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 first need to understand the physics behind circular and projectile motion.

1. Understand the Circular Motion:
   • The particle is moving in a circle of radius $R$ with uniform speed. It takes time $T$ to complete one revolution.
   • Thus, the speed $v$ of the particle is given by the circumference divided by time period: $$ v = \frac{2\pi R}{T} $$
2. Understand the Projectile Motion:
   • The particle is then projected at an angle $\theta$ with the same speed $v$.
   • The vertical component of this velocity is $v\sin\theta$.
   • The maximum height $H$ attained by a projectile is given by: $$ H = \frac{(v\sin\theta)^2}{2g} $$
   • According to the problem, the maximum height is $4R$. Therefore, $$ \frac{(v\sin\theta)^2}{2g} = 4R $$
   • Substitute $v = \frac{2\pi R}{T}$ in the equation: $$ \frac{\left(\frac{2\pi R}{T} \sin\theta\right)^2}{2g} = 4R $$
3. Simplification:
   • Simplifying, we get: $$ \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} = 4R $$
   • Cancel out $R$ from both sides: $$ \frac{4\pi^2 R \sin^2\theta}{2gT^2} = 4 $$
   • Simplify further: $$ \pi^2 R \sin^2\theta = 2gT^2 $$
   • Divide both sides by $ \pi^2 R $: $$ \sin^2\theta = \frac{2gT^2}{\pi^2 R} $$
4. Find the Angle:
   • Take the square root of both sides: $$ \sin\theta = \left(\frac{2gT^2}{\pi^2 R}\right)^{1/2} $$
   • Therefore, the angle $\theta$ is given by: $$ \theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2} $$

Hence, the correct option is $\theta=\sin ^{-1}\left(\frac{2 gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$.