Version 001 Rotational Motion ramadoss (171) 1

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1 Version 001 Rotational Motion ramadoss (171) 1 This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Please do the assignment in this distribution. You will get board credit for solving 2 of these the next day. ( 1 problem/person). I will choose the questions to be solved on the board HW for Sat and Sun. C on Monday 02/ HW for Mon. C ontuesday 02/ HWfor Tue. C onwednesday 02/ HW for Wed. C onthursday 02/ HW for Thu. C on Friday 02/ HW for Sat and Sun. C on Monday 02/12 Ang accel conversion points The angular velocity of a rotating wheel increases 2 revolutions per second every minute. The angular acceleration, in rad/s 2, of this wheel is: 1. 4π 2. π/15 correct 3. 2π 4. 4π /30 Ang vel of watch points The angular velocity in rad/s of the minute hand of a watch is: 1. π/1800 correct 2. π/ /π /π 5. π Holt SF 07A points A beetle sits at the top of a bicycle wheel and flies away just before it would be squashed. Assuming that the wheel turns clockwise, the beetle s angular displacement is π rad, which corresponds to an arc length of 1.2 m. What is the wheel s radius? Correct answer: m. Holt SF 07A 04short 004 (part 1 of 4) 10.0 points Consider the following table: θ s r a m 0.97 m rad b 8.9 m c 4.7 m 0.54 m m d What is the value of a? Correct answer: rad. 005 (part 2 of 4) 10.0 points What is the value of b? Correct answer: m. 006 (part 3 of 4) 10.0 points What is the value of c? Correct answer: (part 4 of 4) 10.0 points What is the value of d? Correct answer: m. Holt SF 07Rev points When a wheel is rotated through an angle of 34, a point on the circumference travels through an arc length of 3.7 m. When the wheel is rotated through angles of 34 rad and 34 rev, the same point travels through arc lengthsof 212 m and m, respectively. What is the radius of the wheel?

2 Version 001 Rotational Motion ramadoss (171) 2 Correct answer: m. Rotating Paper Disks points The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 94 cm apart on the same axle. From the angular displacement 48.6 of the two bullet holes in the disks and the rotational speed 1156 rev/min of the disks, we can determine the speed of the bullet. v t = C 7. Only at t = 0 8. t = C 9. t = 3C Spinning Disk (part 1 of 2) 10.0 points A disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and magnitude of the accelerationofabugclingingtotherimofthe disk? m/s and 1 m/s rev/min 94 cm What is the speed of the bullet? Correct answer: m/s. Zero Angular Accel points The position of a point on a rotating rigid bodyisgivenbyθ(t) = A+t 2 Ct 4,where A, and C are positive constants. Whatistheearliestt > 0atwhichthebody instantaneously has no angular acceleration? 1. t = 3 4C 2. 1 m/s and 0 m/s 2 (Disk spins at constant speed.) 3. 1 m/s and 10 m/s 2 correct m/s and 10 m/s (part 2 of 2) 10.0 points Let the turntable spin faster and faster, with constant angular acceleration α. Which sketch qualitatively shows the direction of the acceleration vector a of the bug? 1. a 2. t = 3 4C 3. t = 6C correct 4. t = C 5. t = 2 C 2. a correct

3 Version 001 Rotational Motion ramadoss (171) a a Correct answer: 7.7 rad. Holt SF 07D points The wheel on an upside-down bicycle moves through 18.3 rad in 5.10 s. What is the wheel s angular acceleration if its initial angular speed is 2.8 rad/s? Correct answer: rad/s 2. Spin Dry Cycle points The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.7 rev/s in 4.7 s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 17.5 s. During the whole process of starting up and stopping, through how many revolutions does the tub turn? Assume constant angular acceleration while it is starting and stopping. Correct answer: rev. ug Acceleration points A bug is on the rim of a 78 rev/min, 12 in. diameter record. The record moves from rest to its final angular speed in 3.37 s. Find the bug s centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm). Rotation Vector Direction points If the angular velocity vector of a spinning bodypointsoutfrom thefront ofapagethen, when you are looking at the front of the page, the body is spinning: 1. counterclockwise about an axis that is perpendicular to the page correct 2. clockwise about an axis that is parallel to the page 3. about an axis that is at 45 out of the page 4. counterclockwise about an axis that is parallel to the page 5. clockwise about an axis that is perpendicular to the page angular acceleration vector points Correct answer: m/s 2. Angular Position of a Wheel points Att = 0,awheelrotatingaboutafixedaxisat a constant angular deceleration of0.84 rad/s 2 has an angular velocity of 2.4 rad/s and an angular position of 6.2 rad. What is the angular position of the wheel after 5 s?

Version 001 Rotational Motion ramadoss (171) 4 4. None of these 5. Spinning Disk 020 10.

4 Version 001 Rotational Motion ramadoss (171) 4 4. None of these 5. Spinning Disk points A disk is spinning about an axis through its center of mass, and at a given moment has an angular velocity vector ω as shown. ω If the disk is slowing down, in what direction is its angular acceleration vector α relative to ω? 1. α at right angles (90 ) to ω. 2. Not enough information is provided. 3. α is in the same direction as ω. Angular Velocity points Seen from outer space, every point on the surface of the earth moves from West to East (which, of course, is why the sun appears to rise in the East and set in the West). C A D Which vector represents the angular velocity ω of the planet? 1. D 2. C 3. A correct 4. α is in the opposite direction to that shown for ω. correct Ang acc to lin acc points A particle moves in a circular path of radius 0.1 m with a constant angular velocity of 5 rev/s. The acceleration of the particle is: 1. (1000 m/s 2 )π 2 2. (10 m/s 2 )π 2 correct 3. (0.1 m/s 2 )π 4. (500 m/s 2 )π m/s 2 eetle on a Pendulum

5 Version 001 Rotational Motion ramadoss (171) (part 1 of 3) 10.0 points A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 172 cm long. If the beetle rides through a swing of 40, how far has he traveled along the path of the pendulum? Correct answer: cm. 023 (part 2 of 3) 10.0 points What is the displacement experienced by the beetle while moving theough the same angle 40? Correct answer: cm. 024 (part 3 of 3) 10.0 points If the pendulum at some instant is swinging at 1.8 rad/s, how fast is the beetle traveling? Correct answer: cm/s. ond in N points The length of each of the bonds between the atoms of a molecule of nitrogen (N 2 ) is nm. The mass of a nitrogen atom is pg. Determine the moment of inertia of the molecule about an axis passing through the center of mass perpendicular to the line joining the two atoms. Correct answer: pg nm 2. Disk Inertia points A non-uniform disk of mass M and radius R has its mass distributed in such a way that the mass per unit area is a function of the radial distance r from the center of the disk: σ(r) = br, where b is a constant to be determined. What is the rotational inertia of this disk about an axis through the center of mass and perpendicular to the plane of the disk? The area differential can be written as a ring of radius r and thickness dr: da = 2πrdr M R2 2. M R M R M R M R M R M R2 correct M R2 Four Point Masses points Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible mass. The distance between any two adjacent masses is l. l l l m m m m What is the rotational inertia I cm of this system about its center of mass? 1. 5ml 2 correct 2. 4ml ml ml ml2 6. 2ml ml 2 8. ml 2

Version 001 Rotational Motion ramadoss (171) 6 Hinged Rod With Mass 07 028 10.0 points Consider a thin rod of length L which is pivoted at one end.

6 Version 001 Rotational Motion ramadoss (171) 6 Hinged Rod With Mass points Consider a thin rod of length L which is pivoted at one end. A uniformly dense spherical object (with mass m and radius r = 1 ) 5 L is attached to the free end of the rod. The moment of inertia of the rod about an end is I rod = 1 3 ml2 and the moment of inertia of the sphere about its center of mass is I sphere = 2 5 mr2. C L m θ r= 1 5 L m Find the moment of inertia of the rod plus mass system with respect to the pivot point. The acceleration of gravity is 9.8 m/s I = ml2 2. I = ml2 3. I = ml2 correct 4. I = ml2 5. I = ml2 6. None of these moment of inertia (concept) points Inertia of Disk Ring Square points Consider three objects of equal masses but different shapes: a solid disk, a thin ring, and a thinhollow square. Thering and thesquare are hollow and their perimeters carry all the mass, but the disk is solid and has uniform mass density over its whole area. disk ring square R R 2R Compare the three objects moments of inertia when rotated around their respective centers of mass. 1. I cm disk > Icm ring > Icm square 2. I cm square > Icm ring > Icm disk correct 3. I cm ring > Icm square > Icm disk 4. I cm square > Icm disk > Icm ring

7 Version 001 Rotational Motion ramadoss (171) 7 5. I cm disk > Icm square > I cm ring 6. I cm ring > Icm disk > Icm square Spoked Wheel 031 (part 1 of 2) 10.0 points A wheel is formed from a hoop of mass 4.3 kg and five equally spaced spokes, each of mass kg. The hoop s radius is the length 0.22 m of each spoke kg 4.3 kg 0.22 m Find the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel. Correct answer: kg m (part 2 of 2) 10.0 points Determine the moment of inertia of the wheel about an axis through its rim and perpendicular to the plane of the wheel. Correct answer: kg m 2. AP 1998 MC points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force F is applied at the other end, at an angle θ to the rod. If F were to be applied perpendicular to the rod, at what distance d from the axis L m θ F of rotation should it be applied in order to produce the same torque τ? 1. d = L cosθ 2. d = 2 L 3. d = L 4. d = L tanθ 5. d = L sinθ correct AP M 1998 MC points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels, as shown below. 2F F 3R 2R What isthemagnitude ofthenet torqueon the system about the axis? 1. τ = F R 2. τ = 14F R 3. τ = 2F R correct 4. τ = 0 5. τ = 5F R eam Instability points A non-uniform beam with a mass 3M and length L is in stable equilibrium when placed at the edge of a table with half its length F F

8 Version 001 Rotational Motion ramadoss (171) 8 sticking off the table edge. ut when a pointlikemassm isplacedatthefarendofthepart ofthebeam off thetable, the beam isbrought to unstable equilibrium; i.e., it is right on the verge of tipping off the table. 3M M At what distance to the left of the table edgeisthecenterofmassofthebeamlocated? 1. L 6 correct 2. 4L 5 3. L L L 5 6. L 7 7. L 4 8. L L L 4 Net Torque on a Wheel points A circular-shaped object of mass 15 kg has an inner radius of 11 cm and an outer radius of 30 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 26 N, and 15 N act on the object, as shown. The force of magnitude 26 N acts 33 below the horizontal. L 2 12 N 33 ω 26 N 15 N Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 5.24 N m. Problems (part 1 of 2) 10.0 points To tighten a bolt, you push with a force of 60 N at the end of a wrench handle that is 0.22 m from the axis of the bolt. What torque are you exerting? Correct answer: 13.2 N m. 038 (part 2 of 2) 10.0 points If you move your hand inward to be only 0.13 m from the bolt, what force do you have to exert to achieve the same torque? Correct answer: N. PS points It is specified that a certain nut be tightened to a torque of 51 N m. If the mechanic is using a 61 cm long wrench, how much force must he apply to the end of the wrench to meet specs? Correct answer: N. PS points A mother anddaughter areonaseesaw inthe park. How far from the center must the 117 lb mothersitinordertobalancethe47lbdaughter sitting 5 ft from the center?

9 Version 001 Rotational Motion ramadoss (171) 9 Correct answer: ft. Swinging Rod points A uniform, rigid rod of mass M and length L is pivoted frictionlessly at its upper end. If the rod is dropped from an initially horizontal position, it swings freely through a vertical position, as shown. D In which of the four positions illustrated is the net torque about the pivot on the rod greatest, and in which of the four positions is it the least? C 1. Greatest at D; least (zero) at A 2. The same in all four positions, since the force M g acts with lever arm L 2 positions. A in all four 3. Greatest at A; least (zero) at D correct 4. Not enough information is provided. Wheel Torque points Four forces of the same magnitude but differing directions act at and tangent to the rim of a uniform wheel free to spin about its center of mass (CM). Which statement about the wheel is correct? 1. Neither the net force on the wheel, nor the net torque on the wheel about the CM is zero. 2. None of these 3. The net force on the wheel is zero but the net torque about the CM is not zero. 4. The net force on the wheel and the net torque on the wheel about the CM are zero. 5. The net force on the wheel is not zero but thenettorqueaboutthecmiszero. correct Cable Over a Pulley points A cable passes over a pulley. ecause the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 127 N, and the force on the other side is 88 N. Assuming that the pulley is a uniform disk of mass 1.82 kg and radius 0.85 m, find the magnitude of its angular acceleration. [For a uniform disk, I = (1/2)mr 2.] Correct answer: rad/s 2. Contacting a Surface C points A disk has mass m and outer radius R with a radial mass distribution (which may not be uniform) so that its moment of inertia

10 Version 001 Rotational Motion ramadoss (171) 10 is 9 10 mr2. The disk is rotating at angular speed ω 0 around its axis when it touches the surface, as shown. The disk is carefully lowered onto a horizontal surface and released at time t 0 with zero initial linear velocity along the surface. Assume that when the disk lands on the surface it does not bounce. The coefficient of friction between the disk and the surface is µ. The kinetic friction force between the surface and the disk slows down the rotation of the disk and at the same time gives it a horizontal acceleration. Eventually, the disk s linear motion catches up with its rotation, and the disk begins to roll (at time t rolling ) without slipping on the surface. m ω t = 9 19 Rω 0 µg correct Rope Around Cylinder points A massless rope is wrapped around a uniform cylinder of radius R and mass M. M What is the linear acceleration of the cylinder? Assume the unwrapped portion of the rope is vertical and the axis of the cylinder remains horizontal. 1. a = 2g 5 2. a = g 5 ω R µ How long after the first contact with the surfacedoesittakeforthedisktorollwithout slipping? 1. t = 5 Rω 0 12 µg 2. t = 2 Rω 0 5 µg 3. t = 5 Rω 0 14 µg 4. t = 3 Rω 0 7 µg 5. t = 7 Rω 0 16 µg 6. t = 4 Rω 0 11 µg 7. t = 1 Rω 0 3 µg 8. t = 4 Rω 0 9 µg 9. t = 8 Rω 0 17 µg 3. a = g 3 4. a = 2g 3 correct 5. a = 2g 6. a = 3g 4 7. a = g 4 8. a = g 9. a = 3g a = g 2 Rope Rotates a Cylinder points An 8 kg mass is attached to a string, which is wrapped several times around a uniform solid cylinder of radius 9 m and moment of inertia of 7 kg m 2.

11 Version 001 Rotational Motion ramadoss (171) 11 r Correct answer: m/s 2. I = 7 kg m 2 8 kg Find the torque on the cylinder. Assume the cylinder can rotate freely and the acceleration of gravity is 9.8 m/s 2. Correct answer: N m. Ferris Wheel Motor points A motor keep a Ferris wheel (with moment of inertia kg m 2 ) rotating at 12 rev/hr. When the motor is turned off, the wheel slows down (because of friction) to 9.6 rev/hr in 22 s. What wasthepower of themotor thatkept the wheel rotating at 12 rev/hr despite friction? Correct answer: W. Complex Atwood Machine points An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). Theleft wheel hasamassof2.1kg and radius cm. The right wheel has a mass of 2.7 kg and radius cm. The hanging mass on the left is 1.85 kg and on the right 1.29 kg. m 1 m 2 m 3 m 4 What is the acceleration of the system? The acceleration of gravity is 9.8 m/s 2.

Webreview Torque and Rotation Practice Test

Webreview 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 0.30-m-radius automobile