Question

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t = 0$ is $4 \, \text{m/s}$, the time taken to complete the first revolution will be$\frac{1}{\alpha} \left[ 1 - e^{2\pi} \right] \, \text{s},$where $\alpha = \, \underline{\hspace{2cm}}$.

Answer 1

**Given: **
- The radius $r = 0.5 \, \text{m}$,
- The initial velocity $v_0 = 4 \, \text{m/s}$.

The particle’s normal and tangential accelerations are equal, so:
$\frac{v^2}{r} = \frac{dv}{dt}$

Rearrange the equation to separate variables:
$v \, dv = r \, dt$

Integrating both sides from $t = 0$ to $t = T$ and $v = 4 \, \text{m/s}$ to $v = v_T$, we get:
$\int_4^{v_T} v \, dv = r \int_0^T dt$

Now calculating this we get:
$\int_4^{v_T} v \, dv = rT$

This can be simplified as:
$v = \frac{4}{1 - 8t} \, \frac{ds}{dt}$

Where $r = 0.5 \, \text{m}$ and $s = 2\pi r = \pi$ for a complete revolution.

Now we integrate the expression for $s$:
$4 \times \int_0^T \frac{1}{(1 - 8t)} \, dt = \pi$

$\int (\ln(1 - 8t)) = \frac{1}{8} \, \text{[for complete revolution at} \, s = 2\pi]$

From this, solving for $\alpha$, we get:
$\alpha = 8$

Thus, the value of $\alpha$ is 8.

The Correct Answer is: 8