are a and d. Answer d, however, would cause the plywood to rotate so the answer is a.

a

9. A door is free to rotate about its hinges. A force F applied normal to the door a distance x from the hinges produces an angular acceleration of α. What angular acceleration is produced if the same force is applied normal to the door at a distance of 2x from the hinges?

a) ¼ α b) ½ α c) α d) 2 α

2nd Law of Motion for rotational variables

ατ I=Σ and τ = Fr sinθ

)2()2( α

α

door

door

IxFthen

IFxif

=

=

d

10. A bowling ball is thrown down the bowling lane so that it is initially spinning with back-spin and sliding forward at the same time. As it moves, how does the force of friction affect the ball’s spin rate and the speed of the ball’s center of mass?

Spin rate Speed of center of mass a. spins faster decreases b. spins faster increases c. no change no change d. spins slower decreases 2nd Law of Motion for linear and

d



rotational variables

mF

a

maFmaF

f

f

−=

=−=Σ

Negative acceleration means the ball’s linear speed will slow down since it is opposite the ball’s linear velocity.

IRFIRF

I

f

f

−=

=−=Σ

α

αατ

Negative angular acceleration means the ball’s spin rate will slow down.

11. A uniform rigid bar of weight W is supported in a horizontal orientation as shown above by a rope that makes a 30° angle with the horizontal. The force exerted on the bar at point O, where it is pivoted, is best represented by a vector whose direction is which of the following?

a)

b) c)

d)

Statics and torque τ = r⊥F = rF sinθ

b



12. A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown above. The magnitude of the net torque on the system about the axis is

a) FR b) 2FR c) 5FR d) 14FR

Statics and torque

τ = r⊥F = rF sinθ

b



Free Response Question 1

2008M2 The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30° with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.

A. On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.

(4 points)

1 point for correctly drawing and labeling T, the tension in the cord 1 point for correctly drawing and labeling Mg, the weight of the rod 1 point for correctly drawing and labeling mg, the weight of the block 1 point for correctly drawing or labeling H, the force exerted on the rod by the hinge One earned point was deducted if one or more of the following were present: a correct vector not starting on the body, a component if the total force was also



shown, or any extraneous vectors

B. Calculate the reading on the spring scale. (4 points) The simplest method is to take the torque about the hinge, directly yielding an equation that can be solved for the tension in the cord which is equal to the reading on the spring scale. Στ = 0τ = r⊥F = rF sinθ

TLsin30°−mgL −MG L2

⎛⎝⎜

⎞⎠⎟= 0

T =mgL +Mg L

2⎛⎝⎜

⎞⎠⎟

Lsin30°

T =0.50kg( ) 9.8ms2

⎛⎝⎜

⎞⎠⎟0.60m( )+ 2.0kg( ) 9.8ms2

⎛⎝⎜

⎞⎠⎟0.30m( )

0.30mT = 29N

1 point for an indication that the sum of the torques is equal to zero 1 point for a correct expression for the torque exerted by the cord 1 point for a correct expression for both the torque due to the weight of the rod and the torque due to the weight of the hanging block 1 point for the correct answer including units and a reasonable number of significant digits

C. If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the rod-block system about the hinge. The rotational inertia of the rod-block system about the hinge is 0.42 kgm2. (3 points)

α = Σ

τI

=τ net

I

Στ = mgL+ Mg L

2

Στ = 0.50kg( ) 9.8 m

s2

⎛⎝⎜

⎞⎠⎟

0.60m( ) + 2.0kg( ) 9.8 ms2

⎛⎝⎜

⎞⎠⎟

0.30m( )Στ = 8.82N i mα = Σ

τI

= 8.82N i m0.42kgm2 = 21rad

s2

1 point for indicating that the sum of the torques is equal to Iα 1 point for the correct summation of the torques about the hinge due to the block and rod 1 point for the correct answer including units and a reasonable number of significant digits



Question 2 Guadalupe has a motorized globe on her desk that has a 0.16 m radius. She turns on the 4.25-watt motor and the globe begins to spin. The globe starts from rest and has a final rotational speed of 0.628 rad/s (1 revolution per 10 seconds). The globe reaches this speed in 25 seconds.

A. What is the angular displacement during the 25-second time interval in radians?

(2 points)

Given and Unknown:

?25

/628.00

====

θ

ωω

stsrad

o

Equation

to )(21 ωωθ +=

Solution

1 (0 0.628 / )(25 )27.85 rad

rad s sθ

θ

= +

=

1 point for correct rotational kinematics equation 1 point for correct answer in terms of the given quantities

B. What is the linear speed of a point on the globe’s equator?

(1 point) Given and

Unknownmr

srad16.0

/628.0==ω

Equation ωrv = Solution

(0.16 )(0.628 / )

m0.10 s

v m rad s

v

=

=

1 point for correct answer in terms of the given quantities



C. What is the average torque provided by the motor during the 25-second time interval?

(1 point) (a) Given and Unknown

P = 4.25WΔθ = 7.85radt = 25s

Equation P = τΔθ

t

Solution

τ = PtΔθ

τ = (4.25W )(25s)7.85rad

τ =13.5 Nim

1 point for correct answer in terms of the given quantities



Question 3 In the diagram below, two bodies of different masses (M1 and M2 =2M1) are connected by a string which passes over a pulley of negligible friction but has a rotational inertia

equivalent to a solid disc, 12

mr 2 . The pulley has a mass of ½ M1 and a radius of R.

A. On the diagram below, draw and label vectors to represent all the forces acting on M1 and M2.

(4 points)

1 point for correctly drawing and labeling T1, the tension in the cord 1 point for correctly drawing and labeling W1, the weight of M1 1 point for correctly drawing and labeling T2, the tension in the cord 1 point for correctly drawing and labeling W2, the weight of M2 One earned point is deducted for any extraneous vectors

M2 M1



B. Qualitatively describe the forces acting on M1 and M2 and indicate why they cause the pulley to rotate.

(2 points) There are two forces acting on each of the bodies: weight downward and the tension in the string upward. The tension is different on each side of the pulley since there must be a non-zero net torque to cause the pulley to begin to rotate.

1 point for correctly describing the forces 1 point for indicating that the tension is different on each side of the pulley

C. Describe how you could determine the magnitude of the acceleration of the system in terms of the given quantities and fundamental constants? Be sure to explicitly describe the calculations you would make, specifying all equations you would use (but do not actually do any algebra or arithmetic).

(4 points) Using Newton’s 2nd law, for each body sum the forces in the y dimension and set equal to may and/or the x dimension. For M1 the tension is positive and the weight is negative since the acceleration is upward.

aMgMTaMFT

aMF

g

y

111

111

1

=−

=−

=Σ

For M2 the tension is negative and the weight is positive since the acceleration is downward.

aMgMTaMFT

aMF

g

y

222

222

2

=+−

=+−

=Σ

1 point for discussing Newton’s second law for M1

1 point for discussing Newton’s second law for M2

1 point for discussing Newton’s second law for rotation for the pulley 1 point for substituting angular variables with linear/angular relationships and setting the torques equal to Iα



Next, Use Newton’s 2nd law for rotation for the pulley and make substitutes for angular variables with linear/angular relationships. Finally manipulate the equation and solve for the acceleration of the system. The following algebraic manipulation is for the edification of the presenter, but not required for the student.

( )aMTT

RaRMTTR

RaRMRTRT

IRTRTI

p

p

p

⎟⎠⎞⎜

⎝⎛=−

⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛=−

⎟⎠⎞⎜

⎝⎛⎟⎠⎞⎜

⎝⎛=−

=−=Σ

2121)(

21

12

212

212

12 αατ

( ) ( )

ga

aagag

aMaMgMaMgM

aMTT

aMgMTaMgMT

aMgMTaMgMT

134

4122

4122

2/121

2222

11111

112

112

112

111

111

=

=−−−

=−−−

⎟⎠⎞⎜

⎝⎛=−

−==+−

+==−

1 point for using the correct height of CM



Question 4

A solid sphere begins at rest and rolls down the incline and through a 2.0 m radius loop without slipping. The sphere has a mass of 5.0 kg and a radius of 0.125 m. When it is at the top of the loop, the ball has a linear speed of 5.45 m/s. A. Indicate what variable determines if the sphere completes one trip around the loop with the specified speed at the top of the loop and justify your answer qualitatively without using equations. (4 points) The height above the horizontal ground from which the ball starts rolling determines if the specified speed at the top of the loop is met. Consequently, this is a conservation of mechanical energy problem. First, choose where the gravitational potential energy is zero, Ug=0. In this case, let Ug=0 at the horizontal ground level. Use conservation of energy, realizing that the ball will have translational kinetic energy, rotational kinetic energy and gravitational potential energy at the top of the loop. The sphere must have this same amount of energy

1 point for the correct answer, the height above the ground 1 point for work that shows an understanding of Conservation of Mechanical Energy 1 point for work that shows an understanding that at the top of the loop the sphere has translational and rotational kinetic energies and gravitational potential energy 1 point for work that indicates the height that the sphere is released from determines



when released from rest at the top of the incline. Now make substitutions for angular variables with linear/angular relationships and combine like terms and solve for the unknown height.

B. Justify your answer about what variable determines if the sphere completes one trip around the loop with the specified speed at the top of the loop with appropriate equations.

The moment of inertia of a solid sphere is 2

52mrI = .

(4 points)

)2(21

21

_____

22 rRMgIMvMgH

loopoftopatenergyenergyinitial

−++=

=

ω

rRgvH

rRgvvgH

rRgvvMMgH

rRMgvMMvMgH

rRMgrvMrMvMgH

−+=

−++=

⎟⎠⎞⎜

⎝⎛ −++=

−+⎟⎠⎞⎜

⎝⎛+=

−+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞⎜

⎝⎛+=

2107

)2(51

21

)2(51

21

)2(52

21

21

)2(52

21

21

2

22

22

22

2

222

2

2

7 (5.45 / ) 2(2.0 ) (0.125 )10 9.8 /6.0 m

m sH m mm s

H

= + −

=

1 point for a statement of Conservation of Mechanical Energy 1 point for correct translational & rotational kinetic energies and potential energy 1 point for relationship between the linear and angular variables 1 point for correct answer and correct units with reasonable number of significant digits

TIPERS are provided to help students with concepts. They are not in the student study packet, only the presenter packet. Use as you see fit.

NT8A-CRT2: PULLEY AND WEIGHT—ANGULAR VELOCITY AND ACCELERATION GRAPHS A weight is tied to a rope that is wrapped around a pulley. The pulley is initially rotating counterclockwise and is pulling the weight up. The tension in the rope creates a torque on the pulley that opposes this rotation.

a) On the axes below, draw a graph of the angular velocity versus time for the period from the initial instant shown until the weight comes back down to the same height. Take the initial angular velocity as positive.

b) Draw a graph of the angular acceleration versus time for the same time period.

ω

Time

α

Time

Explain.

Answer: ω

Time

α

Time

The tension in the string resulting from the weight of the hanging block produces a constant torque on the pulley. So the pulley will rotate counterclockwise but slow down to stop at an instant, and then start rotating clockwise at an increasing rate. If we take the initial angular velocity as positive, then the angular acceleration has to be constant and negative.

ω

Axis ofrotation

v

NT8A-CRT4: ANGULAR VELOCITY VS. TIME GRAPH—ANGULAR ACCELERATION VS. TIME GRAPH

Sketch an angular acceleration versus time graph given the angular velocity versus time graph shown for the same time interval.

ω

Time

α

Time

Explain.

Answer: The slope of the line in the angular velocity graph tell us the angular acceleration. The first segment has a positive slope for two time units, then a zero slope for one time unit, a negative slope for one time unit, and finally a zero slope at zero velocity. The positive slope has a smaller magnitude than the negative slope, so the acceleration-time graph should look like the following:

ω

Time

α

Time

NT8B-RT8: RIGID 3-DIMENSIONAL POINT OBJECTS—MOMENT OF INERTIA ABOUT X-AXIS Shown below are six small brass and aluminum spheres connected by three stiff lightweight rods to form a rigid object. The rods are joined at their centers, are mutually perpendicular, and lie along the axes of the coordinate system shown. All spheres are the same distance from the connection point of the three rods at the origin of the coordinate axis. The brass spheres are shaded in the diagram and are identical. The aluminum spheres are identical, have less mass than the brass spheres, and are unshaded in the diagram. For this problem, treat the metal spheres as point masses and ignore the mass of the connecting rods.

y

x

zA

y

x

zB

y

x

zC

y

x

zE

y

x

zD

y

x

zF

Rank these objects on the basis of the moment of inertia about the x-axis.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The moment of inertia about the x-axis is the same but not zero for all these objects. ___

OR, The moment of inertia about the x-axis is zero for all these objects. ___

OR, The ranking for the moments of inertia about the x-axis cannot be determined. ___

Explain your reasoning.

Answer: F > C = D = E > A = B. The moment of inertia depends on the product of the mass and the square of the distance from that mass to the axis of rotation. For a rigid object, we can find the moment of inertia by adding this product together for all the mass in the object. Since the axis of rotation in this case is the x-axis, we can ignore all masses along this axis since the distance to them is zero. To rank the moments of inertia we need to rank the total mass that is not along the x-axis, since all of this mass is at the same distance. Case F has 4 brass spheres not on the x-axis, Cases C, D, and E each have two, and A and B have none.

NT8B-RT10: FLAT OBJECTS—MOMENT OF INERTIA PERPENDICULAR TO SURFACE Three flat objects (circular ring, circular disc, and square loop) have the same mass M and the same outer dimension (circular objects have diameters of 2R and the square loop has sides of 2R). The small circle at the center of each figure represents the axis of rotation for these objects. This axis of rotation passes through the center of mass and is perpendicular to the plane of the objects.

A Circularring

2R

B Circulardisc

2R

C Squareloop

2R

Rank the moment of inertia of these objects about this axis of rotation.

Greatest 1 _______ 2 _______ 3 _______ Least

OR, The moment of inertia of these objects is the same. ___

OR, We cannot determine the ranking for the moment of inertia of these objects. ___

Please explain your reasoning.

Answer: C > A > B, based on the distribution of mass. Mass farther from axis contributes more to the moment of inertia than mass closer to the axis. For the circular ring, all of the mass is at a distance R from the axis of rotation; and for the square loop almost all of the mass is at a distance that is greater than R. All of the mass of the disc is at a distance R or less.

NT8C-QRT14: THREE EQUAL FORCES APPLIED TO A RECTANGLE—TORQUE

Three forces of equal magnitude are applied to a 3-m by 2-m rectangle. Forces

F 1

and

F 2 act at 45° angles to the vertical as shown, while

F 3 acts horizontally.

a) Is the net torque about point A clockwise, counterclockwise, or zero?

Explain how you determined your answer.

Answer: Clockwise. Since the line of action for F2 is through point A, F2 creates no torque about point A. The line of action for force F3 passes through point B, 1 meter away from P. The shortest distance from A to the line of action for force F1 is 1.41 meters, the distance from point A to point C. Force F1 will produce a clockwise torque about point A, and force F3 will produce a counterclockwise torque about point A. Since all of the forces are equal, the force acting at the largest perpendicular distance

b) Is the net torque about point B clockwise, counterclockwise, or zero?


Answer: Zero. The line of action for F3 is through point B, so F3 creates no torque about point B. The line of action for forces F2 and F1 both pass a distance of 0.707 meters from point B, but one creates a clockwise torque about B and one a counterclockwise torque. So the torques due to these two forces will cancel, and the net torque about B is zero.

c) Is the net torque about point C clockwise, counterclockwise, or zero?


Answer Zero: The line of action for all forces pass through point C, so the torque due to each force is zero, and the net torque is also zero.

B

A

45° 45°

F3

F2F1

C

NT8C-CT15: FISHING ROD—WEIGHT OF TWO PIECES An angler balances a fishing rod on her finger as shown.

If she were to cut the rod along the dashed line, would the weight of the piece on the left-hand side be greater than, less than, or equal to the weight of the piece on the right-hand side?

Explain your reasoning.

Greater than. The net torque about the balancing point of the rod is zero, since the rod has no angular acceleration. The weight of the rod to the left of the balancing point creates a counterclockwise torque about that point, and the weight of the rod to the right creates a clockwise torque. The magnitudes of these torques must be equal for the net torque to be zero. The weight of the left side of the rod (the handle) acts at the center of mass of the left side, and the weight of the right side acts at the center of mass of the right side. Since the right side is longer than the left, the center of mass of the right side is further away from the dashed line than the center of mass of the left side. For the torques to be equal, the weight of the left side must be greater to compensate for the smaller perpendicular distance.

NT8C-RT18: SUSPENDED SIGNS—TORQUE The figures show six signs that are suspended from equal length rods on the side of a building. For each case, the mass of the sign compared to the mass of the rod is small and can be ignored. The mass of the sign is given in each figure. In cases B and D, the rod is horizontal; in the other cases, the angle that the rod makes with the vertical is given. The rope supporting the signs is 50 cm long in cases A, B, and C and 1 meter long in cases D, E, and F.

A B C

D E F

60°

80 kg100 kg

30°

60 kg

60°

70 kg

30°

50 kg

90 kg

50 cm 50 cm 50 cm

1 m 1 m1 m

Rank these situations on the basis of the magnitude of the torque the signs exert about the point at which the rod is attached to the side of the building.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The torque is the same but not zero for all these arrangements. ___

OR, The torque is zero for all these arrangements. ___

OR, We cannot determine the ranking for the torques in these arrangements. ___


Answer: B > D > C > E > F > A; the torque depends on the weight of the sign (mg) times the distance from the line of action of this weight to the point of attachment. The line of action of the weight is along the rope that is suspending the sign, which is one rod length (L) away from the attachment point in cases B and D, closer than that in cases C and E (0.866 L), and closer yet (0.5L) in cases A and F. Since A and F have the smallest masses and the smallest distances, they will be smallest with A smaller than F. Since B and D have the largest masses and the largest distances, they will have the largest torques, with B > D. C and E have intermediate distances and intermediate masses, with C > E.

NT8C-RT19: HEXAGON—TORQUE ABOUT CENTER Four forces act on a plywood hexagon as shown in the diagram. The sides of the hexagon each have a length of 1 meter.

4 N4 N

4 NCB

AD 6 N

Rank the magnitude of the torque applied about the center of the hexagon by each force.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ Least

OR, The magnitude of the torque due to each force is the same, but not zero. ___

OR, The magnitude of the torque due to each force is zero. ___

OR, We cannot determine the ranking of the magnitude of the torques. ___


Answer: A > C > B = D. The magnitude of the torque due to each force is equal to the magnitude of the force times the perpendicular distance between the line of action of that force and the pivot point. The lines of action of forces B and D pass through the center of the hexagon, so the torques due to forces B and D are both zero. The perpendicular distance between the line of action and the pivot point is equal to the height of one of the triangles shown for force C, and is equal to the side of a triangle for force A. The side of the triangle is longer than the height, and so the torque due to force A is greatest

NT8E-RT32: ROLLING OBJECTS RELEASED FROM REST—TIME DOWN RAMP Four objects are placed in a row at the same height near the top of a ramp and are released from rest at the same time. The objects are (A) a 1-kg solid sphere; (B) a 1-kg hollow sphere; (C) a 2-kg solid sphere; and (D) a 1-kg thin hoop. All four objects have the same diameter, and the hoop has a width that is one-quarter its diameter. The time it takes the objects to reach the finish line near the bottom of the ramp is recorded. The moment of inertia for an axis passing through its center of mass for a solid

sphere is 2

5MR2; for a hollow sphere it is

2

3MR2; and for

a hoop it is MR2 .

Rank the four objects from fastest (shortest time) down the ramp to slowest.

Fastest 1 _______ 2 _______ 3 _______4 _______ Slowest

OR, The time is the same for these objects. ___

OR, We cannot determine the ranking for the times for these objects. ___


Answer: A = C > B > D Each of these objects begins with gravitational potential energy at the top of the ramp that is comverted to kinetic energy at the bottom. The objects will have both translational and rotational KE at the bottom. Energy is conserved so we can set mgh = 1/2mv2 + 1/2Iω2. Assuming that all of the objects roll without slipping the translational KE will be proportional to the rotational KE. After some algebra we find that the masses cancel for each case and that the determining factor is the fraction in the moment of inertia for each object.

1-kg solid sphereA

2-kg solid sphereC1-kg hollow sphereB

1-kg hoopD

Start line

Finish line

NT8G-BCT45: HOOP ROLLING UP A RAMP—BASIC BAR CHART A thin hoop or ring with a radius of 2 m is moving so that its center of mass is initially moving at 20 m/s while also rolling without slipping at 10 rad/s along a horizontal surface. It rolls up an incline, coming to rest as shown below.

ω = 10 rad/s

ω = 0

v = 0

hv = 20 m/s

Complete the qualitative energy bar chart below for the earth-hoop system for the time between when the hoop is rolling on the horizontal surface and when it has rolled up the ramp and is momentarily at rest. Put the zero point for the gravitational potential energy at the height of the center of the hoop when it is rolling on the horizontal surface.

PEgrav PEspring

Translationalkinetic energy

PEgravGravitationalpotential energy

PEspringSpringpotential energy

WextWork done byexternal forces

Bar chart keyKErot PEgrav PEspring Wext

DuringInitial system energy Final system energy

Use g = 10 m/s2

for simplicity

0

KEtrans KErotKEtrans

KEtrans

Rotationalkinetic energyKErot

Explain.

In the initial situation there has to be rotational kinetic energy in additon to the translational kinetic energy. For the final situation the hoop is instantaneously at rest so there is no kinetic energy, but the center of mass is now above the zero level for the gravitational potential energy, so that is all of the final energy.

PEgrav PEspring

Translationalkinetic energy

PEgravGravitationalpotential energy

PEspringSpringpotential energy

WextWork done byexternal forces

Bar chart keyKErot PEgrav PEspring Wext

DuringInitial system energy Final system energy

Use g = 10 m/s2

for simplicity

0

KEtrans KErotKEtrans

KEtrans

Rotationalkinetic energyKErot

NT8G-RT55: OBJECTS MOVING DOWN RAMPS—SPEED AT BOTTOM In each case below, a 1-kg object is released from rest on a ramp at a height of 2 meters from the bottom. All of the spheres roll without slipping, and the blocks slide without friction. The ramps are identical in cases A, D, and F. The ramps in cases B, C, and E are identical and are not as steep as the others.

A B C

D

2 m

E F

Solid sphereSolid sphere

Hollow sphereHollow sphere

2 m2 m

2 m 2 m

2 m

Rank these cases on the basis of the speed of the objects when they reach the horizontal surface at the bottom of the ramp.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The maximum speed is the same for all cases. ___

OR, We cannot determine the ranking for the maximum speed of these objects. ___


E = F > D = C > A = B. Since all of the masses are the same a large speed corresponds to a large kinetic energy. At the bottom of the ramp all of the objects have converted all of their initial potential energy into kinetic energy. However the spheres will have converted some of the potential energy into rotational kinetic energy as well as translational kinetic energy, whereas all of the initial potential energy of the blocks will have been converted into translational kinetic energy. Since the height of the objects initially is all the same, all of the initial energyies are the same. The shape of the ramp does not matter. Since a hollow sphere has agreater moment of inertia than a solid sphere, the hollow spheres will have a greter fraction of their energy as rotational energy when they are rolling withour slipping.

NT8H-RT56: STATIONARY HORIZONTAL BOARDS—FORCE BY LEFT POST ON BOARD In each case below, uniform boards with different masses are supported by two vertical posts. Each board holds either a 10-kg box or a 30-kg box. The boards all have the same lengths. The masses of the boards and boxes as well as the distances from the left post to the center of mass of each box are given.

A B C

D E F80cm

30 kg

50 cm30 kg

50 cm10 kg

80 cm10 kg

mboard = 30 kg

100 cm10 kg

mboard = 30 kg mboard = 30 kg

100 cm30kg

mboard = 10 kg mboard = 30 kg mboard = 10 kg

Rank these situations on the basis of the force that the left post exerts on the board.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The force is the same but not zero for all these arrangements. ___

OR, The force is zero for all these arrangements. ___

OR, We cannot determine the ranking for the forces in these arrangements. ___


Answer: B > C > A > E > F > D; Each board is in equilibrium, so the net torque about any point must be zero. If we choose as our pivot point the right end of the board then the force the right post exerts acts at a zero distance from the pivot point and exerts no torque. The weight of the board and the weight of the box each create counterclockwise torques about this pivot and the sum of these two torques will be balanced by the torque exerted by the left post. Since those posts are all at the same distance from the pivot the force will be proportional to the sum of the two torques.

NT8H-RT57: TILTED PIVOTED RODS WITH VARIOUS LOADS—FORCE TO HOLD RODS Six identical massless rods are all supported by a fulcrum and are tilted at the same angle to the horizontal. A mass is suspended from the left end of the rod, and the rods are held motionless by a downward force on the right end. The locations of the rod with respect to the fulcrum and the values of the suspended masses are shown for each case. Each rod is marked at 1-meter intervals.

A

100 kg

B C

100 kg

F

100 kg

FF

D

200 kg

E F

200 kg

F

200 kg

FF

Rank the magnitude of the vertical force F applied to the end of the rod.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The vertical force applied to the end of the rod for all of the systems is the same. ___

OR, We cannot determine the ranking for the vertical force for these systems. ___


D > B > E > C > F > A All of the forces that act on the rod are vertical forces. Choosing the point of contact between the fulcrum and the rod as the pivot point, the perpendicular distance between the line of action of each force and the pivot is proportional to the distance along the rod from the pivot to that force. In cases A and F, the ratio of the distance between the applied force F and the pivot point and the distance from the pivot to the line of action of the force on the rod by the string attached to the mass is 3:1. In cases E and C, this ratio is 1:1, and in cases B and D, this ratio is 1:3. The net torque on each rod is zero (there is no rotational acceleration) so in each case the clockwise torque due to the applied force F must have the same magnitude as the counterclockwise torque due to the weight of the hanging mass (or the tension in the string). So the ratio of applied force to weight of the hanging mass must be the inverse of the distance ratios above. In cases A and F, the applied force must be one-third the weight of the hanging mass, in cases E and C, the applied force is equal to the weight, and in cases B and D it must be three times the weight.

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Advanced Placement PHYSICS 1