Question 548: Centre of Mass and Rotational Motion

Question

A thin uniform rod of mass '$$m$$' and length '$$P$$' 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

Accepted Answer

The kinetic energy of the rod while passing through the mean position will be,
$$\begin{aligned}
	ext { K.E. } & =\frac{1}{2} \mathrm{I} \omega^2 \
& =\frac{1}{2} \frac{\mathrm{m} l^2}{3} 	imes \omega^2=\frac{\mathrm{m} l^2 \omega^2}{6}
\end{aligned}$$

Answer

$$\mathrm{ml}{ }^2 \omega^2$$

Answer

$$\frac{\mathrm{m} l^2 \omega^2}{4}$$

Answer

$$\frac{\mathrm{m} l^2 \omega^2}{6}$$

Answer

$$\frac{\mathrm{m} \mathrm{l}^2 \omega^2}{12}$$

Centre of Mass and Rotational Motion

Choose the correct answer: