Midterm 2 Solutions - PHYS 3201 TEST 2 April 14, 2010 1.)...

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Unformatted text preview: PHYS 3201 TEST 2 April 14, 2010 1.) Disk in a Bowl-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.1 R r A disk of mass M , moment of inertia I , and radius r is placed in a bowl of radius R , where R r . Assume the disk rolls without slipping. (a) Write the Lagrangian for the disk in terms of the angle and the constants listed above. Acceleration due to gravity is g everywhere. The potential energy of the disk is V = MgR cos( ) the translational kinetic energy is that of a mass moving in polar coordinates at fixed radius, which is T trans = 1 2 MR 2 2 We can find the rotational kinetic energy by defining an angle that the disk has rotated T rot = 1 2 I 2 but if the disk moves without slipping, the distance the disk has covered on the surface of the bowl must equal the distance along the surface of the disk. R = r = R r This answer makes sense in the limiting cases- a very skinny disk will spin very fast, where as a wide disk will rotate at a slower rate. So we have total kinetic energy T = 1 2 MR 2 + 1 2 I parenleftbigg R r parenrightbigg 2 2 and so our Lagrangian is L = 1 2 MR 2 + 1 2 I parenleftbigg R r parenrightbigg 2 2 + MgR cos(...
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This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Institute of Technology.

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Midterm 2 Solutions - PHYS 3201 TEST 2 April 14, 2010 1.)...

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