Question 53: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

To solve for the angle of projection $	heta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.
The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:
$$ v = \frac{2\pi R}{T} $$
When the particle is projected with the same speed $v$ at an angle $	heta$ to the horizontal, its vertical component of velocity is given by $v_y=v\sin	heta$.
The maximum height $H$ reached by the projectile can be found from the kinematic equation:
$$ H = \frac{v_y^2}{2g} = \frac{(v\sin	heta)^2}{2g} $$
We are given that the maximum height attained $H$ is equal to $4R$, so:
$$ 4R = \frac{(v\sin	heta)^2}{2g} $$
Substitute $v$ from the first equation into the second one:
$$ 4R = \frac{((\frac{2\pi R}{T})\sin	heta)^2}{2g} $$
$$ 4R = \frac{(2\pi R\sin	heta)^2}{2gT^2} $$
$$ 4R = \frac{4\pi^2 R^2 \sin^2	heta}{2gT^2} $$
$$ 2gT^2 = \pi^2 R \sin^2	heta $$
Now solve for $\sin	heta$:
$$ \sin	heta =\left[\frac{2gT^2}{\pi^2 R}ight]^{1/2} $$
Finally, to solve for $	heta$, take the inverse sine of both sides:
$$ 	heta = \sin^{-1}\left(\left[\frac{2gT^2}{\pi^2 R}ight]^{1/2}ight) $$
Therefore, the correct answer is:
Option A
$$ \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}ight]^{1/2} $$

Answer

$\sin ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}$

Answer

$\sin ^{-1}\left[\frac{\pi^2 \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}$

Answer

$\cos ^{-1}\left[\frac{\pi \mathrm{R}}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}$

Answer

$\cos ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}$

Work, Energy and Power & Circular Motion

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