Classical Mechanics - Rotational Kinetic Energy

Three equal masses m are rigidly connected to each other by massless rods of length l forming an equilateral triangle. The assembly is to be given an angular velocity ω about an axis perpendicular to the triangle. For fixed ω, the ratio of the kinetic energy of the assembly for an axis through B compared with that for an axis through A is equal to

A. 3
B. 2
C. 1
D. 1/2
E. 1/3

(GR9677 #32)

Solution:

KE = ½Iω²

For fix ω, the ratio KE_B/KE_A = I_B/I_A

I = ∑_i m_ir_i²

From A:
m₁= m₂ = m₃ = m
x₁= x₂ = x₃ = l/√3 (see notes)

I_A = m₁x₁² + m₂x₂² + m₃x₃²
= 3m( l/√3)² = ml²

From B:
I_B = m₁x₁² + m₂x₂² 
= ml² + ml² = 2ml²

KE_A/KE_B = 2ml²/ml² = 2

Answer: B

Notes:

Cos 30^o = ½l / x₁ 
Cos 30^o = ½√3
½l / x₁ = ½√3
x₁ = x₂ = x₃ = l/√3