Question

A particle of mass $m$ moves along a circle of radius $R$ with a normal acceleration varying with time as $w_n=a{t}^2$, where $a$ is a constant. If P denotes the time dependence of the power developed by all the forces acting on the particle, and $P_{avg}$ denotes the mean value of this power averaged over the first $t$ seconds after the beginning of motion. Find $\frac{P}{P_{avg}}$

Answer 1

We have $w_n=\frac{v^2}{R}=a{t}^2$, or, $v=\sqrt{aR}t$,
$t$ is defined to start from the beginning of motion from rest.
So, $w_t=\frac{dv}{dt}=\sqrt{aR}$
Instantaneous power, $P=\overrightarrow{F}.\overrightarrow{v}=m(w_t\widehat{u_t}+w_n\widehat{u_n})\cdot \left(\sqrt{aR}t\widehat{u_t}\right)$,

(where $\widehat{u_t}$ and $\widehat{u_n}$ are unit vectors along the direction of tangent (velocity) and normal respectively)

So, $P=m{w}_t\sqrt{aR}t=maRt$

Hence, the sought average power
$⟨P⟩=\frac{{\int }_0^{t}Pdt}{{\int }_0^{t}dt}=\frac{{\int }_0^{t}maRtdt}{t}$
Hence $⟨P⟩=\frac{maR{t}^2}{2t}=\frac{maRt}{2}$

$\frac{P}{P_{avg}}=2$