A metre stick is positioned vertically with one end resting on a rough horizontal floor. It is gently released to fall onto the floor. Assuming the end on the floor does not slip, determine the angular speed of the rod

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Quantrex Academy · Accepted Answer

Correct answer: D. 5.4 rad/s

Using conservation of mechanical energy:

Loss in potential energy = Gain in rotational kinetic energy

mg\dfrac{L}{2} = \dfrac{1}{2}I\omega^2

Since the rod rotates about its lower end, its moment of inertia is

I = \dfrac{mL^2}{3}

.

mg\dfrac{L}{2} = \dfrac{1}{2}\left(\dfrac{mL^2}{3}\right)\omega^2

\omega^2 = \dfrac{3g}{L}

Given

L = 1 \text{ m}

and

g = 9.8 \text{ m/s}^2

:

\omega = \sqrt{\dfrac{3 \times 9.8}{1}} = \sqrt{29.4} \approx 5.4 \text{ rad/s}

A metre stick is positioned vertically with one end resting on a...

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