Question

A particle is moving in a circle of radius $R$ in such a way that any instant the total acceleration makes an angle of ${45}^o$ with radius. The initial speed of the particle is $v_0$. The time taken to complete the first revolution is

• A. $\frac{R}{v_0}{e}^{-2\pi }$
• B. $\frac{R}{v_0}(1-e^{-2\pi })$
• C. $\frac{R}{v_0}$
• D. $\frac{2R}{v_0}$

Answer 1

The correct option is B $\frac{R}{v_0}(1-e^{-2\pi })$

The tangential acceleration is along the tangent to the circle and centripetal acceleration is along the radius of the circle. Since the total acceleration makes $45°$ angle with the radius, this concludes that tangential acceleration magnitude is equal to the magnitude of centripetal acceleration.

$\therefore a_t=a_c\Rightarrow \frac{dv}{dt}=\frac{v^2}{R}\Rightarrow \int \frac{R}{v^2}dv=\int dt\Rightarrow \frac{R}{v}=t+C$

To evaluate $C$, put $V=V_0$ at $\therefore C=-\frac{R}{V_0}\therefore \frac{R}{v}=t-\frac{R}{v_0}\Rightarrow t=R\left[\frac{v-v_0}{v_{0}v}\right]$

also $v=\frac{ds}{dt}$ where $s$ is the arc length covered by the particle.

$\therefore t=\frac{R}{v_0}-\frac{R}{\frac{ds}{dt}}=\frac{R}{v_0}-R\frac{dt}{ds}\Rightarrow tds=\frac{R}{v_0}-Rdt$

$\Rightarrow \int \frac{ds}{R}=\int \frac{dt}{\frac{R}{v_0}-t}\Rightarrow \frac{s}{R}=-\ln (\frac{R}{v_0}-t)+c^{{}^{\prime }}$

Putting $s=0$ at $t=0$

$c^{{}^{\prime }}=\frac{\ln R}{v_0}\therefore \frac{s}{R}=\frac{\ln R}{v_0}-\ln (\frac{R}{v_0}-t)=\ln \left(\frac{R}{R-v_{0}t}\right)$

For one revolution $s=2\pi R$, $t=T$

$\frac{2\pi R}{R}=\ln \left(\frac{R}{R-v_{0}t}\right)\Rightarrow T=\frac{R}{v_0}(1-e^{-2\pi })$