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
A body of mass 100 g is moving in a circular path of radius 2 m on a vertical plane as shown in the figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s^2)
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: