Question

A meter stick is held 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 m/s
• B. 5.4 m/s
• C. 2.5 m/s
• D. none of these

Answer 1

Answer: (B)

Potential energy is converted into kinetic energy of rotation.
From law of conservation of energy, we have the potential energy of rod when it is vertical is converted to kinetic energy of rotation.

$\therefore \frac{M}{2} g l=\frac{1}{2} I \,\omega^{2}$
Moment of inertia of a thin rod is $I=\frac{1}{3} \,M l^{2}$
$\frac{M}{2} \,g \,l=\frac{1}{2} \frac{M l^{2}}{3} \,\omega^{2}$
$\Rightarrow \omega=\sqrt{\frac{3 g}{l}}$
Given, $l=1 \,m$
$\omega=\sqrt{3 g}$
Also, $v=r \,\omega$
$\Rightarrow v=1 \times \sqrt{3 \times 9.8}=5.4 \,m / s$