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## Q. 20.6

A uniform thin rod AB of mass m and length 2a has a particle of mass M attached at B and is hinged at its centre O so that it rotates freely in a vertical plane. The rod is held horizontally and released.
i) Show the moment of inertia of the system about the hinge is $\frac{4}{3}Ma^{2}$.
ii) Find the angular speed when the rod is vertical. ## Verified Solution

i) The moment of inertia of the rod alone is $\frac{1}{3}Ma^{2}$ (see page 458).
The moment of inertia of the particle about the axis is Ma².
So the total moment of inertia of the system about the axis is

$I = Ma² + \frac{1}{3}Ma^{2}$

$= \frac{4}{3}Ma^{2}.$

ii) When in the vertical position, the gain in kinetic energy is:

$\frac{1}{2}Iω² = \frac{2}{3}Ma^{2}ω².$

The loss in potential energy arises from the descent of the mass at B and is given by Mga.
Hence, by conservation of energy:

$\begin{matrix} \frac{2}{3}Ma^{2}\omega ^{2} = Mga & \boxed{\text{Note that there is no change in the potential energy of the rod during the rotation because the axis is through its centre of mass. In the subsequent sections you will meet the case where this is not so.}} \\ \Rightarrow \omega = \sqrt{\frac{3g}{2a} } & \end{matrix}$