Question

An object moves along the circle with normal acceleration proportional to $t^{\alpha }$, where $t$ is the time and $\alpha$ is a positive constant. The power developed by all the forces acting on the object will have time dependence proportional to

• A. $t^{\alpha -1}$
• B. $t^{\alpha /2}$
• C. $t^{\frac{1+\alpha }{2}}$
• D. $t^{2\alpha }$

Answer 1

Answer: (A)

Given, $a_c\propto t^{\alpha }$
or$a_c=k{t}^{\alpha }$
Now,$a_c=\frac{v_t^2}{r}$
(where, $v_t=$ tangential component of velocity)
$\Rightarrow v_t=\sqrt{kr.t^{\alpha }}$
and $a_t=\frac{d{v}_t}{dt}=\sqrt{(kr)}\cdot \frac{\alpha }{2}\cdot t^{\frac{\alpha }{2}-1}$
Now, power developed is
$=P=Fv=m{a}_t{v}_t$
$=m\sqrt{kr}\cdot \frac{\alpha }{2}\cdot t^{\frac{\alpha }{2}-1}$
$P\propto t^{\alpha -1}$