Circular Motion

Given, radius of circular path, $r=2 \mathrm{~m}$

Speed of circular path, $v=4 \mathrm{~m} / \mathrm{s}$

Acceleration of electron moving on circular path with constant speed is equal to centripetal acceleration.

$\therefore$ Acceleration of electron $(a)=$ Centripetal acceleration

$a=\frac{v^2}{r}=\frac{4^2}{2}=8 \mathrm{~m} / \mathrm{s}^2$

If v is instantaneous velocity, centripetal acceleration

$\begin{aligned} a_c & =\frac{v^2}{r} \\ \therefore \quad \frac{v^2}{r} & =k^2 r t^2 \\ \Rightarrow \quad v & =k r t \quad\left[\because a_c=k^2 r t^2 \text { is given }\right] \end{aligned}$

In circular motion, work done by centripetal force is always zero and work is done only by tangential force.

$\therefore$ Tangential acceleration

$a_t=\frac{d v}{d t}=\frac{d}{d t}(k r t)=k r$

$\therefore$ Tangential force $F_t=m a_t=m k r$

Power $P=F_t v=(m k r)(k r t)=m k^2 r^2 t$

At the lowest point of a vertical circle, the minimum velocity at the bottom $u_{\min }=\sqrt{5 g r}$

At $u=\sqrt{5 g r}$

Velocity at highest point $v=\sqrt{g r}$