Question

A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by

• A. \( \sin^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right] \)
• B. \( \sin^{-1} \left[ \sqrt{\frac{\pi^2R}{2gT^2}} \right] \)
• C. \( \cos^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right] \)
• D. \( \cos^{-1} \left[ \sqrt{\frac{\pi R}{2gT^2}} \right] \)

Answer 1

The Correct Option is A

Given:

\[ V = \frac{2\pi R}{T} \]

The maximum height attained by the particle is given by:

\[ H = \frac{v^2 \sin^2 \theta}{2g} \]

We are given that:

\[ 4R = \frac{4\pi^2 R^2 \sin^2 \theta}{T^2 \cdot 2g} \]

Simplifying:

\[ \sin^2 \theta = \frac{2gT^2}{\pi^2 R} \]

Taking the square root:

\[ \sin \theta = \sqrt{\frac{2gT^2}{\pi^2 R}} \]

Thus:

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

Answer 2

The problem describes a particle first moving in a uniform circular motion and then projected as a projectile with the same speed. We are given the parameters of both motions and asked to find the angle of projection.

Concept Used:

1. Uniform Circular Motion: The speed (\(v\)) of a particle moving in a circle of radius \(R\) with a time period \(T\) is the circumference of the circle divided by the time period.

\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{2\pi R}{T} \]

2. Projectile Motion: When a particle is projected with an initial speed \(u\) at an angle \(\theta\) to the horizontal, the maximum vertical height (\(H\)) it attains is given by the formula:

\[ H = \frac{u^2 \sin^2\theta}{2g} \]

where \(g\) is the acceleration due to gravity.

Step-by-Step Solution:

Step 1: Determine the speed of the particle from its circular motion.

The particle completes one revolution (a distance of \(2\pi R\)) in time \(T\). Therefore, its uniform speed \(v\) is:

\[ v = \frac{2\pi R}{T} \]

Step 2: Use the information about the projectile motion.

The particle is projected with the same speed, so the initial speed of projection is \(u = v = \frac{2\pi R}{T}\).

The maximum height attained by the projectile is given as \(H = 4R\).

Step 3: Substitute the known values into the formula for the maximum height of a projectile.

The formula is \(H = \frac{u^2 \sin^2\theta}{2g}\). Substituting \(H = 4R\) and \(u = \frac{2\pi R}{T}\):

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

Step 4: Simplify the equation to solve for \(\sin\theta\).

First, expand the squared term:

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

Combine the terms in the numerator:

\[ 4R = \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} \]

We can cancel \(4R\) from both sides (assuming \(R \neq 0\)):

\[ 1 = \frac{\pi^2 R \sin^2\theta}{2gT^2} \]

Now, rearrange the equation to isolate \(\sin^2\theta\):

\[ \sin^2\theta = \frac{2gT^2}{\pi^2 R} \]

Step 5: Find the angle \(\theta\).

Take the square root of both sides to find \(\sin\theta\):

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

Finally, express \(\theta\) using the inverse sine function:

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

The angle of projection \(\theta\) is given by \(\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}\).