Question

A uniform rod of length $2.0 \, m$ is suspended through its endpoint about which it performs small angular oscillations in the vertical plane. Its time period is nearly

• A. $1.6 \, s$
• B. $1.8 \, s$
• C. $2.0 \, s$
• D. $2.3 \, s$

Answer 1

Answer: (D)

It is a physical pendulum, and its time period $T = 2 \pi \sqrt{\frac{I}{m g x}}$
here I = moment of inertia of the pendulum about point of suspension
$I = \frac{M L^{2}}{12} + \frac{M L^{2}}{4} = \frac{M L^{2}}{3}$
and x = distance of point of suspension from its COM.
x = L/2
$\Rightarrow \, T = 2 \pi \sqrt{\frac{M L^{2} / 3}{m g L / 2}} = 2 \pi \sqrt{\frac{2 L}{3 g}}$
$\Rightarrow T=2\pi \sqrt{\frac{2 \times 2}{3 \times 10}}=2.3\text{ sec}\text{.}$