Question 102: Work, Energy and Power & Circular Motion

Question

A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is $$\alpha$$ with respect to point P. Which of the following graphs represent the correct relation between A and $$\alpha$$ when ball goes from Q to R?

Accepted Answer

At Q ;
The equation of motion of the ball is
$$
\begin{aligned}
& \frac{m v^{2}}{R}=N-m g \sin \alpha \\
& N=\frac{m v^{2}}{R}+m g \sin \alpha .......(i)
\end{aligned}
$$
From figure, $$h = R\sin \alpha $$
From law of conservation of energy,
$$
(mg)R \sin \alpha=\frac{1}{2} m v^{2}
$$
$	herefore \quad m v^{2} / R=2 m g \sin \alpha$
So, Eq. (i) becomes,
$N=2 m g \sin \alpha+m g \sin \alpha=3 m g \sin \alpha$
$	herefore$ Ratio, $\frac{m v^{2}}{R N}=\frac{2}{3}=$ constant
$A=2 / 3$, which is independent of $\alpha$.

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Work, Energy and Power & Circular Motion

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