Question

A particle of mass $m$ moves in a circle of radius $R$. The distance $s$ described by it varies with time $t$ as $s=\alpha t^2$, where $\alpha$ is a constant.

What is the force acting on the particle?

• A. $2\alpha m\sqrt{1+\frac{4s^2}{R^2}}$
• B. $\frac{2\alpha ms}{R}$
• C. $4\alpha m\sqrt{1+\frac{s^2}{R^2}}$
• D. $\frac{\alpha ms}{R}$

Answer 1

Answer: (A)

$2\alpha m\sqrt{1+\frac{4s^2}{R^2}}$

Given $s=\alpha t^2$;

as $\frac{ds}{dt}=v$

so, $v=2\alpha t$

Tangential acceleration = $\frac{dv}{dt}$ = $2\alpha$

Normal acceleration = $\frac{v^2}{r}$ = $4\alpha s/R$

Total acceleration$=\sqrt{a_r^2+a_t^2}$

$=\sqrt{4{\alpha }^2+\frac{16{\alpha }^2{s}^2}{R^2}}$
Force$=$ mass $\times$ total acceleration
$=m\sqrt{4{\alpha }^2+\frac{16{\alpha }^2{s}^2}{R^2}}$
$=2\alpha m\sqrt{1+\frac{4s^2}{R^2}}$