Question

A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?

Answer 1

It is given that:
Radius of the hollow sphere, r = 2 cm
Length of the long thread, l = 18 cm = $\frac{18}{100}=0.18\mathrm{m}=0.2\mathrm{m}$.


Let I be the moment of inertia and $\omega$ be the angular speed.
Using the energy equation, we can write:
$mgl(1-\cos \theta \mathit{)}+\frac{1}{2}I{\omega }^2=\mathrm{constant}$
$$\begin{gathered} mg\left(0.20\right)\left(1-\cos \theta \right)+\frac{1}{2}I{\omega }^2=C...\left(1\right) \\ \mathrm{Moment}\mathrm{of}\mathrm{inertia}\mathrm{about}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{suspension}\mathrm{A}\mathrm{is}\mathrm{given}\mathrm{by}, \\ I=\frac{2}{3}m{r}^2+m{l}^2 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}l\mathit{}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ I=\frac{2}{3}m{\left(0.02\right)}^2+m{\left(0.2\right)}^2 \\ =\frac{2}{3}m\left(0.0004\right)+m\left(0.04\right) \\ =m\left[\frac{0.0008}{3}+0.04\right] \\ =m\left(\frac{0.1208}{3}\right) \end{gathered}$$`

On substituting the value of I in equation (1) and differentiating it, we get:
$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}t}\left[mg\left(0.2\right)\left(1-\cos \theta \right)+\frac{1}{2}\frac{0.1208}{3}m{\mathrm{\omega }}^2\right]=\frac{\mathrm{d}}{\mathrm{d}t}\left(c\right) \\ \Rightarrow mg\left(0.2\right)\sin \theta \frac{\mathrm{d}\theta }{\mathrm{d}t}+\frac{1}{2}\left(\frac{0.1208}{3}\right)m\times 2\mathrm{\omega }\frac{\mathrm{d\omega }}{\mathrm{d}t}=0 \\ \Rightarrow 2\sin \theta =\frac{0.1208}{3}\alpha \left[\mathrm{because},g=10\mathrm{m}/{\mathrm{s}}^2\right] \\ \Rightarrow \frac{\alpha }{\theta }=\frac{6}{0.1208} \\ \Rightarrow {\omega }^2=49.66 \\ \Rightarrow \omega =7.04 \\ \mathrm{Thus},\mathrm{time}\mathrm{period}\left(T\right)\mathrm{will}\mathrm{be}: \\ T=\frac{2\pi }{\omega }=0.89\mathrm{s} \end{gathered}$$

For a simple pendulum, time period (T) is given by,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g}}$
$\Rightarrow T=2\mathrm{\pi }\sqrt{\frac{0.18}{10}}$= 0.86 s

$\%\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{time}\mathrm{period}=\frac{0.89-0.86}{0.89}\times 100=0.3$
∴ It is about 0.3% greater than the calculated value.