# hmwk07 - Su (ycs73) HW07 Tsoi (58020) 1 This print-out...

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Unformatted text preview: Su (ycs73) HW07 Tsoi (58020) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points Consider a rod of length L and mass m which is pivoted at one end. An object with mass m is attached to the free end of the rod. C L m m 26 Determine the moment of inertia of the system with respect to the pivot point. The acceleration of gravity g = 9 . 8 m / s 2 . Consider the mass at the end of the rod to be a point particle. 1. I = 5 4 m L 2 2. I = 5 3 m L 2 3. I = L 2 4. I = 3 2 m L 2 5. None of these. 6. I = 13 12 m L 2 7. I = 4 3 m L 2 correct Explanation: The moment of inertia of the rod with re- spect to the pivot point is I rod = 1 3 mL 2 , and the moment of inertia of the mass m with respect to the pivot point is I mass = mL 2 . Thus the moment of inertia of the system is I = I rod + I m = 1 3 mL 2 + mL 2 = 4 3 mL 2 . 002 (part 2 of 4) 10.0 points The length C in the figure represents the lo- cation of the center-of-mass of the rod plus mass system. Determine the position of the center of mass from the pivot point. 1. C = 3 4 L correct 2. C = 5 8 L 3. C = L 4. None of these 5. C = 7 8 L 6. C = 1 2 L Explanation: The center of mass of the rod is 1 2 L , so the center of mass of the rod-plus-mass system is C = ( 1 2 suppressL + L ) m m + m = 1 4 L + 1 2 L = 3 4 L. 003 (part 3 of 4) 10.0 points The unit is released from rest in the horizontal position. What is the kinetic energy of the unit when the rod momentarily has a vertical orienta- tion? 1. K = 5 2 mg L 2. K = 3 2 mg L correct 3. K = mg L Su (ycs73) HW07 Tsoi (58020) 2 4. None of these 5. K = 1 2 mg L 6. K = 2 mg L Explanation: The potential energy released can be com- puted in two ways: U = U vextendsingle vextendsingle vextendsingle cm of rod + U vextendsingle vextendsingle vextendsingle mass = mg L 2 + mg L = 3 2 mg L, or U | rod+m = U | cm of rod+m = (2 m ) g parenleftbigg 3 4 L parenrightbigg = 3 2 mg L. 004 (part 4 of 4) 10.0 points Let the kinetic energy be K V at the vertical position and the moment of inertia of the system be I . Find the angular velocity of the system in terms of K V and I . 1. = radicalbigg 2 K V I correct 2. = K V I 3. = radicalbigg K V 2 I 4. None of these 5. = 2 K V I 6. = radicalbigg K V I Explanation: The rotational kinetic energy of the system is given by K V = 1 2 I 2 = radicalbigg 2 K V I . 005 10.0 points A uniform solid disk of radius 9 . 9 m and mass 7 . 7 kg is free to rotate on a frictionless pivot through a point on its rim. 9 . 9 m Pivot If the disk is released from rest in the position shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? The acceleration of gravity is 9 . 8 m / s 2 ....
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## This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas at Austin.

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hmwk07 - Su (ycs73) HW07 Tsoi (58020) 1 This print-out...

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