Ex_moi - I = I i Example 1. A 7.5 Kg thin rod and a small 5...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
2D rigid body dynamics Moment of inertia 2 I rdm = – Special cases for some uniform rigid bodies: (1) Slender rod (mass = M and length = L ) pivoted at the center: I = ML 2 /12 (2) Slender rod (mass = M and length = L ) pivoted at the end: I = ML 2 /3 (3) Circular disk or cylinder (mass = M and radius = R ) pivoted at the center: I = MR 2 /2 (4) A point mass M of negligible size: I = Mr 2 , r = distance from the pivot to the particle – Parallel-axis theorem: I O = I cm + md 2 , d = distance between the pivot point O and cm. Note: For a uniform rigid body, center of mass coincides with its geometric center. – Moment of inertia of a composite body (assembly of several parts):
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: I = I i Example 1. A 7.5 Kg thin rod and a small 5 Kg sphere are mounted on a 25 Kg disk as shown. The size of the sphere can be ignored and the disk is pivoted at its center. Find the moment of inertia of the assembly. ( Ans : 3.2 Kgm 2 ) R = 0.4 m O 2. Two small spheres A ( m A = 2.5 Kg) and B ( m A = 2 Kg) are attached to a 3 Kg slender rod as shown. The sizes of the spheres are negligible and the assembly is pivoted at point O . Find the moment of inertia of the system. ( Ans : 4 Kgm 2 ) O B A 0.6 m 0.5 m 1.5 m...
View Full Document

Ask a homework question - tutors are online