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Question
A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor.
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Solution
Let the mass of the rod and its angular velocity be m and ω (when hits the ground), respectively.
It is given that the rod has rotational motion only.
On applying the law of conservation of energy, we get
\[\frac{1}{2}I \omega^2 = mg\frac{l}{2}\]
\[ \Rightarrow \frac{m l^2}{3} \times \omega^2 = mgl\]
\[ \Rightarrow \omega^2 = \frac{3g}{l}\]
\[ \Rightarrow \omega = \sqrt{\frac{3g}{l}} = \sqrt{\left( 3 \times \frac{9 . 8}{1} \right)}\]
\[ \Rightarrow \omega = 5 . 42\text{ rad/s}\]
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