Question

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

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

Answer 1

Answer: (D)

It is a physical pendulum, and its time period $T=2\pi \sqrt{\frac{I}{mgx}}$
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
$\Rightarrow T=2\pi \sqrt{\frac{M{L}^2/3}{mgL/2}}=2\pi \sqrt{\frac{2L}{3g}}$
$\Rightarrow T=2\pi \sqrt{\frac{2\times 2}{3\times 10}}=2.3\text{ sec}\text{.}$