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Serway and Vuille - College Physics AP 11/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 6 / 52

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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  • Instructions

    College Physics AP edition, 11th edition, by Raymond A. Serway and Chris Vuille, published by Cengage Learning helps students master physical concepts, improve their problem-solving skills, and enrich their understanding of the world around them. Serway/Vuille provides a consistent problem-solving strategy and an unparalleled array of worked examples to help students develop a true understanding of physics. The WebAssign component for this title engages students with immediate feedback, tutorials, videos, and an interactive eBook.

    Question 1 is an Active Example which guides students through the process needed to master a concept. A new "Review" question at the end provides a twist on the in-text Example to test student understanding.

    Question 2 is a Training Tutorial that offers students another training tool to assist them in understanding how to apply certain key concepts presented in a given chapter.

    Questions 3 is a Pre-Lecture Exploration, which combines Active Figures with conceptual and analytical questions that guide students to a deeper understanding and help promote a robust physical intuition.

    Question 4 is a problem with an optional Master It tutorial.

    Question 5 is a complete Master It tutorial.

    Question 6 is an end-of-chapter question with a hint button and detailed solution.

    Question 7 is an Interactive Video Vignette (IVV) question. IVVs encourage students to address their alternate conceptions outside of the classroom. They include online video analysis and interactive individual tutorials to address learning difficulties identified by PER (Physics Education Research).

    Question 8 has a Watch It video tutorial.

    Question 9 is a Conceptual Question (CQ).

    Questions 10 and 11 are AP Multiple-Choice Review Questions. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

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1. 1/4 points  |  Previous Answers SerCPAP11 8.AE.002. My Notes
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EXAMPLE 8.2 The Swinging Door
(a) Top view of a door being pushed by a 300-N force. (b) The components of the 300-N force.
Goal Apply the more general definition of torque.

Problem   (a)   A man applies a force of F = 3.00 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.   (b)   Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?

Strategy Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force.
SOLUTION
(a) Compute the torque due to the applied force exerted at 60.0°.
Substitute into the general torque equation.
τF = rFsin θ = (2.00 m)(3.00 ✕ 102 N) sin 60.0°
 = (2.00 m)(2.60 ✕ 102 N) = 5.20 102 N · m
(b) Calculate the force exerted by the wedge on the other side of the door.
Set the sum of the torques equal to zero.
τhinge + τwedge + τF = 0
The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite the upward 260 N component.
0 + Fwedge(1.50 m) sin (−90.0°) + 5.20 ✕ 102 N · m = 0
Fwedge = 347 N
LEARN MORE
Remarks Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque.

Question To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob?
     Correct: Your answer is correct.

PRACTICE IT
Use the worked example above to help you solve this problem.
(a) A man applies a force of F = 3.00 102 N at an angle of 60.0° to a door, x = 2.40 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.
Incorrect: Your answer is incorrect. seenKey

624
Your response differs from the correct answer by more than 10%. Double check your calculations. N · m

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?
Incorrect: Your answer is incorrect. seenKey

416
Your response differs from the correct answer by more than 10%. Double check your calculations. N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
A man ties one end of a strong rope 7.52 m long to the bumper of his truck, 0.545 m from the ground, and the other end to a vertical tree trunk at a height of 3.81 m. He uses the truck to create a tension of 8.76 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
Incorrect: Your answer is incorrect. seenKey

3010
Draw a picture of the situation. This will help you sort out the trig needed to find the component of the tension force causing the torque N · m
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Tutorial 8.2: Applying the Rotational Second Law The rotational analog of Newton's second law is
τ
 = Iα,
where all torques
τ,
 moments of inertia
I,
 and angular accelerations α, are computed around a fixed axis. This equation states that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. The moment of inertia
I
 is the rotational analog of translational mass.

For a collection of point particles, the moment of inertia is given by
I = 
mr2.
 For rigid objects of uniform composition, this table in the textbook lists moments of inertia about different axes.

Recall that the magnitude of torque generated by a force of magnitude
F
 is
τ = rF sin θ,
where, as before,
r
 is the length of the position vector,
F
 is the magnitude of the applied force, and θ is the angle between the vectors
r
 and
F.
 By convention, when an applied force causes an object to rotate counterclockwise, the torque on the object is positive; when the force causes the object to rotate clockwise, the torque on the object is negative.

The Guided Problem guide you through the following Problem-Solving Strategy.
•   Read the problem carefully at least once.
•   Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
•   Label each force in the picture.
•   Draw a free-body diagram of the object of interest.
•   Apply the rotational analog of Newton's second law.
•   Solve for the desired unknown quantity and substitute the numbers.
Guided Problem
A small sphere of mass
m = 2.60 kg
 is attached to the end of a very light, rigid rod of length
L = 1.22 m
 as shown in the figure. The mass and weight of the rod are negligible, and the sphere is small enough to be treated as a point particle. The left end of the rod is attached to the top of the pivot but free to rotate without friction around an axis perpendicular to the plane of the figure, and the sphere is held so the rod is horizontal and at rest. At the instant the sphere is released, determine its angular acceleration about an axis perpendicular to the plane of the figure and passing through the pivot point.
Part 1 of 6
Read the problem carefully at least once.
Be sure to notice which quantities are known and which must be found. The known quantities are the sphere's mass and the length of the rod. The rod's mass and weight are negligible.

The unknown quantity to be determined is the angular acceleration of the sphere at the instant it is dropped.

Identify all of the forces acting on the rod and the sphere shown in the figure.
     Correct: Your answer is correct.
Correct.
Part 2 of 6
Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
Together, the rod and the sphere form a system that is the object of interest. The figure below shows a picture of this system.
Label each force in the picture.
Each force in this figure is labeled in a way that brings to mind the associated physical quantity. The normal force
n
 is exerted by the pivot on the rod and
Fg
 is the gravity force acting on the sphere.
Part 3 of 6
Draw a free-body diagram of the object of interest.
Which of the following free-body diagrams for the sphere-rod system is correct?
     Correct: Your answer is correct.
Correct.
Part 4 of 6
Draw a free-body diagram of the object of interest. (cont.)
The correct FBD that describes all of the forces acting on the sphere-rod system is shown below.
Apply the rotational analog of Newton's second law.
Applying the rotational analog of Newton's second law means taking
τ
 = Iα,
 determining the axis of rotation, and writing out each term in the equation.

Consider the FBD above for the sphere-rod system. Which of the following is the correct substitution for the net torque
τ
 acting on the system? Remember that the rod is assumed to be massless. All torques are computed about an axis perpendicular to the screen and passing through the left end of the rod (the pivot). Positive torques tend to produce a counterclockwise rotation about this axis and negative torques tend to produce a clockwise rotation.
    



Additional Problems

Question 8.2a:
The figure shows a block of mass
m = 2.30 kg
 hanging from a thin rope wrapped around a solid, disk-shaped pulley. The pulley has mass
M = 7.10 kg,
 radius
R = 0.129 m,
 and is free to rotate on a frictionless, horizontal axle. If the block is released from rest, determine the magnitude of its downward acceleration.
Incorrect: Your answer is incorrect. seenKey

3.85
Incorrect. See m/s2

Question 8.2b:
A simple yo-yo is made by wrapping a thin string of negligible mass around the perimeter of a solid disk of radius
R = 0.105 m
 and mass
M = 2.53 kg.
 The free end of the string is held stationary by a person's hand, as in the figure, and the yo-yo is released from rest. Determine the magnitude of the disk's downward acceleration.
Incorrect: Your answer is incorrect. seenKey

6.53
Incorrect. See m/s2
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3. –/16 points SerCPAP11 8.PLE.002. My Notes
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–/16
 
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Prelecture Exploration: Energy Conservation for Rolling Objects
Darcel and Chandra are sitting in the park after physics class, and they notice some children rolling various objects down a slight incline in the sidewalk. Darcel is curious how the masses and shapes of the objects affect their motion, and Chandra says that it might be fun to think about the energy of motion associated with rolling objects. The friends decide to use the simulation to explore the motion and energies of various objects rolling down an incline.

They can click on an object (solid sphere, spherical shell, hoop, or cylinder) to position it at the top of the incline and click "roll" to release it. The translational and rotational speeds of the object and the time interval the object is on the incline are shown, together with a graph of the percentages of the different energy types as the object rolls down the incline. The "roll" button turns into a "pause" button while the object is in motion to allow them to examine different parameters at various points. For the object–Earth system in the simulation, the zero configuration of gravitational potential energy is defined to occur when the object is at the bottom of the incline. For the simulation, air resistance and any rolling friction effects are neglected, and the masses of all the objects are equal.

Click here to open the simulation in a new window.
Part 1 of 11 - Comparison of Similar Objects of Different Sizes
Darcel clicks on the large solid sphere, then clicks "roll." He pays attention to the time the sphere takes to roll down the incline. Then he clicks on the small solid sphere and repeats. Which of Darcell's statements about his observations of the time it takes for the two objects to reach the bottom is correct?
    
Part 2 of 11 - Comparison of Similar Objects of Different Sizes
Now Chandra clicks on the large hollow sphere, and clicks "roll." She pays attention to the time the sphere takes to roll down the incline. Then she clicks on the small hollow sphere and repeats. Which statement about her observations of the times is correct?
    
Part 3 of 11 - Comparison of Similar Objects with Different Mass Distributions
Now Darcel compares the times it takes for the large solid sphere and the large hollow sphere to roll down the incline. What can he say about his observations of the time it take for the two objects to reach the bottom?
    
Part 4 of 11 - Speeds of Objects at the Bottom of the Incline
The speed of the object at the bottom of the incline is equal to the translational speed once the object reaches the bottom. Which of Chandra's statements is supported by her observations?
    
Part 5 of 11 - Moments of Inertia
Darcel pulls out his physics book and opens it to the page that has a table of the moments of inertia for the objects in this simulation.
Object Solid Sphere Hollow Sphere Hoop Cylinder
Moment of Inertia
2
5
MR2
2
3
MR2
MR2
1
2
MR2
From these expressions for ICM, how can he describe how the mass distribution about the axis of rotation of an object relates to the moment of inertia?
    
Part 6 of 11 - Rolling Motion
With respect to rolling motion, neglecting any rolling friction, which of the following statements Chandra makes to Darcel is false?
    
Part 7 of 11 - Energy of the System
Darcel runs the simulation for the large version of each object. What can he say in general about the energies of the objects in the simulation as they roll down the incline? (Select all that apply.)

Part 8 of 11 - Rolling Motion
Chandra reminds Darcel that the rotational kinetic energy of a rolling object is equal to
KR = 
1
2
ICMω2,
 where I is the moment of inertia of the object, the translational kinetic energy is equal to
KT = 
1
2
MvCM2,
 and for a circular-shaped object that rolls without slipping, vCM = Rω. Which of the following is the conservation of energy equation that Darcel can write down for the object–Earth system?
    
Part 9 of 11 - Moment of Inertia and Speed
From their observations, Darcel and Chandra found that the speed of an object at the bottom of an incline depends on how the mass is distributed in the object. Now they consider the moments of inertia of the objects in the simulation, which are related to their mass distributions. Darcel asks Chandra what effect the moment of inertia has on the final speed. Which is the correct response?
    
Part 10 of 11 - Moment of Inertia and Speed
Darcel challenges Chandra to use the conservation of mechanical energy equation,
1
2
ICM
R2
 + M
vCM2 = Mgh,
 to write an expression for the speed of a rolling object at the bottom of an incline of vertical height, h. Which is the correct expression?
    
Part 11 of 11 - Analyze
Now Chandra and Darcel decide to try a problem.

Suppose that the height of the incline is h = 14.8 m. Find the speed at the bottom for each of the following objects.
solid sphere     (No Response) seenKey

14.4

 m/s
spherical shell     (No Response) seenKey

13.2

 m/s
hoop     (No Response) seenKey

12

 m/s
cylinder     (No Response) seenKey

13.9

 m/s

In a race, which object would win?
    
Going Further: Water and Ice
Darcel and Chandra decide to try a little experiment. They completely fill two identical cylindrical bottles with water. One is placed in a freezer until the water has turned to ice; then both bottles are placed on a short incline and allowed to roll to the bottom. Assume that the ice and bottle expand along the axis, so that the radii of both bottles are the same. Darcel asks Chandra to predict which will reach the bottom of the incline first, and to explain her reason. Which response is correct?
    
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4. –/3 points SerCPAP11 8.P.025.MI. My Notes
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A 600-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 150-N rod as indicated in the figure below. The left end of the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.)
(a) Find the (magnitude of the) tension T in the cable.
N

(b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions. Indicate the direction with the sign of your answer.)
horizontal component N
vertical component N

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5. –/11 points SerCPAP11 8.P.045.MI.SA. My Notes
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–/11
 
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Tutorial Exercise
A 212-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s?
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6. –/1 points SerCPAP11 8.P.016. My Notes
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Spectators watch a bicycle stunt rider travel off the end of a 52.5° ramp, rise to the top of his trajectory and, at that instant, suddenly push his bike horizontally away from him so that he falls vertically straight down, reaching the ground 0.600 s later. How far (in m) from the rider does the bicycle land if the rider has a mass
mrider = 75.0 kg
 and the bike has a mass
mbike = 13.0 kg?
 Neglect air resistance and assume the ground is level.
m

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Watch the video vignette below, and be sure to follow any instructions when prompted during the videos. Then answer the questions below about the concepts discussed in the video.
(a)
As you saw in the video, the two bullets in the experiment were fired from the same rifle. Assume the bullets also had the same mass and were fired from identically prepared cartridges. From this, what can we conclude about the initial momentum of the bullets just before they struck the blocks?
     Incorrect: Your answer is incorrect.
(b)
As you saw in the video, the two blocks had about the same mass, and both were initially at rest. Which of the following is true about the two bullet–block systems, just after the bullets collided with their respective blocks?
    
(c)
Which of the following statements provides correct reasoning for why both bullet–block systems reached the same maximum height?
    
(d)
Which of the following statements are true about the energy of the two bullet–block systems immediately after the collisions? (Select all that apply.)

(e)
Imagine comparing the two blocks after the experiment. In which block would we expect the bullet to be embedded the greatest distance?
     Incorrect: Your answer is incorrect.
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A uniform plank of length 2.00 m and mass 28.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a 720–N person is d = 0.500 m from the left end.
magnitude of
T1
     		N
magnitude of
T2
     		N
magnitude of
T3
     		N

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A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first?
     Correct: Your answer is correct.
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A lever is used as a manual brake for a large, spinning, 3.00 m diameter flywheel, as shown below. If a 150 N force is applied to the end of the handle and the coefficient of friction between the brake and the flywheel is 0.5, what stopping torque will be provided to the flywheel?
     Correct: Your answer is correct.


Solution or Explanation
(C). First, determine the force the lever delivers to the brake pad by recognizing the torques in the handle must be in equilibrium:
τ = τpush + τbreak pad = 0,
 or
Fpushrpush + Fbrake padrbrake pad = 0.
 Substituting,
Fbrake pad = 
(150 N)(1 m)
(0.25 m)
 = 600 N.
 Since the torque on the flywheel is due to the friction acting perpendicular to the radius of the wheel, the magnitude of the friction force must then be calculated using the brake pad force as the normal to the surface of the flywheel:
|Ff| ≤ μ|Fn|
 or
|Ff| ≤ (0.5)(600 N) = 300 N.
 Finally, the torque delivered to the flywheel is found with the following equation.
τwheel = Ff rwheel
τwheel = (300 N)(1.5 m) = 450 N · m
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Students wanted to investigate the changes in the angular momentum of an object and asked their teacher for some equipment. She gave them a wheel-and-axle, support rods, and clamps, and told them to get what else they needed from the standard equipment in the room.
They measured the radius of the largest wheel on the wheel-and-axle as 6.10 cm and set up the experiment to have the slotted masses drop a distance of 2.10 m from the center of the wheel-and-axle to the floor. Using 1.00 g slotted masses on the mass hanger, they obtained the data below.
Force
(10−2 N) 		Torque
(10−3 N · m) 		Time
(s) 		Linear
acceleration
(m/s2) 		Angular
acceleration
(rad/s2)
1.96 1.12 6.35 0.104 1.70
2.94 1.79 5.03 0.166 2.72
3.92 2.39 4.37 0.220 3.63
4.90 2.99 3.89 0.277 4.54
5.88 3.59 3.30 0.333 5.46
Using this data, they can (select two answers)
Correct: Your answer is correct.



Solution or Explanation
(A), (C). Answer (A) will show an increase in the angular acceleration of a body occurs and will indicate that the angular momentum increased for the wheel-and-axle but not give a mathematical answer. Answer (C) will give the students a mathematical answer.
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