This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Physics 105B Problem Set 3 April 22, 2008 Jeff Schonert: schonert (at) physics.ucsb.edu Taylor 10.2 First consider the kinetic energy that describes the linear motion of the center of mass as well as rotation about the center of mass. This quantity T 1 is then the sum of translational and rotational motion T 1 = M 2 v 2 + 1 2 Iω 2 , (1) where I is the moment of inertia about the center of mass. We are told in the problem that I = MR 2 / 2, and we know that v is related to ω by ω = v/R . Plugging both of these in gives T 1 = M 2 v 2 + 1 4 MR 2 v R 2 = 3 4 Mv 2 . (2) Now consider the kinetic energy about the point on the wheel that is instantaneously at rest. The linear velocity of this point is 0, since at each instant of time, it is at rest. Therefore, there is no translational contribution to the kinetic energy here; it is all rotation. Then the kinetic energy T 2 about the new point is T 2 = 1 2 I ω 2 , (3) where I is the moment of inertia about the new reference point. Plugging in I = 3 MR 2 / 2 as given in the problem and using ω = v/R , we get T 2 = 3 4 MR 2 v R 2 = 3 4 Mv 2 . (4) So the two kinetic energies agree. Taylor 10.9 For a cylinder of radius R and length L , the volume is πR 2 L , which implies ρ = M/πR 2 L ....

View
Full
Document