phys1443-spring02-2nd-term-exam-solutions

# phys1443-spring02-2nd-term-exam-solutions - PHYS 1443-501,...

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PHYS 1443-501, Spring 2002, 2 nd -Term Exam, Wednesday, Apr. 10, 2002 Name: ID: [ 1 – 20 points] A uniform rod of length 150 cm and mass 900g is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown in the figure. Answer the following series of questions, assuming that the rod is released from rest in the horizontal position, and the magnitude of the gravitational acceleration g is 2 / 80 . 9 s m . a) What is the line density of the rod? 1. 6.00kg/cm 3 2. 0.600kg/m 3. 0.600kg/cm 4. 6.00kg/m 3 Solution: Since line density is the density per unit length, the line density for this rod is  m/L=0.900/1.50=0.600kg/m. b) What is the moment of inertia of the rod in this motion? 1. 2 267 . 0 m kg 2. m kg 35 . 1 3. 2 35 . 1 m kg 4. 2 675 . 0 m kg Solution: Moment of inertia of this rod when rotates about the axis at one end is I=ML 2 /3=0.900x(1.50) 2 =0.675kg.m 2 . c) What is the potential energy of the rod before the release? 1. s m kg / 2 . 13 2. J 2 . 13 3. J 62 . 6 4. mgh Solution: Potential energy is mgh. The height of the CM when it is in its equilibrium position is L/2. Therefore the potential energy U=mgh=mgL/2=0.9x9.8x1.5/2=6.62J. d) What is the angular speed of the rod when the rod reaches the bottom? 1. s / 43 . 4 2. s / 38 . 3 3. s m / 62 . 6 4. s / 3 . 16 Solution: Since potential energy is transferred to rotational energy, at the bottom the angular speed  becomes: 1 Turn over

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PHYS 1443-501, Spring 2002, 2 nd -Term Exam, Wednesday, Apr. 10, 2002 e) What is the linear velocity of the center of mass when the rod is at its vertical position? 1. s m / 66 . 7 2. s m / 32 . 3 3. s / 64 . 6 4. s / 38 . 3 Solution: Since linear velocity is v=r  L  x4.43/2=3.32m/s. f) What is the linear speed of the lowest point of the rod when the rod is in its vertical position? 1. s m / 65 . 6 2. s / 33 . 2 3. s m / 78 . 6 4. s m / 32 . 3 Solution: Since linear velocity is v=r  L  x4.43=6.64m/s. g) What is the magnitude of torque? 1. N 8 . 9 2. m N 96 . 3 3. m N 62 . 6 4. m N 3 . 12 Solution: Since torque is t=r F L/2xmg=  x0.9x9.8=6.62N.m. h) What is the direction of torque in this motion? 1. Upward 2. Out of the page 3. Into the page 4. Downward Solution: Using the right-hand rule, the direction of the torque vector is into the page. i)What is the angular momentum of the rod when it is at its vertical position? 1. s m kg / 99 . 2 2 2. s m kg / 23 . 3 3. 2 / 80 . 9 s m kg 4. s m kg / 00 . 6 2 Solution: Since angular momentum is L=I  0.675x4.43=2.99kg.m2/s. j) What is the direction of angular momentum in this motion? 1. Upward 2. Out of the page 3. Into the page 4. Downward Solution: Using the right-hand rule, the direction of the angular momentum vector is into the page. 2 Turn over s I mgL mgL mgh I / 43 . 4 675 . 0 2 . 13 2 / 2 1 2
PHYS 1443-501, Spring 2002, 2 nd -Term Exam, Wednesday, Apr. 10, 2002 [ 2 – 25 points] •A solid sphere of mass 1.00kg and radius R=50.0cm was released at the top of the incline from rest, as shown in the picture, and is rolling down the incline without slipping. . The magnitude of the gravitational acceleration is g=9.8m/s 2 . The surface of the incline has friction. Answer the following series of questions.

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## This note was uploaded on 03/25/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.

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phys1443-spring02-2nd-term-exam-solutions - PHYS 1443-501,...

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