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{k r . t^{\alpha}} $
and $ a_{t}=\frac{d v_{t}}{d t}=\sqrt{(k r)} \cdot \frac{\alpha}{2} \cdot t^{\frac{\alpha}{2}-1}$
Now, power developed is
$=P=F v=m \,a_{t} \,v_{t}$
$=m \sqrt{k r} \cdot \frac{\alpha}{2} \cdot t^{\frac{\alpha}{2}-1}$
$P \propto t^{\alpha-1}$