This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 TEAL Fall Term 2004 InClass Problems 3032: Moment of Inertia, Torque, and Pendulum: Solutions Problem 30 Moment of Inertia of a Uniform disc. A uniform disc of mass m and radius R is mounted on an axis passing through the center of the disc, perpendicular to the plane of the disc. In this problem, you will calculate the moment of inertia about two different axes that pass perpendicular to the disc. One passes through the center of mass of the disc and the second passes through a point on the rim of the disc a distance R from the center. As a starting point, consider the contribution to the moment of inertia from the mass element dm show in the figure below. Step 1: We can take the point S to be the center of mass of the disc. The axis of rotation passes through the center of the disc, perpendicular to the plane of the disc. Step 2: We choose cylindrical coordinates with the coordinates ( r , θ ) in the plane and the z axis perpendicular to the plane. Step 3: The area element da = rdrd θ can be thought of as the product of arc length rd θ with the radial width dr , (Figure 9). Since the disc is uniform, the mass per unit area is a constant, m σ = dm m total = = 2 . da Area π R Therefore the mass in an infinitesimal area element a distance r from the axis of rotation is given by dm = σ rdrd θ ....
View
Full
Document
This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.
 Spring '08
 WHITE
 Physics, Inertia, Mass

Click to edit the document details