ME 274: Basic Mechanics II Spring April 18, Problem 1 (24 points):

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Whitney James
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At the instant shown the block has zero velocity and the bar has an angular velocity!

1 Problem 1 (24 points): Given: The block shown in the figure slides on a smooth surface. A thin homogenous bar is attached to the block and is free to rotate about a pin joint at A. At the instant shown the block has zero velocity and the bar has an angular velocity!. Find: (a) Determine the angular acceleration of the thin bar. (b) Determine the acceleration of the block. Exam 3 Page 2 of 9

2 This page is for extra work related to Problem 1. Exam 3 Page 3 of 9

Problem 2 (20 points): Given: The constant tensions of F 1 and F 2 are applied to the hoisting cable as shown. The velocity of the block A is ~v A1 upwards and the angular velocity of the pulley is ~!

3 Problem 2 (20 points): Given: The constant tensions of F 1 and F 2 are applied to the hoisting cable as shown. The velocity of the block A is ~v A1 upwards and the angular velocity of the pulley is ~! 1 counterclockwise at time t = 0. The pulley rotates about the smooth pin joint at and the cable does not slip on the pulley. The pulley has radius of gyration k o with respect to point, the mass of the pulley is m and that of the block is M. Find: Determine, at time t: (a) The velocity of the block ~v A2 (b) The angular velocity of the pulley ~! 2 Exam 3 Page 4 of 9

4 This page is for extra work related to Problem 2. Exam 3 Page 5 of 9

5 Problem 3 (16 points): Written justifications for your answers are NT required, nor will they be graded. Part A (6 points): Given: A cable is wrapped around a homogeneous disk, as shown, with one end of the cable attached to a fixed wall and with a constant force acting on the free end of the cable. A constant force P =1.5F acts to the left at the center of the disk. During the subsequent motion, the disk is known to not slip on the cable. At Position 1, the system is released from rest with the cable taut. At Position 2, has moved through a distance of d. Let U (P ) (F ) 1!2 and U 1!2 represent the work done by the forces P and F as the disk moves! from! Position 1 to Position 2. orces P and F as the disk moves from Position 1 to Position 2. A F A F P = 1.5F R no slip P = 1.5F R F Find: What direction does point on the disk move? (a) Left (b) Right (c) It does not move: d=0 M = F ( 2R) +1.5F ( R) = 0.5F α = W moves to right Which of the following best represents the sign of the work done by F in moving from Position 1 to Position 2? (a) U (F ) 1!2 < 0 (b) U (F ) 1!2 =0 (c) U (F ) 1!2 > 0 Since A moves to the right and F acts to the right, then the work done by F is PSITIVE. Which of the following best represents the magnitude of the work done by P as compared to the magnitude of the work done by F in moving from Position 1 to Position 2? (a) U (P ) (F ) 1!2 < U (b) U (P ) (F ) 1!2 = U (c) U (P ) 1!2 1!2 1!2 > U (F ) 1!2 U ( P) 1 2 ( )( R) ( ) U ( F ) 1 2 = 1.5F U ( F ) 1 2 = ( F ) 2R > U ( P) 1 2

6 Part B (4 points): Given: System I is made up of a thin, homogeneous bar A of mass m. Bar A is pinned to ground at end A, and a homogeneous disk of mass m and radius R is pinned to end. System II is identical to System I, except that the radius of the disk is 2R. Each system is released from rest with bar A being horizontal (Position 1). At Position 2, is directly below A. Let! 2 d I and! 2 d represent the angular speed of the disk at Position 2. Also, let!a II 2 I represent the angular speed of bar A at Position 2. onsider all pins to be smooth. and!a 2 II y A m L R m g A m L 2R m mg x SYSTEM I SYSTEM II Find: Which of the following best represents the angular speed of the disk in System I as compared to the angular speed of the disk in System II at Position 2? (a)! d 2 I <!d 2 II (b)! d 2 I =!d 2 II (c)! d 2 I >!d 2 II M d d ( ω 2 ) = ω 2 I = 0 α d = 0 ω d = 0 for all time for both systems ( ) II = 0 Which of the following best represents the angular speed of bar A in System I as compared to the angular speed of bar A in System II at Position 2? (a) (b) (c)! A 2 I <!A 2 II! A 2 I =!A 2 II! A 2 I >!A 2 II T I +V I = T II +V II 0 = 1 2 I A ω A ( 2 ) mv 2 mg L 2 mgl Since v 2 = Lω 2 A, the above shows: ω 2 A = 3mgL = independent of size of disk 2 I A + ml

7 Part (4 points): Given: A circular disk having its geometric center at and mass center at G rolls freely without slipping on a rough horizontal surface. is the point on the disk in contact with the surface on which it rolls. At Position I, G is directly above with having a speed of (v ) I, whereas in Position II, G is directly below with having a speed of (v ) II. Let (N ) I and (N ) II represent the normal contact force acting on the disk at for Positions I and II, respectively. Assume that the disk remains in contact with the surface on which it rolls at all times. mg y G no slip x R G R / 2 ( v ) I no slip R R / 2 G ( v ) II G mg f I f II N I Posi%on (I) Posi%on (II) N II Find: Which of the following best represents the relative sizes of the speed of for Positions I and II? T I +V I = T II +V II 1 2 I I ω 2 I + mg R 2 = 1 2 I I ω 2 II mg R 2 (a) (v ) I < (v ) II (b) (v ) I =(v ) II (c) (v ) I > (v ) II ω II = I I I II ω I 2 + 2mgR > ω I since I I > 1 I II v II = Rω II > Rω I = v I Which of the following best represents the relative sizes of the normal contact force on the disk at for Positions I and II? (a) (N ) I < (N ) II (b) (N ) I =(N ) II (c) (N ) I > (N ) II Position I : a! GI = a! I + α! I r! G/ ω 2! R I rg/ = a I α I 2 î ω 2 R I 2 ĵ 2 R F y = N I mg = ma GIy N I = m g ω I 2 Position II : a! GII = a! II + α! II r! 2! R G/ ω IIrG/ = a II α II 2 î + ω 2 R II 2 ĵ 2 R F y = N II mg = ma GIIy N II = m g + ω II 2 N II > N I

Part D (2 points): Given: The center of the disk shown below is known to have a downward acceleration. Assume that the cable does not slip on the homogeneous disk.

8 Part D (2 points): Given: The center of the disk shown below is known to have a downward acceleration. Assume that the cable does not slip on the homogeneous disk. T D T AB y mg x Find: Which of the following most accurately describes the tension in sections AB and D of the cable? (a) The tension in section AB is smaller than the tension in section D (b) The tension in section AB is equal to the tension in section D (c) The tension in section AB is larger than the tension in section D M = ( T AB T D ) R = I α Since is accelerating downward and the disk is rotating about, then α is W ( α < 0 ). Therefore: T AB T D < 0 T AB < T D

Exam 3 April 16, 2014

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