Question

$A$ particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle ' $\theta$ ' to the horizontal, the maximum height attained by it equals $4 R$. The angle of projection, $\theta$, is then given by :

• A. $\theta=\cos ^{-1}\left(\frac{ gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$
• B. $\theta=\cos ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}$
• C. $\theta=\sin ^{-1}\left(\frac{\pi^{2} R}{ g T ^{2}}\right)^{1 / 2}$
• D. $\theta=\sin ^{-1}\left(\frac{2 gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the motion of a particle in a circular path and its subsequent projectile motion.

Step 1: Understanding Uniform Circular Motion

A particle moving in a circle of radius \(R\) with a uniform speed completes one revolution in time \(T\). The velocity \(v\) of the particle is given by the formula of circular motion:

\(v = \frac{2\pi R}{T}\)

Step 2: Transition to Projectile Motion

The particle is projected with the same speed \(v\) at an angle \(\theta\) to the horizontal. The components of this velocity are:

• Horizontal velocity: \(v_x = v \cos\theta\)
• Vertical velocity: \(v_y = v \sin\theta\)

Step 3: Maximum Height of Projectile

The maximum height \(H_{\text{max}}\) reached by a projectile is given by:

\(H_{\text{max}} = \frac{v_y^2}{2g} = \frac{(v \sin\theta)^2}{2g}\)

It is given that this maximum height is equal to \(4R\):

\(\frac{(v \sin\theta)^2}{2g} = 4R\)

Step 4: Substituting the Known Values

Substitute \(v = \frac{2\pi R}{T}\) into the equation above:

\(\frac{\left(\frac{2\pi R}{T} \sin\theta \right)^2}{2g} = 4R\)

Simplify the equation:

\(\frac{(2\pi R)^2 \sin^2\theta}{2g T^2} = 4R\)

Rearranging gives:

\(\sin^2\theta = \frac{8g T^2}{(2\pi)^2 R}\)

Simplify this to:

\(\sin^2\theta = \frac{2g T^2}{\pi^2 R}\)

Step 5: Solving for \(\theta\) 

Finally, take the square root of both sides:

\(\sin\theta = \left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}\)

Therefore, the angle of projection is:

\(\theta = \sin^{-1}\left(\frac{2g T^2}{\pi^2 R}\right)^{1/2}\)

Thus, the correct option is:

\(\theta=\sin ^{-1}\left(\frac{2 gT ^{2}}{\pi^{2} R }\right)^{1 / 2}\)