Exam 3 Practice Solutions


 Lauren Hodge
 2 years ago
 Views:
Transcription
1 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 the same instant, in which order will they reach the bottom? (a) disk, sphere, hoop (b) sphere, disk, hoop (c) hoop, disk, sphere (d) hoop, sphere, disk (e) sphere, hoop, disk At the top of the ramp, each object has the same gravitational potential energy. The gravitational potential energy is converted to translational and rotational kinetic energy. More of the energy will show up as rotational kinetic eneryg for objects with higher moments of inertia. Therefore, the objects with lower moments of inertia will reach the bottom first: sphere ( 2 5 MR2 ), disk ( 1 2 MR2 ), hoop (MR 2 ). 2. A student is sitting on a heavy stool with their feet off of the ground. The stool s seat is free to rotate. The student holds a spinning wheel with a heavy rim, as shown below. As viewed from the front, the student tilts the wheel to the right (the student tilts the wheel to their left). What happens? (a) the stool rotates clockwise as viewed from above When the student tips the wheel to the right, their is an angular momentum change of L as shown above. Since angular momentum is conserved, there must be a corresponding angular momentum change of L for the stool. This corresponds to a clockwise rotation when viewed from above. 1
2 3. The vectors shown below represent forces of equal magnitude applied to boxes of equal mass (shown in gray). All forces are either vertical, horizontal, or at a 45 angle. Which of the boxes are in translational and rotational equilibrium? (a) I only (b) II only (c) III only (d) I and II (e) I and III (f) II and III (g) I, II, and III Box II will rotate clockwise. Boxes I and III are in translation equilibrium: for every force vector, there is a force vector pointing in the opposite direction. Boxes I and III are also in rotational equilibrium: for every torque, there is an equal and opposite torque. 4. A child stands of mass m at the edge of a rotating disk of mass M (like a merrygoround). The disk rotates with angular velocity ω 0. Suppose that she starts walking opposite the direction of rotation so that she is at rest with respect to the ground. What is the final angular velocity of the disk? ) (a) ω 0 (1 2m M M (b) ω 0 m ( (c) ω m ) M ) (d) ω 0 (M 2m M (e) ω 0 m M Angular momentum must be conserved. The angular momentum before and after she starts walking are equal. Since she is walking so that she is motionless with respect to the ground, her angular velocity must be zero. She no longer carries any angular momentum, so the angular velocity of the 2
3 disk must increase so that it carries more angular momentum. L = L I disk ω 0 + I child ω 0 = I disk ω 1 2 MR2 ω 0 + mr 2 ω 0 = 1 2 MR2 ω 2 M ( 1 2 M + m ) ω 0 = ω ( 1 + 2m M ) ω 0 = ω 5. The moment of inertia of a disk of mass m and radius R about an axis through the edge of the disk and perpendicular to the disk s surface is (a) 1 2 mr2 (b) mr 2 (c) 3 2 mr2 (d) 2mR 2 (e) none of the above Use the parallel axis theorem to find the new moment of inertia. I = I cm + mh 2 I = 1 2 mr2 + mr 2 I = 3 2 mr2 6. A particle of mass m moves in the +î direction with velocity v as shown below. The ĵ direction is up and the ˆk direction is out of the page. The angular momentum of the particle about the point P is (a) mrv ˆk (b) mrv ˆk (c) mrv sin θ ˆk 3
4 (d) mrv sin θ ˆk (e) none of the above The angular momentum is L = r p. The vector r points from P to the particle, so the direction of r p is into the page, or ˆk. The magnitude of the cross product is rp sin θ = mrv sin θ. 7. A thin rod of mass M and length L spins clockwise with angular velocity ω about an axis that passes through one end of the rod, perpendicular to its length. An object of mass m moving to the left with velocity v strikes the rod at its midpoint and sticks to it. The rod stops spinning. What was v? (a) MLω 3m (b) MLω 6m (c) MLω 12m (d) 2MLω 3m (e) there is not enough information to solve this problem Angular momentum is conserved in the collision, so the angular momenta of the object and the rod must be equal and opposite. The angular momentum of the object is mvl 2, out of the page. The angular momentum of the rod is Iω 2 = 1 3 ML2 ω. mvl 2 = 1 3 ML2 ω v = 2MLω 3m 4
5 8. A nonuniform rod of length L is suspended horizontally by two cords as shown below. The left cord makes an angle of 30 with the vertical and the right cord makes an angle of 60 with the vertical. The rod is in equilibrium. Measuring from the left, where is the center of mass of the rod? (a) L 4 (b) 2L 3 (c) 3L 4 (d) 7L 8 (e) there is not enough information to solve this problem Since the rod is in equilibrium and shifted to the right, we expect that the center of mass will be closer to the right. The tension in the right string is T R, in the left string is T L, and the location of the center of mass is x. The rod is in translational equilibrium, so: Fx = 0 T L sin 60 + T R sin 30 = 0 (.866)T L + (.5)T R = 0 T L = (.577)T R and Fy = 0 Mg + T L cos60 + T R cos30 = 0 Mg + (.5)T L + (.866)T R = 0 Mg + (.5)(.577)T R + (.866)T R = 0 Mg + (1.155)T R = 0 T R = (.866)Mg The rod is also in rotational equilibrium. Setting the pivot point at the left end of the rod: τ = 0 Mgx (.866)T R L = 0 Mgx (.866)(.866)MgL = 0 x =.75L 5
6 9. A mass is attached to a cord which is wound about a pulley that is fixed in placed and free to rotate. If the mass is allowed to drop, which set of graphs best describes the angular displacement, angular velocity, and angular acceleration of the pulley? The horizontal axis on each graph is time, while the vertical axes are θ, ω, or α, as labeled. Angular acceleration is the derivative (with respect to time) of angular velocity, which is the derivative of angular displacement. Furthermore, for a falling mass, we expect a constant acceleration, a linearly increasing velocity, and a parabolic displacement. The only choice that correctly displays these characteristics is (b). 6
7 Problems 1. A cart of mass m (cart A) slides with velocity v to the right on a frictionless surface. Another cart (cart B), also of mass m is at rest a distance x 1 ahead of it. A distance x 2 beyond the second cart, a spring of spring constant k is attached to a wall. Assume that all collisions are perfectly elastic. What are the final velocities of carts A and B? Explain briefly. Cart A is moving to the right with velocity v, so it will collide with cart B. Since both carts have the same mass and the collision is elastic, they will simply exchange velocities. Cart A stops while cart B moves to the right with velocity v. Cart B then collides with and compresses the spring. Since the spring force is conservative, no energy is lost in the compression and subsequent expansion of the spring. Cart B is sent back to the left with the same velocity v. It then collides again with cart A. Once again, since the carts have the same mass and the collision is elastic, they simply exchange velocities. The final velocity of cart A is v to the left and cart B is at rest. 7
8 2. A block of mass m=1 kg is released from rest at the top of a frictionless inclined plane. The plane is h=5 m high at an angle of 45. When the block reaches the end of the plane, it transitions smoothly to a rough surface where friction cannot be neglected. The block slides a distance d before coming to a stop. (a) Sketch a graph that shows the block s potential energy, kinetic energy, and total mechanical energy as a function of position x (see figure). It is convenient to set the zero of gravitational potential energy to be the bottom of the ramp (that definition will be used for the remainder of this problem). The total mechanical energy, the potential energy, and the kinetic energy of the block are graphed below. The blocks initially has only gravitation potential energy. As it slides down the ramp, it gains kinetic energy but loses gravitational potential energy. When it reaches the bottom of the ramp, all of the gravitational potential energy has been converted into kinetic energy. While the block is on the ramp, the total mechanical energy remains constant because only conservative forces (gravity) are acting. When the block begins to slide horizontally, friction starts to do work. Friction is a nonconservative force, so the mechanical energy drops until the work done by friction has removed all of the mechanical energy from the system and converted it to heat. Thus, the mechanical energy drops to zero over some distance (and the kinetic energy, of course, which is equal to the mechanical energy). 8
9 (b) If the frictional force is a constant 5 N, what is d? When the block stops, it has zero mechanical energy. The initial mechanical energy was entirely gravitational potential energy. The mechanical energy was lost to work done by friction, or F F d, where F F is the force of friction and d is the distance the block slides on the horizontal surface. We can use conservation of total energy to determine how far the block slides: mgh = F F d d = mgh F F d = (1)(9.8)(5) 5 d = 9.8 m (c) Take a look at your graph. Is mechanical energy conserved? Why or why not? What about total energy? No, mechanical energy is not conserved. Mechanical energy is lost to work done by friction. Total energy is conserved, since the work done by friction is the same as the mechanical energy lost. Total energy is always conserved; unfortunately it is most often converted into unusable forms such as heat (via friction in this case). 9
10 3. A block of mass m=.306 kg is dropped from height h=1.33 m. Directly underneath the block is a spring in its equilibrium position with spring constant k=2 N/m. The spring has a massless platform that is attached to its top (the top of the spring with platform is even with the ground level). The block lands on and compresses the spring by a distance x. This is not a homework problem  feel free to round your answers as you go. Hint: The following relation may be useful: if ax 2 + bx + c = 0, then x = b± b 2 4ac 2a. (a) What is the speed of the block just before it hits the spring? Set the zero of gravitational potential energy as the ground level from which h is measured (and keep this definition throughout the problem). Mechanical energy is conserved in this system since we have only conservative forces (gravity and spring forces). The block starts out at rest with only gravitational potential energy mgh. When the block is about to hit the spring, the gravitational potential energy is entirely converted to kinetic energy: mgh = 1 2 mv2 v 2 = 2gh v = 2gh = 2(9.8)(1.33) v = 5.1 m/s (b) When the block returns to height h after bouncing off of the spring, what is its speed? Explain. Since mechanical energy is conserved, the block will always have the same mechanical energy. The block starts with total mechanical energy mgh and speed v=0. When the block returns to height h, it has gravitational potential energy mgh which is the same as the total mechanical energy. Thus, the speed of the block must be zero again. 10
11 (c) Write an expression for the conservation of energy when the spring is at maximum compression (in terms of the givens m, g, h, x, or k). At maximum compression, the spring is compressed a distance x by the block. Thus, the spring will have potential energy 1 2 kx2. The block stops moving at this point and has no kinetic energy. However, the block is now located a distance x below the ground level where we set the zero of gravitation potential energy. Thus, the block has gravitational potential energy mgx. The block started out with gravitational potential energy mgh. Conservation of energy says that E initial = E final mgh = 1 2 kx2 mgx (d) Find x, the maximum distance the spring is compressed. To find x, we need to solve the equation we determined in (c). This is a quadratic equation: 0 = 1 2 kx2 mgx mgh 0 = 1 2 (2)x2 (.306)(9.8)x (.306)(9.8)(1.33) 0 = x 2 3x 4 Using the quadratic formula with a = 1, b = 3, and c = 4, we obtain x = 3 ± x = 3 ± 5 2 x = 4 or 1 Since x is positive based on my definition of the gravitational potential energy, the solution is x=4 m. 11
12 4. Two identical spools of thread (solid cylinders with several windings of thread) are held the same height above the floor. The thread from spool A is tied to a support, while spool B is not connected to a support. The thread has negligible mass. Both spools are released from rest at the same instant. (a) Draw a freebody diagram for each spool corresponding to an instant after they are released, but before they hit the ground. Be sure to draw each force exactly where it acts. Compare the magnitude of all forces on your diagrams. The tension T < F g. (b) Which spool will reach the floor first? Explain how your answer is consistent with your freebody diagrams. Spool B will reach the ground first, as it has only one force acting on it, F g, in the downward direction. Spool A has two forces acting on it, one of which is in the upward direction, the tension T. Thus the downward acceleration of spool B will be larger. 12
13 (c) Will spool A strike the floor directly below the point where it was released, to the right of this point, or to the left of this point? Explain. Spool A will strike the floor directly below where it was released. There are no horizontal forces on spool A (it is in equilibrium in the horizontal direction) so there is no acceleration in the horizontal direction. (d) What is the ratio of the accelerations of the two spools ( a A ab )? The acceleration of spool B is simply a B = g, the acceleration due to gravity. For spool A, we need to find the acceleration by taking the tension into account. From Newton s second law for translational motion, we have T + F g = ma A where m is the mass of the spool. From Newton s second law for rotational motion, we have τ = I α. Working on the rotational motion first, we can relate the linear acceleration a A to the angular acceleration α by α = aa R where R is the radius of the spool. This is because the string constrains the rotation of the spool. τ = Iα TR = 1 2 mr2a A R T = 1 2 ma A Now we can substitute our expression for T into the equation for the translational motion. T mg = ma A 1 2 ma A mg = ma A 3 2 a A = g a A = 2 3 g Where we were careful to explicitly note the directions of tension, acceleration, and acceleration due to gravity in the first step. The ratio of the accelerations is thus a A a B = 2g 3g = 2 3 (e) What is the ratio of the velocities of the two spools just before each hits the floor ( v A vb )? The easiest way to solve this problem is using conservation of energy. Using the kinematic equations is challenging because each spool is in the air for a different amount of time. Both spools start with the same gravitational potential energy, mgh, where h is the height of the spool above the floor. This energy is shared between rotational and translational motion as the spool falls. Since spool B will not rotate, it will have a higher velocity. 1 2 mv2 B = mgh v B = 2gh 13
14 Spool A will rotate: 1 2 mv2 A Iω2 A = mgh 1 2 mv2 A mr2 ( v A R )2 = mgh 1 2 mv2 A mr2 ( v A R )2 = mgh 3 4 v2 A = mgh 4gh v A = 3 The ratio of the velocities is v A v B = 4gh (3)2gh =
Rolling, Torque & Angular Momentum
PHYS 101 Previous Exam Problems CHAPTER 11 Rolling, Torque & Angular Momentum Rolling motion Torque Angular momentum Conservation of angular momentum 1. A uniform hoop (ring) is rolling smoothly from the
More informationConcept 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 informationPotential Energy & Conservation of Energy
PHYS 101 Previous Exam Problems CHAPTER 8 Potential Energy & Conservation of Energy Potential energy Conservation of energy conservative forces Conservation of energy friction Conservation of energy external
More informationEndofChapter Exercises
EndofChapter Exercises Exercises 1 12 are conceptual questions that are designed to see if you have understood the main concepts of the chapter. 1. Figure 11.21 shows four different cases involving a
More informationReview questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.
Review questions Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.
More information11. (7 points: Choose up to 3 answers) What is the tension,!, in the string? a.! = 0.10 N b.! = 0.21 N c.! = 0.29 N d.! = N e.! = 0.
A harmonic wave propagates horizontally along a taut string of length! = 8.0 m and mass! = 0.23 kg. The vertical displacement of the string along its length is given by!!,! = 0.1!m cos 1.5!!! +!0.8!!,
More informationPHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011
PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a righthanded Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this
More informationRotation review packet. Name:
Rotation review packet. Name:. A pulley of mass m 1 =M and radius R is mounted on frictionless bearings about a fixed axis through O. A block of equal mass m =M, suspended by a cord wrapped around the
More informationA) 4.0 m/s B) 5.0 m/s C) 0 m/s D) 3.0 m/s E) 2.0 m/s. Ans: Q2.
Coordinator: Dr. W. AlBasheer Thursday, July 30, 2015 Page: 1 Q1. A constant force F ( 7.0ˆ i 2.0 ˆj ) N acts on a 2.0 kg block, initially at rest, on a frictionless horizontal surface. If the force causes
More informationAP 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= 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 informationPhysics I (Navitas) FINAL EXAM Fall 2015
95.141 Physics I (Navitas) FINAL EXAM Fall 2015 Name, Last Name First Name Student Identification Number: Write your name at the top of each page in the space provided. Answer all questions, beginning
More informationPHYSICS 220. Lecture 15. Textbook Sections Lecture 15 Purdue University, Physics 220 1
PHYSICS 220 Lecture 15 Angular Momentum Textbook Sections 9.3 9.6 Lecture 15 Purdue University, Physics 220 1 Last Lecture Overview Torque = Force that causes rotation τ = F r sin θ Work done by torque
More informationAP Physics 1: Rotational Motion & Dynamics: Problem Set
AP Physics 1: Rotational Motion & Dynamics: Problem Set I. Axis of Rotation and Angular Properties 1. How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? 2. How many degrees are
More informationIt will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points? A) I B) II C) III D) IV
AP Physics 1 Lesson 16 Homework Newton s First and Second Law of Rotational Motion Outcomes Define rotational inertia, torque, and center of gravity. State and explain Newton s first Law of Motion as it
More informationSolution 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 informationPY205N Spring The vectors a, b, and c. are related by c = a b. The diagram below that best illustrates this relationship is (a) I
PY205N Spring 2013 Final exam, practice version MODIFIED This practice exam is to help students prepare for the final exam to be given at the end of the semester. Please note that while problems on this
More informationI pt mass = mr 2 I sphere = (2/5) mr 2 I hoop = mr 2 I disk = (1/2) mr 2 I rod (center) = (1/12) ml 2 I rod (end) = (1/3) ml 2
Fall 008 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 informationYour Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon
1 Your Name: PHYSICS 101 MIDTERM October 26, 2006 2 hours Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon Problem Score 1 /13 2 /20 3 /20 4
More informationCHAPTER 8: ROTATIONAL OF RIGID BODY PHYSICS. 1. Define Torque
7 1. Define Torque 2. State the conditions for equilibrium of rigid body (Hint: 2 conditions) 3. Define angular displacement 4. Define average angular velocity 5. Define instantaneous angular velocity
More informationPH1104/PH114S MECHANICS
PH04/PH4S MECHANICS SEMESTER I EXAMINATION 0607 SOLUTION MULTIPLECHOICE QUESTIONS. (B) For freely falling bodies, the equation v = gh holds. v is proportional to h, therefore v v = h h = h h =.. (B).5i
More informationVersion 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 informationPHYSICS 149: Lecture 21
PHYSICS 149: Lecture 21 Chapter 8: Torque and Angular Momentum 8.2 Torque 8.4 Equilibrium Revisited 8.8 Angular Momentum Lecture 21 Purdue University, Physics 149 1 Midterm Exam 2 Wednesday, April 6, 6:30
More informationis acting on a body of mass m = 3.0 kg and changes its velocity from an initial
PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block
More informationPreAP Physics Review Problems
PreAP Physics Review Problems SECTION ONE: MULTIPLECHOICE QUESTIONS (50x2=100 points) 1. The graph above shows the velocity versus time for an object moving in a straight line. At what time after t =
More information31 ROTATIONAL KINEMATICS
31 ROTATIONAL KINEMATICS 1. Compare and contrast circular motion and rotation? Address the following Which involves an object and which involves a system? Does an object/system in circular motion have
More informationPHYSICS 221 SPRING EXAM 2: March 30, 2017; 8:15pm 10:15pm
PHYSICS 221 SPRING 2017 EXAM 2: March 30, 2017; 8:15pm 10:15pm Name (printed): Recitation Instructor: Section # Student ID# INSTRUCTIONS: This exam contains 25 multiplechoice questions plus 2 extra credit
More informationBig Ideas 3 & 5: Circular Motion and Rotation 1 AP Physics 1
Big Ideas 3 & 5: Circular Motion and Rotation 1 AP Physics 1 1. A 50kg boy and a 40kg girl sit on opposite ends of a 3meter seesaw. How far from the girl should the fulcrum be placed in order for the
More informationChapter 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 MFMcGrawPHY 45 Chap_10HaRotationRevised
More informationName SOLUTION Student ID Score Speed of blocks is is decreasing. Part III. [25 points] Two blocks move on a frictionless
Name SOLUTION Student ID Score last first Speed of blocks is is decreasing. Part III. [25 points] Two blocks move on a frictionless v o incline with initial speed v o, as shown, while a hand pushes with
More informationPhysics 53 Summer Final Exam. Solutions
Final Exam Solutions In questions or problems not requiring numerical answers, express the answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s
More informationTest 7 wersja angielska
Test 7 wersja angielska 7.1A One revolution is the same as: A) 1 rad B) 57 rad C) π/2 rad D) π rad E) 2π rad 7.2A. If a wheel turns with constant angular speed then: A) each point on its rim moves with
More informationRolling, Torque, Angular Momentum
Chapter 11 Rolling, Torque, Angular Momentum Copyright 11.2 Rolling as Translational and Rotation Combined Motion of Translation : i.e.motion along a straight line Motion of Rotation : rotation about a
More informationRotational motion problems
Rotational motion problems. (Massive pulley) Masses m and m 2 are connected by a string that runs over a pulley of radius R and moment of inertia I. Find the acceleration of the two masses, as well as
More informationPhysics 5A Final Review Solutions
Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone
More informationRotation Quiz II, review part A
Rotation Quiz II, review part A 1. A solid disk with a radius R rotates at a constant rate ω. Which of the following points has the greater angular velocity? A. A B. B C. C D. D E. All points have the
More informationAP Mechanics Summer Assignment
20122013 AP Mechanics Summer Assignment To be completed in summer Submit for grade in September Name: Date: Equations: Kinematics (For #1 and #2 questions: use following equations only. Need to show derivation
More information= o + t = ot + ½ t 2 = o + 2
Chapters 89 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 informationRotation. PHYS 101 Previous Exam Problems CHAPTER
PHYS 101 Previous Exam Problems CHAPTER 10 Rotation Rotational kinematics Rotational inertia (moment of inertia) Kinetic energy Torque Newton s 2 nd law Work, power & energy conservation 1. Assume that
More information= 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 informationSlide 1 / 133. Slide 2 / 133. Slide 3 / How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m?
1 How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? Slide 1 / 133 2 How many degrees are subtended by a 0.10 m arc of a circle of radius of 0.40 m? Slide 2 / 133 3 A ball rotates
More informationSlide 2 / 133. Slide 1 / 133. Slide 3 / 133. Slide 4 / 133. Slide 5 / 133. Slide 6 / 133
Slide 1 / 133 1 How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? Slide 2 / 133 2 How many degrees are subtended by a 0.10 m arc of a circle of radius of 0.40 m? Slide 3 / 133
More informationWebreview Torque and Rotation Practice Test
Please do not write on test. ID A Webreview  8.2 Torque and Rotation Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 0.30mradius automobile
More informationFALL 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 informationPhysics 2210 Homework 18 Spring 2015
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
More informationPhysics 121, Final Exam Do not turn the pages of the exam until you are instructed to do so.
, Final Exam Do not turn the pages of the exam until you are instructed to do so. You are responsible for reading the following rules carefully before beginning. Exam rules: You may use only a writing
More informationAP Physics Free Response Practice Oscillations
AP Physics Free Response Practice Oscillations 1975B7. A pendulum consists of a small object of mass m fastened to the end of an inextensible cord of length L. Initially, the pendulum is drawn aside through
More information1 MR SAMPLE EXAM 3 FALL 2013
SAMPLE EXAM 3 FALL 013 1. A merrygoround rotates from rest with an angular acceleration of 1.56 rad/s. How long does it take to rotate through the first rev? A) s B) 4 s C) 6 s D) 8 s E) 10 s. A wheel,
More informationSt. Joseph s AngloChinese School
Time allowed:.5 hours Take g = 0 ms  if necessary. St. Joseph s AngloChinese School 008 009 First Term Examination Form 6 ASL Physics Section A (40%) Answer ALL questions in this section. Write your
More informationPHYS 1303 Final Exam Example Questions
PHYS 1303 Final Exam Example Questions 1.Which quantity can be converted from the English system to the metric system by the conversion factor 5280 mi f 12 f in 2.54 cm 1 in 1 m 100 cm 1 3600 h? s a. feet
More informationQuiz 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 informationPhysics 218 Exam 3 Spring 2010, Sections
Physics 8 Exam 3 Spring 00, Sections 555 Do not fill out the information below until instructed to do so! Name Signature Student ID Email Section # Rules of the exam:. You have the full class period
More information= W Q H. ɛ = T H T C T H = = 0.20 = T C = T H (1 0.20) = = 320 K = 47 C
1. Four identical 0.18 kg masses are placed at the corners of a 4.0 x 3.0 m rectangle, and are held there by very light connecting rods which form the sides of the rectangle. What is the moment of inertia
More informationWhat is the initial velocity (magnitude and direction) of the CM? Ans: v CM (0) = ( 7 /2) v 0 ; tan 1 ( 3 /2) 41 above horizontal.
Reading: Systems of Particles, Rotations 1, 2. Key concepts: Center of mass, momentum, motion relative to CM, collisions; vector product, kinetic energy of rotation, moment of inertia; torque, rotational
More informationSolution to Problem. Part A. x m. x o = 0, y o = 0, t = 0. Part B m m. range
PRACTICE PROBLEMS: Final Exam, December 4 Monday, GYM, 6 to 9 PM Problem A Physics Professor did a daredevil stunt in his spare time. In the figure below he tries to cross a river from a 53 ramp at an
More informationPhys 270 Final Exam. Figure 1: Question 1
Phys 270 Final Exam Time limit: 120 minutes Each question worths 10 points. Constants: g = 9.8m/s 2, G = 6.67 10 11 Nm 2 kg 2. 1. (a) Figure 1 shows an object with moment of inertia I and mass m oscillating
More informationChapters 10 & 11: Rotational Dynamics Thursday March 8 th
Chapters 10 & 11: Rotational Dynamics Thursday March 8 th Review of rotational kinematics equations Review and more on rotational inertia Rolling motion as rotation and translation Rotational kinetic energy
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Common Quiz Mistakes / Practice for Final Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A ball is thrown directly upward and experiences
More informationAP Physics C: Work, Energy, and Power Practice
AP Physics C: Work, Energy, and Power Practice 1981M2. A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the seat and begins to swing
More informationRolling, Torque, and Angular Momentum
AP Physics C Rolling, Torque, and Angular Momentum Introduction: Rolling: In the last unit we studied the rotation of a rigid body about a fixed axis. We will now extend our study to include cases where
More informationThe net force on a moving object is suddenly reduced to zero. As a consequence, the object
The net force on a moving object is suddenly reduced to zero. As a consequence, the object (A) stops abruptly (B) stops during a short time interval (C) changes direction (D) continues at a constant velocity
More informationLecture 18. Newton s Laws
Agenda: l Review for exam Lecture 18 l Assignment: For Monday, Read chapter 14 Physics 207: Lecture 18, Pg 1 Newton s Laws Three blocks are connected on the table as shown. The table has a coefficient
More informationName: Date: Period: AP Physics C Rotational Motion HO19
1.) A wheel turns with constant acceleration 0.450 rad/s 2. (99) Rotational Motion H19 How much time does it take to reach an angular velocity of 8.00 rad/s, starting from rest? Through how many revolutions
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Monday, 14 December 2015, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 111, INTRODUCTORY PHYSICS I Monday, 14 December 015, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing. There
More informationExam 2 Solutions. PHY2048 Spring 2017
Exam Solutions. The figure shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes
More informationWeek 3 Homework  Solutions
University of Alabama Department of Physics and Astronomy PH 05 LeClair Summer 05 Week 3 Homework  Solutions Problems for 9 June (due 0 June). On a frictionless table, a mass m moving at speed v collides
More informationPHYS 154 Practice Test 3 Spring 2018
The actual test contains 1 multiple choice questions and 2 problems. However, for extra exercise, this practice test includes 4 problems. Questions: N.B. Make sure that you justify your answers explicitly
More informationPhys 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 informationChapter 10: Dynamics of Rotational Motion
Chapter 10: Dynamics of Rotational Motion What causes an angular acceleration? The effectiveness of a force at causing a rotation is called torque. QuickCheck 12.5 The four forces shown have the same strength.
More informationName (please print): UW ID# score last first
Name (please print): UW ID# score last first Question I. (20 pts) Projectile motion A ball of mass 0.3 kg is thrown at an angle of 30 o above the horizontal. Ignore air resistance. It hits the ground 100
More informationAngular Momentum. L r p. For a particle traveling with velocity v relative to a point O, the particle has an angular momentum
Angular Momentum Angular Momentum For a particle traveling with velocity v relative to a point O, the particle has an angular momentum y v θ L r p O r x z When is the angular momentum for a particle 0?
More informationPhysics 121, Sections 1 and 2, Winter 2011 Instructor: Scott Bergeson Exam #3 April 16 April 21, 2011 RULES FOR THIS TEST:
Physics 121, Sections 1 and 2, Winter 2011 Instructor: Scott Bergeson Exam #3 April 16 April 21, 2011 RULES FOR THIS TEST: This test is closed book. You may use a dictionary. You may use your own calculator
More informationUNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics
UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics Physics 111.6 MIDTERM TEST #2 November 16, 2000 Time: 90 minutes NAME: STUDENT NO.: (Last) Please Print (Given) LECTURE SECTION
More informationCentripetal acceleration ac = to2r Kinetic energy of rotation KE, = \lto2. Moment of inertia. / = mr2 Newton's second law for rotational motion t = la
The Language of Physics Angular displacement The angle that a body rotates through while in rotational motion (p. 241). Angular velocity The change in the angular displacement of a rotating body about
More information(a) On the dots below that represent the students, draw and label freebody diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
More informationPHY2020 Test 2 November 5, Name:
1 PHY2020 Test 2 November 5, 2014 Name: sin(30) = 1/2 cos(30) = 3/2 tan(30) = 3/3 sin(60) = 3/2 cos(60) = 1/2 tan(60) = 3 sin(45) = cos(45) = 2/2 tan(45) = 1 sin(37) = cos(53) = 0.6 cos(37) = sin(53) =
More informationRotational Kinematics and Dynamics. UCVTS AIT Physics
Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,
More informationPhysics 131: Lecture 21. Today s Agenda
Physics 131: Lecture 1 Today s Agenda Rotational dynamics Torque = I Angular Momentum Physics 01: Lecture 10, Pg 1 Newton s second law in rotation land Sum of the torques will equal the moment of inertia
More informationPhysics 201 Midterm Exam 3
Physics 201 Midterm Exam 3 Information and Instructions Student ID Number: Section Number: TA Name: Please fill in all the information above. Please write and bubble your Name and Student Id number on
More informationThe Laws of Motion. Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples
The Laws of Motion Newton s first law Force Mass Newton s second law Gravitational Force Newton s third law Examples Gravitational Force Gravitational force is a vector Expressed by Newton s Law of Universal
More informationAP Physics. Harmonic Motion. Multiple Choice. Test E
AP Physics Harmonic Motion Multiple Choice Test E A 0.10Kg 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 informationPlease circle the name of your instructor: EB01: Beamish EB02: Fenrich EB03: Ruhl. EB04: Rahman EB05: Nedie EB06: Ropchan LAST NAME: FIRST NAME: ID#:
Faculty of Engineering and Department of Physics ENPH 131 Final Examination Saturday, April 20, 2013; 2:00 pm 4:30 pm Universiade Pavilion Section EB01 (BEAMISH): Rows 1, 3, 5(seats 145) Section EB02
More informationPhysics 211 Spring 2014 Final Practice Exam
Physics 211 Spring 2014 Final Practice Exam This exam is closed book and notes. A formula sheet will be provided for you at the end of the final exam you can download a copy for the practice exam from
More informationName & Surname:... No:... Class: 11 /...
METU D. F. HIGH SCHOOL 20172018 ACADEMIC YEAR, 1 st SEMESTER GRADE 11 / PHYSICS REVIEW FOR GENERAL EXAM3 UNIFORMLY ACCELERATED MOTION IN TWO DIMENSIONS, ENERGY, IMPULSE & MOMENTUM & TORQUE DECEMBER 2017
More informationOn my honor, I have neither given nor received unauthorized aid on this examination.
Instructor(s): Profs. D. Reitze, H. Chan PHYSICS DEPARTMENT PHY 2053 Exam 2 April 2, 2009 Name (print, last first): Signature: On my honor, I have neither given nor received unauthorized aid on this examination.
More informationPhys101 Second Major173 Zero Version Coordinator: Dr. M. AlKuhaili Thursday, August 02, 2018 Page: 1. = 159 kw
Coordinator: Dr. M. AlKuhaili Thursday, August 2, 218 Page: 1 Q1. A car, of mass 23 kg, reaches a speed of 29. m/s in 6.1 s starting from rest. What is the average power used by the engine during the
More informationExam II. Spring 2004 Serway & Jewett, Chapters Fill in the bubble for the correct answer on the answer sheet. next to the number.
Agin/Meyer PART I: QUALITATIVE Exam II Spring 2004 Serway & Jewett, Chapters 610 Assigned Seat Number Fill in the bubble for the correct answer on the answer sheet. next to the number. NO PARTIAL CREDIT:
More informationWrite your name legibly on the top right hand corner of this paper
NAME Phys 631 Summer 2007 Quiz 2 Tuesday July 24, 2007 Instructor R. A. Lindgren 9:00 am 12:00 am Write your name legibly on the top right hand corner of this paper No Books or Notes allowed Calculator
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity
Chapter 8 Rotational Equilibrium and Rotational Dynamics 1. Torque 2. Torque and Equilibrium 3. Center of Mass and Center of Gravity 4. Torque and angular acceleration 5. Rotational Kinetic energy 6. Angular
More informationProblem Set x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology. 1. Moment of Inertia: Disc and Washer
8.01x Classical Mechanics, Fall 2016 Massachusetts Institute of Technology Problem Set 10 1. Moment of Inertia: Disc and Washer (a) A thin uniform disc of mass M and radius R is mounted on an axis passing
More informationRotational 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 informationAngular 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. d. v A v B. e. none of these.
General Physics I Exam 3  Chs. 7,8,9  Momentum, Rotation, Equilibrium Oct. 28, 2009 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show the formulas you use, the essential
More informationy(t) = y 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!
1. The angle between the vector! A = 3î! 2 ĵ! 5 ˆk and the positive y axis, in degrees, is closest to: A) 19 B) 71 C) 90 D) 109 E) 161 The dot product between the vector! A = 3î! 2 ĵ! 5 ˆk and the unit
More informationPhys 1401: General Physics I
1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS 2426 2. (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this?
More informationRotation and Translation Challenge Problems Problem 1:
Rotation and Translation Challenge Problems Problem 1: A drum A of mass m and radius R is suspended from a drum B also of mass m and radius R, which is free to rotate about its axis. The suspension is
More informationAP Physics II Summer Packet
Name: AP Physics II Summer Packet Date: Period: Complete this packet over the summer, it is to be turned it within the first week of school. Show all work were needed. Feel free to use additional scratch
More information1. An object is dropped from rest. Which of the five following graphs correctly represents its motion? The positive direction is taken to be downward.
Unless otherwise instructed, use g = 9.8 m/s 2 Rotational Inertia about an axis through com: Hoop about axis(radius=r, mass=m) : MR 2 Hoop about diameter (radius=r, mass=m): 1/2MR 2 Disk/solid cyllinder
More informationRotational Dynamics Smart Pulley
Rotational Dynamics Smart Pulley The motion of the flywheel of a steam engine, an airplane propeller, and any rotating wheel are examples of a very important type of motion called rotational motion. If
More information