Question

A thin rod AB is sliding between two fixed right angled surfaces. At some instant its angular velocity is $\omega$. If $I_x$ represent moment of inertia of the rod about an axis perpendicular to the plane and passing through the point $x$(A, B, C or D), the kinetic energy of the rod is:

• A. $\frac{1}{2}{I}_A{\omega }^2$
• B. $\frac{1}{2}{I}_B{\omega }^2$
• C. $\frac{1}{2}{I}_C{\omega }^2$
• D. $\frac{1}{2}{I}_D{\omega }^2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{1}{2}{I}_A{\omega }^2$
B $\frac{1}{2}{I}_B{\omega }^2$
D $\frac{1}{2}{I}_D{\omega }^2$

D is the instantaneous axis of rotation of rod, so KE about D is

$KE=\frac{I_D{\omega }^2}{2}$

Since AC, BC and CD are equal:

$\Rightarrow I_D=I_A=I_B$

So, $KE=\frac{I_D{\omega }^2}{2}=\frac{I_A{\omega }^2}{2}=\frac{I_B{\omega }^2}{2}$

But $KE$ abouts C is $\frac{I_C{\omega }^2}{2}+\frac{m{v}_{cm}^2}{2}$