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

Correct Answer: 8

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