Question

A thin uniform rod of mass 'm' and length 'l' is suspended from one end which can oscillate in a vertical plane about the point of intersection. It is pulled to one side and then released. It passes through the equilibrium position with angular speed '$\omega$'. The kinetic energy while passing through mean position is

• A. $ml^{2}\omega^{2}$
• B. $\frac{ml^{2}\omega^{2}}{4}$
• C. $\frac{ml^{2}\omega^{2}}{6}$
• D. $\frac{ml^{2}\omega^{2}}{12}$

Hint:

Physics Tip: Be careful with the axis of rotation! If the rod were rotating about its center, $I$ would be $ml^2/12$. Since it's suspended from one end, we use $ml^2/3$.

Answer 1

The Correct Option is C

Concept: Physics (Rotational Motion) – Rotational Kinetic Energy.

Step 1: State the formula for rotational kinetic energy. The kinetic energy ($K.E.$) of a rotating body is given by: $$K.E. = \frac{1}{2}I\omega^2$$

Step 2: Determine the Moment of Inertia ($I$). For a thin uniform rod of mass $m$ and length $l$ rotating about an axis passing through one of its ends, the moment of inertia is: $$I = \frac{ml^2}{3}$$

Step 3: Substitute and calculate $K.E.$ $$K.E. = \frac{1}{2} \left( \frac{ml^2}{3} \right) \omega^2$$ $$K.E. = \frac{ml^2\omega^2}{6}$$ $$ \therefore \text{The kinetic energy at the mean position is } \frac{ml^{2}\omega^{2}}{6}. $$