Question

A stone tied to the end of a string of length $1\text{m}$ is whirled in a horizontal circle with a constant speed. If stone makes $22$ revolution in $44$ $\text{second}$, what is the magnitude and direction of the acceleration of stone?

• A. Acceleration will be ${\pi }^2{\text{m/s}}^2$ radially inwards.
• B. Acceleration will be ${\pi }^2{\text{m/s}}^2$ radially outwards.
• C. Acceleration will be ${\pi }^2{\text{m/s}}^2$ along the tangent to the circle.
• D. Acceleration will be $\frac{\pi }{4}{\text{m/s}}^2$ radially inwards.

Answer 1

The correct option is A Acceleration will be ${\pi }^2{\text{m/s}}^2$ radially inwards.

Radial acceleration or centripetal accceleration is given by:
$a_c=\frac{v^2}{r}={\omega }^{2}r$
Frequency of revolution is,
$f=\frac{22}{44}=\frac{1}{2}{\text{s}}^{-1}$
$\Rightarrow \omega =2\pi f=\pi \text{rad/s}$
$\therefore a_c={\omega }^{2}r={\pi }^2\times 1={\pi }^2{\text{m/s}}^2$


Tangential acceleration is the rate of change of speed w.r.t time:
$\Rightarrow a_t=\frac{dv}{dt}=0$
$\because$ speed is constant
Hence net acceleration of stone $\overrightarrow{a}$ given by:
$\overrightarrow{a}={\overrightarrow{a}}_c+{\overrightarrow{a}}_t={\overrightarrow{a}}_c={\pi }^2{\text{m/s}}^2$
$\therefore |\overrightarrow{a}|=|{\overrightarrow{a}}_c|={\pi }^2{\text{m/s}}^2$
Direction is along the radius towards the centre i.e. radially inwards.