Question

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

Answer 1

From geometry,

$AB=\frac{l/2}{cos{30}^{\circ }}=\frac{l}{\sqrt{3}}=l^{\prime }$

Thus moment of inertia about A= $I_A=3\times m{l}^{\prime 2}=m{l}^2$

Moment of inertia about B=$I_B=m{0}^2+m{l}^2+m{l}^2=2m{l}^2$

Thus $\frac{K_B}{K_A}=\frac{\frac{1}{2}{I}_B{\theta }^2}{\frac{1}{2}{I}_A{\theta }^2}$

$=\frac{I_B}{I_A}=2$