University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

359

Pages:

5
Author:

Jabin F.

Introduction: The problem is to find the moment of inertia of a disk about an oblique axis. Coordinate System and Axes: The disk lies in the xy-plane. The unit vector n is defined along the given axis of rotation. The components of n are given as cosθ and sinθ. Moment of Inertia Tensor: The moment of inertia tensor is defined as: I = [Ixx 0 0; 0 Iyy 0; 0 0 Izz] The values of Ixx, Iyy, and Izz are given for a disk: Ixx = Iyy = MR^2 / 6 Izz = MR^2 / 4 Calculation of Moment of Inertia: The moment of inertia about the given axis is calculated using the formula: In = n^T * I * n The final result is: In = MR^2 / 4 * (1 + cos^2θ) Problem 1: A sphere of radius R has mass density ρ = kr, where r is the distance from the center. Calculate the moment of inertia (MI) about its diameter in terms of M (total mass) and R. Solution: The mass element dm is given as dm = ρ dV = kr dV. The volume element dV is expressed in spherical polar coordinates as dV = r^2 sinθ dr dθ dφ. The total mass M is calculated by integrating dm over the entire sphere. The moment of inertia Izz about the z-axis is calculated using the formula Izz = ∫(x^2 + y^2) dm. The final result is Izz = (2/9) MR^2. Problem 2: A solid sphere of uniform density (shaded) with radius R has a spherical cavity scooped out of it, as shown unshaded, which has diameter R. If this object has mass M, its moment of inertia about the tangential axis passing through the point where the sphere touches is to be found. Problem 2: A solid sphere of uniform density (shaded) with radius R has a spherical cavity scooped out of it, as shown unshaded, which has diameter R. If this object has mass M, its moment of inertia about the tangential axis passing through the point where the sphere touches is to be found. Solution: Let M be the mass of the shaded part. Let m be the mass of a sphere of radius R (uniform) and m' be the mass of a sphere of radius R/2. For the same mass density, m = 8m'. Therefore, m - m' = M, which leads to m = 8M/7 and m' = M/7. The moment of inertia of the big sphere (I1) is calculated as (7/5)mR^2. The moment of inertia of the small sphere (I2) is calculated as (7/5)m'(R/2)^2. The final moment of inertia (I) is given by I = I1 - I2. After substituting the values of m and m', the final result is I = (31/90)MR^2.