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 ‘θ’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by

• A. \(\theta=sin^{-1}(\frac{2gT^2}{\pi^2R})^{\frac{1}{2}}\)
• B. \(\theta=cos^{-1}(\frac{gT^2}{\pi^2R})^{\frac{1}{2}}\)
• C. \(\theta=cos^{-1}(\frac{\pi^2R}{gT^2})^{\frac{1}{2}}\)
• D. \(\theta=sin^{-1}(\frac{\pi^2R}{gT^2})^{\frac{1}{2}}\)

Answer 1

The Correct Option is A

To solve the given problem, let's break it down step by step and use the necessary physics concepts and formulae: 

Understanding the Problem: A particle is moving in a circular path with radius \( R \) and uniform speed, completing one full revolution in time \( T \). When projected at an angle \(\theta\) with the initial speed, the particle's maximum height (\( H \)) is given as \( 4R \).

Formulas and Concepts:

• The speed \( v \) of the particle in the circular path can be calculated using the formula for the circumference of the circle: \(v = \frac{2\pi R}{T}\).
• For projectile motion, the vertical component of velocity is \(v_y = v \sin\theta\), and the horizontal component is \(v_x = v \cos\theta\).
• The maximum height \( H \) for a projectile is given by: \(H = \frac{v_y^2}{2g} = \frac{(v \sin\theta)^2}{2g}\).

Substitute Values:

• From the given information, \( H = 4R \). Therefore, we equate it to the expression for height: \(\frac{(v \sin\theta)^2}{2g} = 4R\).
• Substitute \( v = \frac{2\pi R}{T} \) into the equation: \(\frac{\left(\frac{2\pi R}{T} \sin\theta\right)^2}{2g} = 4R\).

Solve for \(\theta\):

• Solve the equation for \(\sin^2\theta\): \(\frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} = 4R\).
• Simplify to find: \(\sin^2\theta = \frac{2gT^2}{\pi^2R}\).
• Thus, the angle of projection \(\theta\) can be expressed as: \(\theta = \sin^{-1} \left(\sqrt{\frac{2gT^2}{\pi^2R}}\right)\).

Conclusion: The correct option is: \(\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{\frac{1}{2}}\). This matches the given correct answer.