Q. A meter stick is hold vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor the stick does not slip, the velocity of the other end when it hits the floor, will be
A) $10.8\ {ms}^{- 1}$
B) $5.4\ {ms}^{- 1}$
C) $2.5\ {ms}^{- 1}$
D) None of these
Explanation
(b) From law of conservation of energy, we have the potential energy of rod when it is vertical is converted to kinetic energy of rotation.
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$\therefore$ $\frac{M}{2}gl = \frac{1}{2}\ I\omega^{2}$
Moment of inertia of a thin rod is
$I = \frac{1}{3}\ Ml^{2}$
$\frac{M}{2}gl = \frac{1}{2}\frac{Ml^{2}}{3}\omega^{2}$
$\Rightarrow$ $\omega = \sqrt{\frac{3g}{l}}$
Given, $l = 1\ m,\ \omega = \sqrt{3g}$
Also, $v = r\omega$
$\Rightarrow$ $v = 1 \times \sqrt{3 \times 9.8} = 5.4\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.