PHY131_F07_M2_Solutions

# PHY131_F07_M2_Solutions - Physics 131, Midterm II &...

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Unformatted text preview: Physics 131, Midterm II & Solutions Abhay Deshpande November 6, 2007 Write your name, SBU ID, and Recitation number or instructor’s name clearly. Please start a new problem on a new page 1. A ball of radius r 1 = 0 . 080 m and mass 1.00 kg is attached by a massless rod 0.400 m in length to a second ball with radius r 2 = 0 . 100 m and mass 2.00 kg. The rod is attached at the surface of both spheres. (a) [ 10 points] Where is the center of gravity of the system? Solution: Take the origin to be at the center of the small ball; then, x cm = (1 . kg ) × (0 m ) + (2 . kg ) × (0 . 58 m ) 3 . kg = 0 . 387 from the center of the small ball. (b) [ 10 points] If the connecting rod between the two masses was 1.5 kg instead of being mass- less. Where would the center of gravity be? What can you say about this new center of gravity position with respect to the solution when the rod was mass-less? Solution: x cm = (1 . kg ) × (0 m ) + (1 . 5 kg ) × (0 . 28 m ) + (2 . kg ) × (0 . 58 m ) 4 . 5 kg = 0 . 351 The result is smaller than the earlier case. (c) [10 points] If you attach a string at the location of the center gravity assuming the mass- less rod as in (a), is the system going to be in equilibrium? (assume that the other end of the string is attached to the ceiling). Support your answer by explicit calculation ofof the string is attached to the ceiling)....
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## This test prep was uploaded on 04/07/2008 for the course PHY 131 taught by Professor Rijssenbeek during the Fall '03 term at SUNY Stony Brook.

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PHY131_F07_M2_Solutions - Physics 131, Midterm II &...

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