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

Answer 2

Given: At every instant the magnitude of tangential acceleration equals the magnitude of centripetal acceleration: \[ |a_t|=|a_c|. \] For motion on a circle of radius \(r\), \[ a_c=\frac{v^2}{r},\qquad a_t=\frac{dv}{dt}. \] Hence \[ \frac{v^2}{r}=\frac{dv}{dt}. \] Initial speed \(v(0)=4\ \text{m s}^{-1}\) and radius \(r=0.5\ \text{m}\).

1) Solve the differential equation for \(v(t)\). \[ \int_{4}^{v}\frac{dv}{v^2}=\int_{0}^{t}\frac{dt}{r} \;\Longrightarrow\; \Big[-\frac{1}{v}\Big]_{4}^{v}=\frac{t}{r}. \] With \(r=0.5\) (so \(1/r=2\)): \[ -\frac{1}{v}+\frac{1}{4}=2t \;\Longrightarrow\; \frac{1}{v}=\frac{1}{4}-2t \;\Longrightarrow\; v(t)=\frac{4}{1-8t}. \] This may be written as \(v(t)=\dfrac{4}{1-\alpha t}\), so by comparison \[ \boxed{\alpha=8\ \text{s}^{-1}}. \]

2) Time for one revolution (check step in figure).
Since \(v=\dfrac{ds}{dt}\), \[ \frac{ds}{dt}=\frac{4}{1-8t} \;\Longrightarrow\; \int_{0}^{s}\!ds=4\int_{0}^{t}\frac{dt}{\,1-8t\,}. \] For one full round, arc length \(s=2\pi r=\pi\). Compute the integral: \[ \pi =4\left[\frac{-1}{8}\ln(1-8t)\right]_{0}^{t} = -\frac{1}{2}\ln(1-8t). \] Thus \[ \ln(1-8t)=-2\pi \;\Longrightarrow\; 1-8t=e^{-2\pi} \;\Longrightarrow\; t=\frac{1-e^{-2\pi}}{8}\ \text{s}. \]

Final: The parameter in the velocity law \(v=\dfrac{4}{1-\alpha t}\) is \[ \boxed{\alpha=8}, \] and the time to complete one revolution is \[ \boxed{t=\dfrac{1-e^{-2\pi}}{8}\ \text{s}}. \]