Hmwk_08_Solutions

Hmwk_08_Solutions - 1 Physics 221 Fall 2008 Homework #8...

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Unformatted text preview: 1 Physics 221 Fall 2008 Homework #8 Solutions Ch. 9:4-6; Ch. 10:1-4 Due Tues, Oct 21, 2008 8.1 N identical point-like particles are equally spaced around the rim of a circular massless wheel of radius R . The total mass of all the particles is M . Derive the formula for the moment of inertia of the wheel for rotation about an axis through the center of the wheel and perpendicular to the plane of the wheel. Why is this formula for the moment of inertia the same as for the thin-walled cylinder in Table 9.2(g) on p. 299 of the text? I = m i r i 2 i = 1 N = R 2 m i i = 1 N = MR 2 This expression is the same as for a thin-walled cylinder because all of the mass is a distance R from the axis of rotation, just as for a thin-walled cylinder. 8.2 State in words what the Parallel-Axis Theorem says. Why is the Parallel-Axis Theorem a useful theorem? The Parallel-Axis Theorem states that if the axis of rotation of a rigid body is a distance d from a parallel axis passing through the center of mass, then the moment of inertia of the rigid body about the axis of rotation is equal to the moment of inertia of the body about the axis through the center of mass, plus the mass of the body times d 2 . This theorem is very useful because one need only do one detailed and possibly complicated calculation of the moment of inertia about an axis through the center of mass in order to simply calculate the moment of inertia about any axis parallel to that axis. 8.3 A solid sphere of mass 1.00 kg and radius R = 0.100 m rotates at angular speed 20.0 rad/s about an axis that touches the edge of the sphere (see figure). (a) What is the value of the moment of inertia of the sphere about its axis of rotation? According to Table 9.2(h), the moment of inertia of the sphere about an axis through its center is I cm = (2/5) MR 2 . Then using the Parallel Axis Theorem, we get I = I cm + M d 2 = (2/5) MR 2 + MR 2 = (7/5) MR 2 = (7/5)(1.00 kg)(0.100 m) 2 = 0.0140 kg m 2 . (b) What is the rotational kinetic energy of the sphere for rotation of the sphere about its axis of rotation? K rot = (1/2) I 2 = (1/2)(0.0140 kg m 2 )(20.0 rad/s) 2 = 2.80 J . R 2 Alternatively, one can view the motion of the sphere as translation of the center of mass of the sphere, plus rotation of the sphere about its center of mass. Viewed this way, the speed of the center of mass is v = R , and the angular speed of the sphere about its center of mass is the same as the angular speed of the center of mass about the axis of rotation stated in the problem. Then the kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy about the center of mass:...
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This note was uploaded on 01/28/2010 for the course PHYSICS 221 taught by Professor Johnson during the Fall '06 term at Iowa State.

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Hmwk_08_Solutions - 1 Physics 221 Fall 2008 Homework #8...

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