Physics 2210 Homework 18 Spring 2015

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Physics 2210 Homework 18 Spring 2015"

Transcription

1 Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30.The sphere has mass M = 8 kg and radius R = 0.19 m. The coefficient of static friction between the sphere and the plane is µ = What is the magnitude of the frictional force on the sphere? See Figure 1. Figure 1: Sphere Incline Solution II.Newton law gives F net = M a 1

2 The net force is composed of the three forces F net = N + F g + f Since the sphere is moving along the incline, only its parallel component will be non-zero, i.e. F net = Mg sin θ f On the other hand, for rotation we have Taking the center as origin, we get 1 No slipping condition requires Combining the above results 2 τ net = I α τ net = Rf a = Rα Mg sin θ f M = F net M = R τnet I = R 2 f (2/5)MR 2 Solving for f f = 2 Mg sin θ 7 f = N Gymnast Wording A gymnast with mass m 1 = 41 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m 2 = 108 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. See Figure 2. Figure 2: Gymnast 1 1 Only the frictional force gives non-zero contribution. 2 Moment of inertia for solid sphere is I = 2 5 MR2. 2

3 Figure 3: Gymnast 2 1. What is the force the left support exerts on the beam? 2. What is the force the right support exerts on the beam? 3. How much extra mass could the gymnast hold before the beam begins to tip? 4. Now the gymnast (not holding any additional mass) walks directly above the right support. What is the force the left support exerts on the beam? See Figure What is the force the right support exerts on the beam? 6. At what location does the gymnast need to stand to maximize the force on the right support? Solution Forces supports push up are F L and F R. Since the system is at rest, we got Since it s not rotating, it s true that 3 0 = F net = F L + F R m 1 g m 2 g 0 = τ net = L 6 F R L 6 F L + L 2 m 1g 1) Solving for the left force F L = 1 2 (4m 1 + m 2 )g F L = N 2) Solving for the right force F R = 1 2 (m 2 2m 1 )g F R = N 3 Taking the center of the beam as the origin. 3

4 3) The beam starts to tip when F R = 0, thus 0 = 1 2 (m 2 2m 1)g m 1 = m 2 /2 And so m 1 = m 1 m 1 = m 2 2 m 1 m 1 = 13 kg 4) Again we have 0 = F net = F R + F L m 1 g m 2 g Site of the force has changed, thus 4 0 = τ net = L 6 F R L 6 F L L 6 m 1g Thus the new left force F L = 1 2 (2m 1 + m 2 )g F L = N 5) Solving for the new right force F R = 1 2 m 2g F R = N 6) Hanging Beam Wording At the right edge of the beam A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of m b = 6.1 kg and the sign has a mass of m s = 16.8 kg. The length of the beam is L = 2.44 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = See Figure 4. 4 Origin at the beam s center. 4

5 Figure 4: Hanging Beam 1. What is the tension in the wire? 2. What is the net force the hinge exerts on the beam? 3. The maximum tension the wire can have without breaking is T = 977 N. What is the maximum mass sign that can be hung from the beam? 4. What else could be done in order to be able to hold a heavier sign? Solution Since the beam is not moving, we have In coordinates 5 0 = F net = F b + T + F s + F h 0 = τ net = τ b + τ t + τ s + τ h 0 = T F hx (1) 0 = F hy m s g m b g (2) 0 = L 2 m bg cos θ 2L 3 T sin θ + Lm sg cos θ (3) 1) Solving for tension(from the Eq. (3)) T = 3(m b + 2m s )g 4 tan θ T = N 5 Taking the position of the hinge as the origin. 5

6 2) Components of the hinge force are F hx = T F hy = (m s + m b )g And thus its magnitude F h = (F hx ) 2 + (F hy ) 2 = T 2 + (m s + m b ) 2 g 2 or 9(mb + 2m s ) F h = g 2 16 tan 2 + (m s + m b ) θ 2 F h = N 3) Turning around the formula for the tension m s,max = 2T max tan θ 3g m s,max = kg m b 2 4) while still keeping it horizontal, attach the wire to the end of the beam keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam attach the sign on the beam closer to the wall Meterstick Wording A meterstick (L = 1 m) has a mass of m = kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark. See Figure 5. Figure 5: Meterstick 1 1. What is the tension in the left string? 6

7 Figure 6: Meterstick 2 2. Now the right string is cut! What is the magnitude of the initial angular acceleration of the meterst about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) 3. What is the tension in the left string right after the right string is cut? 4. After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical? See Figure What is the magnitude of the acceleration of the center of mass of the meterstick when it is vertical? 6. What is the tension in the string when the meterstick is vertical? 7. Where is the angular acceleration of the meterstick a maximum? Solution 1) Since the stick is stable so 6 0 = F net = T L mg + T R 0 = τ net = (L/4)T R (L/4)T L Solving for the tension in the left string T L = 1 2 mg T L = N 2) The torque about the pivot is τ = lf sin φ = (L/4)mg 6 Origin is in the middle of the stick. 7

8 The second Newton law for rotation τ = Iα Moment of inertia of the rod about this pivot can be obtained by parallel axis theorem I = I 0 + md 2 = 1 12 ml2 + m(l/4) 2 = 7 12 ml2 So solving for the angular acceleration α = τ I = 12g 7L α = rad/s 2 3) At this moment, the acceleration of the center of mass is a = (L/4)α. So the second Newton law gives mg T L = F net = ma = m(l/4)α Solving for the tension T L = 4 7 mg T L = N 4) Using conservation of energy Solving for the angular speed Iω2 = mg L 4 mgl ω = = 2I 6g 7L ω = rad/s 5) Acceleration of the center of mass is completely centripetal 8 and hence a c = ω 2 l = 6g 7L L 4 a c = 3 14 g a c = m/s 2 7 Using the moment of inertia about the pivot - see above. 8 The radius of rotation being l = L/4. 8

9 6) Once again, II.Newton law tells us T mg = F net = ma c So the tension is and finally T = m(g + a c ) = m(g + (3/14)g) T = mg T = N 7) IE Sign Wording Right after the string is cut and the meterstick is still horizontal A sign has a mass of 1050 kg, a height h = 1 m, and a width W = 4 m. It is held by a light rod of length 5 m that is perpendicular to a rough wall. A guy wire at 23 to the horizontal holds the sign to the wall. Note that the distance from the left edge of the sign to the wall is 1 m. See Figure 7. Suppose we rely upon friction between the wall and the rod to hold up the sign (there is no hinge attaching the rod to the wall). What is the smallest value of the coefficient of friction µ such that the sign will remain in place? Figure 7: IE Sign 9

10 Solution The rod is not moving, so 0 = F net 0 = τ net The sign is exerting two equal 9 forces of (M/2)g at D and D W, where D is the length of the rod, while the wire pulls with tension T. Taking origin at the point of contact, our requirements translate into 10 0 = T sin θ (M/2)g (M/2)g + f 0 = N T cos θ 0 = D(M/2)g (D W )(M/2)g + DT sin θ And, of course, f = µn, where f is the force of friction and N is the normal force. Solving for the coefficient of static friction Ladder Wording µ = ( W 2D W µ = ) tan θ A ladder of length L = 2.6 m and mass m = 15 kg rests on a floor with coefficient of static friction µ s = Assume the wall is frictionless. See Figure 8. Figure 8: Ladder 1 1. What is the normal force the floor exerts on the ladder? 9 This can be explicitly shown, but but shoud be intuitively clear. 10 The x-axis points to the right, while the y-axis points upwards. That also makes the z-axis 10

11 Figure 9: Ladder 2 2. What is the minimum angle the ladder must make with the floor to not slip? 3. A person with mass M = 65 kg now stands at the very top of the ladder. What is the normal force the floor exerts on the ladder? See Figuer What is the minimum angle to keep the ladder from sliding? Solution There are four forces at play: Normal force from the floor : N 1 - pushing upwards Frictional force: f - pushing to the right Gravitational force (effectively) acting at the center of the ladder: mg - pointing down Normal force from the wall: N 2 - pushing to the left 1) For the ladder not to move, the net force needs to be zero, thus, in components 11 0 = f N 2 (4) 0 = N 1 mg (5) Thus we have N 1 = mg N 1 = N 11 The x-axis to the right, the y-axis upwards. 11

12 2) For the ladder not to rotate, the net torque needs to be zero 12, i.e. and so 0 = L 2 mg cos θ + LN 2 sin θ (6) tan θ = mg 2N 2 Using Eq. (4) with frictional force formula, we get N 2 = f = µn 1 = µmg so now ( ) 1 θ min = arctan 2µ θ min = ) At this, we introduced an additional force - weight of the person, which acts at the top end of the ladder. Thus, we need to modify the Eq. (5) to 0 = N 1 mg Mg which leads to N 1 = (m + M)g N 1 = N 4) Likewise, we need to modify Eq (6) to which yields while Eq. (4) is still applicable 0 = L 2 mg cos θ + LN 2 sin θ LMg cos θ tan θ = (m + 2M)g 2N 2 N 2 = f = µn 1 and thus θ min = arctan [( ) ] m + 2M 1 m + M 2µ θ min = With the origin at the bottom end of the ladder. 12

Exam 3 Practice Solutions

Exam 3 Practice Solutions Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at

More information

Physics 211 Week 10. Statics: Walking the Plank (Solution)

Physics 211 Week 10. Statics: Walking the Plank (Solution) Statics: Walking the Plank (Solution) A uniform horizontal beam 8 m long is attached by a frictionless pivot to a wall. A cable making an angle of 37 o, attached to the beam 5 m from the pivot point, supports

More information

Angular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion

Angular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for

More information

Equilibrium. For an object to remain in equilibrium, two conditions must be met. The object must have no net force: and no net torque:

Equilibrium. For an object to remain in equilibrium, two conditions must be met. The object must have no net force: and no net torque: Equilibrium For an object to remain in equilibrium, two conditions must be met. The object must have no net force: F v = 0 and no net torque: v τ = 0 Worksheet A uniform rod with a length L and a mass

More information

AP Physics Multiple Choice Practice Torque

AP Physics Multiple Choice Practice Torque AP Physics Multiple Choice Practice Torque 1. A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick? (A) 16 cm (B) 36 cm (C) 44

More information

Your Comments. That s the plan

Your Comments. That s the plan Your Comments I love physics as much as the next gal, but I was wondering. Why don't we get class off the day after an evening exam? What if the ladder has friction with the wall? Things were complicated

More information

Rotational N.2 nd Law

Rotational N.2 nd Law Lecture 0 Chapter 1 Physics I Rotational N. nd Law Torque Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi IN THIS CHAPTER, you will continue discussing rotational dynamics Today

More information

Rotational Kinetic Energy

Rotational Kinetic Energy Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body

More information

Physics 53 Exam 3 November 3, 2010 Dr. Alward

Physics 53 Exam 3 November 3, 2010 Dr. Alward 1. When the speed of a rear-drive car (a car that's driven forward by the rear wheels alone) is increasing on a horizontal road the direction of the frictional force on the tires is: A) forward for all

More information

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved: 8) roller coaster starts with a speed of 8.0 m/s at a point 45 m above the bottom of a dip (see figure). Neglecting friction, what will be the speed of the roller coaster at the top of the next slope,

More information

Chapter 10. Rotation

Chapter 10. Rotation Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised

More information

AP Physics C: Rotation II. (Torque and Rotational Dynamics, Rolling Motion) Problems

AP Physics C: Rotation II. (Torque and Rotational Dynamics, Rolling Motion) Problems AP Physics C: Rotation II (Torque and Rotational Dynamics, Rolling Motion) Problems 1980M3. A billiard ball has mass M, radius R, and moment of inertia about the center of mass I c = 2 MR²/5 The ball is

More information

= o + t = ot + ½ t 2 = o + 2

= o + t = ot + ½ t 2 = o + 2 Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the

More information

AP Physics. Harmonic Motion. Multiple Choice. Test E

AP Physics. Harmonic Motion. Multiple Choice. Test E AP Physics Harmonic Motion Multiple Choice Test E A 0.10-Kg block is attached to a spring, initially unstretched, of force constant k = 40 N m as shown below. The block is released from rest at t = 0 sec.

More information

Physics for Scientist and Engineers third edition Rotational Motion About a Fixed Axis Problems

Physics for Scientist and Engineers third edition Rotational Motion About a Fixed Axis Problems A particular bird s eye can just distinguish objects that subtend an angle no smaller than about 3 E -4 rad, A) How many degrees is this B) How small an object can the bird just distinguish when flying

More information

Equilibrium & Elasticity

Equilibrium & Elasticity PHYS 101 Previous Exam Problems CHAPTER 12 Equilibrium & Elasticity Static equilibrium Elasticity 1. A uniform steel bar of length 3.0 m and weight 20 N rests on two supports (A and B) at its ends. A block

More information

= y(x, t) =A cos (!t + kx)

= y(x, t) =A cos (!t + kx) A harmonic wave propagates horizontally along a taut string of length L = 8.0 m and mass M = 0.23 kg. The vertical displacement of the string along its length is given by y(x, t) = 0. m cos(.5 t + 0.8

More information

11-2 A General Method, and Rolling without Slipping

11-2 A General Method, and Rolling without Slipping 11-2 A General Method, and Rolling without Slipping Let s begin by summarizing a general method for analyzing situations involving Newton s Second Law for Rotation, such as the situation in Exploration

More information

= 2 5 MR2. I sphere = MR 2. I hoop = 1 2 MR2. I disk

= 2 5 MR2. I sphere = MR 2. I hoop = 1 2 MR2. I disk A sphere (green), a disk (blue), and a hoop (red0, each with mass M and radius R, all start from rest at the top of an inclined plane and roll to the bottom. Which object reaches the bottom first? (Use

More information

Chapter 6, Problem 18. Agenda. Rotational Inertia. Rotational Inertia. Calculating Moment of Inertia. Example: Hoop vs.

Chapter 6, Problem 18. Agenda. Rotational Inertia. Rotational Inertia. Calculating Moment of Inertia. Example: Hoop vs. Agenda Today: Homework quiz, moment of inertia and torque Thursday: Statics problems revisited, rolling motion Reading: Start Chapter 8 in the reading Have to cancel office hours today: will have extra

More information

Upthrust and Archimedes Principle

Upthrust and Archimedes Principle 1 Upthrust and Archimedes Principle Objects immersed in fluids, experience a force which tends to push them towards the surface of the liquid. This force is called upthrust and it depends on the density

More information

Physics 221. Exam III Spring f S While the cylinder is rolling up, the frictional force is and the cylinder is rotating

Physics 221. Exam III Spring f S While the cylinder is rolling up, the frictional force is and the cylinder is rotating Physics 1. Exam III Spring 003 The situation below refers to the next three questions: A solid cylinder of radius R and mass M with initial velocity v 0 rolls without slipping up the inclined plane. N

More information

t = g = 10 m/s 2 = 2 s T = 2π g

t = g = 10 m/s 2 = 2 s T = 2π g Annotated Answers to the 1984 AP Physics C Mechanics Multiple Choice 1. D. Torque is the rotational analogue of force; F net = ma corresponds to τ net = Iα. 2. C. The horizontal speed does not affect the

More information

Angular Speed and Angular Acceleration Relations between Angular and Linear Quantities

Angular Speed and Angular Acceleration Relations between Angular and Linear Quantities Angular Speed and Angular Acceleration Relations between Angular and Linear Quantities 1. The tires on a new compact car have a diameter of 2.0 ft and are warranted for 60 000 miles. (a) Determine the

More information

Name Student ID Score Last First. I = 2mR 2 /5 around the sphere s center of mass?

Name Student ID Score Last First. I = 2mR 2 /5 around the sphere s center of mass? NOTE: ignore air resistance in all Questions. In all Questions choose the answer that is the closest!! Question I. (15 pts) Rotation 1. (5 pts) A bowling ball that has an 11 cm radius and a 7.2 kg mass

More information

Rotational Motion. Rotational Motion. Rotational Motion

Rotational Motion. Rotational Motion. Rotational Motion I. Rotational Kinematics II. Rotational Dynamics (Netwton s Law for Rotation) III. Angular Momentum Conservation 1. Remember how Newton s Laws for translational motion were studied: 1. Kinematics (x =

More information

Chapter 12. Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx

Chapter 12. Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx Chapter 1 Lecture Notes Chapter 1 Oscillatory Motion Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = -kx When the mass is released, the spring will pull

More information

Chap. 10: Rotational Motion

Chap. 10: Rotational Motion Chap. 10: Rotational Motion I. Rotational Kinematics II. Rotational Dynamics - Newton s Law for Rotation III. Angular Momentum Conservation (Chap. 10) 1 Newton s Laws for Rotation n e t I 3 rd part [N

More information

UNIT 4 NEWTON S THIRD LAW, FORCE DIAGRAMS AND FORCES. Objectives. To understand and be able to apply Newton s Third Law

UNIT 4 NEWTON S THIRD LAW, FORCE DIAGRAMS AND FORCES. Objectives. To understand and be able to apply Newton s Third Law UNIT 4 NEWTON S THIRD LAW, FORCE DIAGRAMS AND FORCES Objectives To understand and be able to apply Newton s Third Law To be able to determine the object that is exerting a particular force To understand

More information

Summer Physics 41 Pretest. Shorty Shorts (2 pts ea): Circle the best answer. Show work if a calculation is required.

Summer Physics 41 Pretest. Shorty Shorts (2 pts ea): Circle the best answer. Show work if a calculation is required. Summer Physics 41 Pretest Name: Shorty Shorts (2 pts ea): Circle the best answer. Show work if a calculation is required. 1. An object hangs in equilibrium suspended by two identical ropes. Which rope

More information

Concept Question: Normal Force

Concept Question: Normal Force Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical

More information

AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force).

AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). 1981M1. A block of mass m, acted on by a force of magnitude F directed horizontally to the

More information

Chapter 8. Rotational Motion

Chapter 8. Rotational Motion Chapter 8 Rotational Motion Rotational Work and Energy W = Fs = s = rθ Frθ Consider the work done in rotating a wheel with a tangential force, F, by an angle θ. τ = Fr W =τθ Rotational Work and Energy

More information

PHYS 1303 Final Exam Example Questions

PHYS 1303 Final Exam Example Questions PHYS 1303 Final Exam Example Questions (In summer 2014 we have not covered questions 30-35,40,41) 1.Which quantity can be converted from the English system to the metric system by the conversion factor

More information

Friction is always opposite to the direction of motion.

Friction is always opposite to the direction of motion. 6. Forces and Motion-II Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:

More information

Use the following to answer question 1:

Use the following to answer question 1: Use the following to answer question 1: On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are uniformly accelerated for 21 seconds to

More information

Physics 6A Winter 2006 FINAL

Physics 6A Winter 2006 FINAL Physics 6A Winter 2006 FINAL The test has 16 multiple choice questions and 3 problems. Scoring: Question 1-16 Problem 1 Problem 2 Problem 3 55 points total 20 points 15 points 10 points Enter the solution

More information

Practice Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question.

Practice Test 3. Multiple Choice Identify the choice that best completes the statement or answers the question. Practice Test 3 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During

More information

General Physics (PHY 2130)

General Physics (PHY 2130) General Physics (PHY 130) Lecture 0 Rotational dynamics equilibrium nd Newton s Law for rotational motion rolling Exam II review http://www.physics.wayne.edu/~apetrov/phy130/ Lightning Review Last lecture:

More information

PHYSICS 231 Laws of motion PHY 231

PHYSICS 231 Laws of motion PHY 231 PHYSICS 231 Laws of motion 1 Newton s Laws First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was

More information

Lecture 4. Newton s 3rd law and Friction

Lecture 4. Newton s 3rd law and Friction Lecture 4 Newton s 3rd law and Friction Newtons First Law or Law of Inertia If no net external force is applied to an object, its velocity will remain constant ("inert"). OR A body cannot change its state

More information

Revolve, Rotate & Roll:

Revolve, Rotate & Roll: I. Warm-UP. Revolve, Rotate & Roll: Physics 203, Yaverbaum John Jay College of Criminal Justice, the CUNY Given g, the rate of free-fall acceleration near Earth s surface, and r, the radius of a VERTICAL

More information

PHYS 101 Previous Exam Problems. Force & Motion I

PHYS 101 Previous Exam Problems. Force & Motion I PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward

More information

Practice Test 3. Name: Date: ID: A. Multiple Choice Identify the choice that best completes the statement or answers the question.

Practice Test 3. Name: Date: ID: A. Multiple Choice Identify the choice that best completes the statement or answers the question. Name: Date: _ Practice Test 3 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A wheel rotates about a fixed axis with an initial angular velocity of 20

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) You are standing in a moving bus, facing forward, and you suddenly fall forward as the

More information

4) Vector = and vector = What is vector = +? A) B) C) D) E)

4) Vector = and vector = What is vector = +? A) B) C) D) E) 1) Suppose that an object is moving with constant nonzero acceleration. Which of the following is an accurate statement concerning its motion? A) In equal times its speed changes by equal amounts. B) In

More information

Description: Using conservation of energy, find the final velocity of a "yo yo" as it unwinds under the influence of gravity.

Description: Using conservation of energy, find the final velocity of a yo yo as it unwinds under the influence of gravity. Chapter 10 [ Edit ] Overview Summary View Diagnostics View Print View with Answers Chapter 10 Due: 11:59pm on Sunday, November 6, 2016 To understand how points are awarded, read the Grading Policy for

More information

第 1 頁, 共 7 頁 Chap10 1. Test Bank, Question 3 One revolution per minute is about: 0.0524 rad/s 0.105 rad/s 0.95 rad/s 1.57 rad/s 6.28 rad/s 2. *Chapter 10, Problem 8 The angular acceleration of a wheel

More information

PHY218 SPRING 2016 Review for Final Exam: Week 14 Final Review: Chapters 1-11, 13-14

PHY218 SPRING 2016 Review for Final Exam: Week 14 Final Review: Chapters 1-11, 13-14 Final Review: Chapters 1-11, 13-14 These are selected problems that you are to solve independently or in a team of 2-3 in order to better prepare for your Final Exam 1 Problem 1: Chasing a motorist This

More information

Lecture 11 - Advanced Rotational Dynamics

Lecture 11 - Advanced Rotational Dynamics Lecture 11 - Advanced Rotational Dynamics A Puzzle... A moldable blob of matter of mass M and uniform density is to be situated between the planes z = 0 and z = 1 so that the moment of inertia around the

More information

Version A (01) Question. Points

Version A (01) Question. Points Question Version A (01) Version B (02) 1 a a 3 2 a a 3 3 b a 3 4 a a 3 5 b b 3 6 b b 3 7 b b 3 8 a b 3 9 a a 3 10 b b 3 11 b b 8 12 e e 8 13 a a 4 14 c c 8 15 c c 8 16 a a 4 17 d d 8 18 d d 8 19 a a 4

More information

Static Equilibrium and Torque

Static Equilibrium and Torque 10.3 Static Equilibrium and Torque SECTION OUTCOMES Use vector analysis in two dimensions for systems involving static equilibrium and torques. Apply static torques to structures such as seesaws and bridges.

More information

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational

More information

Physics Mechanics. Lecture 11 Newton s Laws - part 2

Physics Mechanics. Lecture 11 Newton s Laws - part 2 Physics 170 - Mechanics Lecture 11 Newton s Laws - part 2 Newton s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: Newton s Second Law of

More information

4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight.

4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight. 1 1 wooden block of mass 0.60 kg is on a rough horizontal surface. force of 12 N is applied to the block and it accelerates at 4.0 m s 2. wooden block 4.0 m s 2 12 N hat is the magnitude of the frictional

More information

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003 FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003 NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 14 pages. Make sure none are missing 2. There is

More information

Dynamics of Rotational Motion: Rotational Inertia

Dynamics of Rotational Motion: Rotational Inertia Connexions module: m42179 1 Dynamics of Rotational Motion: Rotational Inertia OpenStax College This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License

More information

TutorBreeze.com 7. ROTATIONAL MOTION. 3. If the angular velocity of a spinning body points out of the page, then describe how is the body spinning?

TutorBreeze.com 7. ROTATIONAL MOTION. 3. If the angular velocity of a spinning body points out of the page, then describe how is the body spinning? 1. rpm is about rad/s. 7. ROTATIONAL MOTION 2. A wheel rotates with constant angular acceleration of π rad/s 2. During the time interval from t 1 to t 2, its angular displacement is π rad. At time t 2

More information

Chapter 12. Rotation of a Rigid Body

Chapter 12. Rotation of a Rigid Body Chapter 12. Rotation of a Rigid Body Not all motion can be described as that of a particle. Rotation requires the idea of an extended object. This diver is moving toward the water along a parabolic trajectory,

More information

Phys101 Lecture 5 Dynamics: Newton s Laws of Motion

Phys101 Lecture 5 Dynamics: Newton s Laws of Motion Phys101 Lecture 5 Dynamics: Newton s Laws of Motion Key points: Newton s second law is a vector equation Action and reaction are acting on different objects Free-Body Diagrams Ref: 4-1,2,3,4,5,6,7. Page

More information

Exam 2: Equation Summary

Exam 2: Equation Summary MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term 2012 Exam 2: Equation Summary Newton s Second Law: Force, Mass, Acceleration: Newton s Third Law: Center of Mass: Velocity

More information

In-Class Problems 30-32: Moment of Inertia, Torque, and Pendulum: Solutions

In-Class Problems 30-32: Moment of Inertia, Torque, and Pendulum: Solutions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 TEAL Fall Term 004 In-Class Problems 30-3: Moment of Inertia, Torque, and Pendulum: Solutions Problem 30 Moment of Inertia of a

More information

Physics x141 Practice Final Exam

Physics x141 Practice Final Exam Physics x141 Practice Final Exam Name: Partial credit will be awarded. However, you must show/explain your work. A correct answer without explanatory material will not receive full credit. Clearly indicate

More information

Chapter 4: Newton s Second Law F = m a. F = m a (4.2)

Chapter 4: Newton s Second Law F = m a. F = m a (4.2) Lecture 7: Newton s Laws and Their Applications 1 Chapter 4: Newton s Second Law F = m a First Law: The Law of Inertia An object at rest will remain at rest unless, until acted upon by an external force.

More information

Chapter 10 Practice Test

Chapter 10 Practice Test Chapter 10 Practice Test 1. At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration of 0.40 rad/s 2 has an angular velocity of 1.5 rad/s and an angular position of 2.3 rad. What

More information

Quiz Number 4 PHYSICS April 17, 2009

Quiz Number 4 PHYSICS April 17, 2009 Instructions Write your name, student ID and name of your TA instructor clearly on all sheets and fill your name and student ID on the bubble sheet. Solve all multiple choice questions. No penalty is given

More information

Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis

Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis... 2 21.1 Introduction... 2 21.2 Translational Equation

More information

UNIVERSITY OF TORONTO Faculty of Arts and Science

UNIVERSITY OF TORONTO Faculty of Arts and Science UNIVERSITY OF TORONTO Faculty of Arts and Science DECEMBER 2013 EXAMINATIONS PHY 151H1F Duration - 3 hours Attempt all questions. Each question is worth 10 points. Points for each part-question are shown

More information

Electric Force and Field Chapter Questions

Electric Force and Field Chapter Questions Electric Force and Field Chapter Questions 1. What happens to a plastic rod when it is rubbed with a piece of animal fur? What happens to the piece of fur? 2. How many types of electric charge are there?

More information

Concept of Force and Newton s Laws of Motion

Concept of Force and Newton s Laws of Motion Concept of Force and Newton s Laws of Motion 8.01 W02D2 Chapter 7 Newton s Laws of Motion, Sections 7.1-7.4 Chapter 8 Applications of Newton s Second Law, Sections 8.1-8.4.1 Announcements W02D3 Reading

More information

Work and kinetic Energy

Work and kinetic Energy Work and kinetic Energy Problem 66. M=4.5kg r = 0.05m I = 0.003kgm 2 Q: What is the velocity of mass m after it dropped a distance h? (No friction) h m=0.6kg mg Work and kinetic Energy Problem 66. M=4.5kg

More information

Newton s 3 Laws of Motion

Newton s 3 Laws of Motion Newton s 3 Laws of Motion 1. If F = 0 No change in motion 2. = ma Change in motion Fnet 3. F = F 1 on 2 2 on 1 Newton s First Law (Law of Inertia) An object will remain at rest or in a constant state of

More information

Physics 23 Exam 3 April 2, 2009

Physics 23 Exam 3 April 2, 2009 1. A string is tied to a doorknob 0.79 m from the hinge as shown in the figure. At the instant shown, the force applied to the string is 5.0 N. What is the torque on the door? A) 3.3 N m B) 2.2 N m C)

More information

A uniform rod of length L and Mass M is attached at one end to a frictionless pivot. If the rod is released from rest from the horizontal position,

A uniform rod of length L and Mass M is attached at one end to a frictionless pivot. If the rod is released from rest from the horizontal position, A dentist s drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of 2.51 10 4 rev/min. (a) Find the drill s angular acceleration. (b) Determine the angle (in radians)

More information

Physics for Scientists and Engineers. Chapter 6 Dynamics I: Motion Along a Line

Physics for Scientists and Engineers. Chapter 6 Dynamics I: Motion Along a Line Physics for Scientists and Engineers Chapter 6 Dynamics I: Motion Along a Line Spring, 008 Ho Jung Paik Applications of Newton s Law Objects can be modeled as particles Masses of strings or ropes are negligible

More information

Find the value of λ. (Total 9 marks)

Find the value of λ. (Total 9 marks) 1. A particle of mass 0.5 kg is attached to one end of a light elastic spring of natural length 0.9 m and modulus of elasticity λ newtons. The other end of the spring is attached to a fixed point O 3 on

More information

AP Physics 1 Review. On the axes below draw the horizontal force acting on this object as a function of time.

AP Physics 1 Review. On the axes below draw the horizontal force acting on this object as a function of time. P Physics Review. Shown is the velocity versus time graph for an object that is moving in one dimension under the (perhaps intermittent) action of a single horizontal force. Velocity, m/s Time, s On the

More information

Exercise Torque Magnitude Ranking Task. Part A

Exercise Torque Magnitude Ranking Task. Part A Exercise 10.2 Calculate the net torque about point O for the two forces applied as in the figure. The rod and both forces are in the plane of the page. Take positive torques to be counterclockwise. τ 28.0

More information

Fall 2007 RED Barcode Here Physics 105, sections 1 and 2 Please write your CID Colton

Fall 2007 RED Barcode Here Physics 105, sections 1 and 2 Please write your CID Colton Fall 007 RED Barcode Here Physics 105, sections 1 and Exam 3 Please write your CID Colton -3669 3 hour time limit. One 3 5 handwritten note card permitted (both sides). Calculators permitted. No books.

More information

University of Houston Mathematics Contest: Physics Exam 2017

University of Houston Mathematics Contest: Physics Exam 2017 Unless otherwise specified, please use g as the acceleration due to gravity at the surface of the earth. Vectors x, y, and z are unit vectors along x, y, and z, respectively. Let G be the universal gravitational

More information

Written Homework problems. Spring (taken from Giancoli, 4 th edition)

Written Homework problems. Spring (taken from Giancoli, 4 th edition) Written Homework problems. Spring 014. (taken from Giancoli, 4 th edition) HW1. Ch1. 19, 47 19. Determine the conversion factor between (a) km / h and mi / h, (b) m / s and ft / s, and (c) km / h and m

More information

3. A bicycle tire of radius 0.33 m and a mass 1.5 kg is rotating at 98.7 rad/s. What torque is necessary to stop the tire in 2.0 s?

3. A bicycle tire of radius 0.33 m and a mass 1.5 kg is rotating at 98.7 rad/s. What torque is necessary to stop the tire in 2.0 s? Practice 8A Torque 1. Find the torque produced by a 3.0 N force applied at an angle of 60.0 to a door 0.25 m from the hinge. What is the maximum torque this force could exert? 2. If the torque required

More information

Practice Problems from Chapters 11-13, for Midterm 2. Physics 11a Fall 2010

Practice Problems from Chapters 11-13, for Midterm 2. Physics 11a Fall 2010 Practice Problems from Chapters 11-13, for Midterm 2. Physics 11a Fall 2010 Chapter 11 1. The Ferris wheel shown below is turning at constant speed. Draw and label free-body diagrams showing the forces

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics IC-W09D3-2 Group Problem Person on a Ladder Solution

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics IC-W09D3-2 Group Problem Person on a Ladder Solution MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 IC-W09D3- Group Problem Person on a Ladder Solution A person of mass m p is standing on a rung, one third of the way up a ladder

More information

Slide 1 / 37. Rotational Motion

Slide 1 / 37. Rotational Motion Slide 1 / 37 Rotational Motion Slide 2 / 37 Angular Quantities An angle θ can be given by: where r is the radius and l is the arc length. This gives θ in radians. There are 360 in a circle or 2π radians.

More information

Page 1. Chapters 2, 3 (linear) 9 (rotational) Final Exam: Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272

Page 1. Chapters 2, 3 (linear) 9 (rotational) Final Exam: Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272 Final Exam: Wednesday, May 11, 10:05 am - 12:05 pm, BASCOM 272 The exam will cover chapters 1 14 The exam will have about 30 multiple choice questions Consultations hours the same as before. Another review

More information

Circular Motion & Gravitation FR Practice Problems

Circular Motion & Gravitation FR Practice Problems 1) A mass m is attached to a length L of string and hung straight strainght down from a pivot. Small vibrations at the pivot set the mass into circular motion, with the string making an angle θ with the

More information

Chapter 4. Oscillatory Motion. 4.1 The Important Stuff Simple Harmonic Motion

Chapter 4. Oscillatory Motion. 4.1 The Important Stuff Simple Harmonic Motion Chapter 4 Oscillatory Motion 4.1 The Important Stuff 4.1.1 Simple Harmonic Motion In this chapter we consider systems which have a motion which repeats itself in time, that is, it is periodic. In particular

More information

Lagrangian Dynamics: Generalized Coordinates and Forces

Lagrangian Dynamics: Generalized Coordinates and Forces Lecture Outline 1 2.003J/1.053J Dynamics and Control I, Spring 2007 Professor Sanjay Sarma 4/2/2007 Lecture 13 Lagrangian Dynamics: Generalized Coordinates and Forces Lecture Outline Solve one problem

More information

Physics A - PHY 2048C

Physics A - PHY 2048C Physics A - PHY 2048C and 11/15/2017 My Office Hours: Thursday 2:00-3:00 PM 212 Keen Building Warm-up Questions 1 Did you read Chapter 12 in the textbook on? 2 Must an object be rotating to have a moment

More information

Physics 6A Lab Experiment 6

Physics 6A Lab Experiment 6 Biceps Muscle Model Physics 6A Lab Experiment 6 APPARATUS Biceps model Large mass hanger with four 1-kg masses Small mass hanger for hand end of forearm bar with five 100-g masses Meter stick Centimeter

More information

Phys 106 Practice Problems Common Quiz 1 Spring 2003

Phys 106 Practice Problems Common Quiz 1 Spring 2003 Phys 106 Practice Problems Common Quiz 1 Spring 2003 1. For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed

More information

Rotational Motion. 1 Purpose. 2 Theory 2.1 Equation of Motion for a Rotating Rigid Body

Rotational Motion. 1 Purpose. 2 Theory 2.1 Equation of Motion for a Rotating Rigid Body Rotational Motion Equipment: Capstone, rotary motion sensor mounted on 80 cm rod and heavy duty bench clamp (PASCO ME-9472), string with loop at one end and small white bead at the other end (125 cm bead

More information

Concept of Force Challenge Problem Solutions

Concept of Force Challenge Problem Solutions Concept of Force Challenge Problem Solutions Problem 1: Force Applied to Two Blocks Two blocks sitting on a frictionless table are pushed from the left by a horizontal force F, as shown below. a) Draw

More information

1301W.600 Lecture 16. November 6, 2017

1301W.600 Lecture 16. November 6, 2017 1301W.600 Lecture 16 November 6, 2017 You are Cordially Invited to the Physics Open House Friday, November 17 th, 2017 4:30-8:00 PM Tate Hall, Room B20 Time to apply for a major? Consider Physics!! Program

More information

Chapter 4 Dynamics: Newton s Laws of Motion

Chapter 4 Dynamics: Newton s Laws of Motion Chapter 4 Dynamics: Newton s Laws of Motion Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal Force Applications

More information

Chapter 5 Newton s Laws of Motion. Copyright 2010 Pearson Education, Inc.

Chapter 5 Newton s Laws of Motion. Copyright 2010 Pearson Education, Inc. Chapter 5 Newton s Laws of Motion Force and Mass Units of Chapter 5 Newton s First Law of Motion Newton s Second Law of Motion Newton s Third Law of Motion The Vector Nature of Forces: Forces in Two Dimensions

More information

Simple and Physical Pendulums Challenge Problem Solutions

Simple and Physical Pendulums Challenge Problem Solutions Simple and Physical Pendulums Challenge Problem Solutions Problem 1 Solutions: For this problem, the answers to parts a) through d) will rely on an analysis of the pendulum motion. There are two conventional

More information

HATZIC SECONDARY SCHOOL

HATZIC SECONDARY SCHOOL HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT VECTOR DYNAMICS MULTIPLE CHOICE / 45 OPEN ENDED / 75 TOTAL / 120 NAME: 1. Unless acted on by an external net force, an object will stay at rest

More information

Name: AP Physics C: Kinematics Exam Date:

Name: AP Physics C: Kinematics Exam Date: Name: AP Physics C: Kinematics Exam Date: 1. An object slides off a roof 10 meters above the ground with an initial horizontal speed of 5 meters per second as shown above. The time between the object's

More information