Midterm Solutions 2006

# Midterm Solutions 2006 - Solutions to 105 Midterm 2 Ed...

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Unformatted text preview: Solutions to 105 Midterm 2 Ed Groth, 9-Dec-2006 1. (a) The gravitational force and the normal force act through the center of mass and do not make a torque about the center of mass. The friction force acts at the edge of the ring, perpendicular to the plane of the ring, so it makes a torque whose magnitude is fb and whose direction, at the instant shown in the figure in the exam, is in the plane of the ring, out of the paper . (b) The torque computed in part (a) is the rate of change of the (horizontal) component of angular momentum of the ring: fb = dL h dt . The angular momentum of the ring (for the rotation about its axis) points towards the apex of the cone (parallel to the side of the cone). As the ring goes around, the vertical component of the angular momentum is constant and the horizontal component has constant magnitude but rotates with angular frequency . So, 1 L h dL h dt = . We know the time derivative, but what is L h ? If the ring is rotating about its axis at angular speed , then L = mb 2 . and are related since = v R = v h tan , (1) and = v b . If this is hard to see, note that the number of meters laid down by the rim in time dt is b dt and the number of meters traversed on the cone in the same time is Rdt . So = R /b and 1 mRb sin fb = , or f = mh 2 tan sin . (2) Copyright c circlecopyrt 2006, Princeton University Physics Department, Edward J. Groth 3. (continued) The ring is also instantaneously rotating about its diameter perpendicular to the...
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## This note was uploaded on 05/03/2010 for the course PHY 105 at Princeton.

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Midterm Solutions 2006 - Solutions to 105 Midterm 2 Ed...

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