Question

For a particle moving along circular path, the radial acceleration $a_r$ is proportional to time $t$. If $a_t$ is the tangential acceleration, then which of the following will be independent of time?

• A. $a_t$
• B. $a_r.a_t$
• C. $\frac{a_r}{a_t}$
• D. $a_r(a_t{)}^2$

Answer 1

The correct option is D $a_r(a_t{)}^2$

$a_r\propto t\therefore a_r=kt$
$\Rightarrow \frac{v^2}{R}=kt\Rightarrow v^2=kRt$-----(1)
$\Rightarrow v=\sqrt{kRt}$
Also differentiating equation 1 on both sides w.r.t time
$2v\frac{dv}{dt}=kR$
$\therefore \frac{dv}{dt}=\frac{kR}{2v}=\frac{kR}{2\sqrt{kRt}}=a_t$
$\therefore a_t^2\times a_r=\frac{k^2{R}^2}{4\times kRt}\times kt=\frac{k^{2}R}{4}$ = constant
i.e., independent of time.