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
- \( \sin^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right] \)
- \( \sin^{-1} \left[ \sqrt{\frac{\pi^2R}{2gT^2}} \right] \)
- \( \cos^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right] \)
- \( \cos^{-1} \left[ \sqrt{\frac{\pi R}{2gT^2}} \right] \)
The Correct Option is A
Approach Solution - 1
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:
\[ \theta = \sin^{-1} \left( \sqrt{\frac{2gT^2}{\pi^2 R}} \right) \]
Approach Solution -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 \(\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}\).
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