Question

What is the time period of a pendulum formed by pivoting a meter stick so that it is free to rotate about a horizontal axis passing through the $75\text{cm}$ mark?

• A. $1.00\text{sec}$
• B. $1.25\text{sec}$
• C. $1.53\text{sec}$
• D. $2.5\text{sec}$

Answer 1

The correct option is C $1.53\text{sec}$


Let $m$ be the mass and $l$ be the length of the stick.
$l=100\text{cm}$
The distance of the point of suspension from centre of gravity is $d=25\text{cm}$
Moment of inertia about a horizontal axis through O is
$I=I_C+m{d}^2\Rightarrow I=\frac{m{l}^2}{12}+m{d}^2$
Now, time period for physical pendulum is
$\Rightarrow T=2\pi \sqrt{\frac{I}{mgd}}=2\pi \sqrt{\frac{\frac{m{l}^2}{12}+m{d}^2}{mgd}}$
$\Rightarrow T=2\pi \sqrt{\frac{l^2+12d^2}{12gd}}=2\pi \sqrt{\frac{1^2+12(0.25{)}^2}{12\times 9.8\times 0.25}}=1.53\text{s}$