Laws of Motion 5 Question 9

9. A smooth semicircular wire track of radius $R$ is fixed in a vertical plane (figure). One end of a massless spring of natural length $3 R / 4$ is attached to the lowest point $O$ of the wire track. A small ring of mass $m$ which can slide on the track is attached to the other end of the spring. The ring is held stationary at point $P$ such that the spring makes an angle $60^{\circ}$ with the vertical. The spring constant $k=m g / R$. Consider the instant when the ring is making an angle $60^{\circ}$ with the vertical. The spring is released, (a) Draw the free body diagram of the ring. (b) Determine the tangential acceleration of the ring and the normal reaction.

$(1996,5 M)$

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Solution:

  1. $C P=C O$

Radius of circle $(R)$

$\therefore \angle C P O=\angle P O C=60^{\circ}$

$\therefore \angle O C P$ is also $60^{\circ}$.

Therefore, $\triangle O C P$ is an

equilateral triangle.

Hence, $O P=R$

Natural length of spring is $3 R / 4$.

$\therefore$ Extension in the spring

$\Rightarrow$ Spring force,

$$ x=R-\frac{3 R}{4}=\frac{R}{4} $$

$$ F=k x=\frac{m g}{R} \quad \frac{R}{4}=\frac{m g}{4} $$

The free body diagram of the ring will be as shown below.

Here, $F=k x=\frac{m g}{4}$ and $N=$ Normal reaction.

(b) Tangential acceleration, $\boldsymbol{a}_{\boldsymbol{T}}$ The ring will move towards the $x$-axis just after the release. So, net force along $x$-axis

$$ \begin{aligned} F_{x} & =F \sin 60^{\circ}+m g \sin 60^{\circ} \\ & =\frac{m g}{4} \frac{\sqrt{3}}{2}+m g \frac{\sqrt{3}}{2} \\ F_{x} & =\frac{5 \sqrt{3}}{8} m g \end{aligned} $$

The free body diagram of the ring is shown below.

Therefore, tangential acceleration of the ring,

$$ a_{T}=a_{x}=\frac{F_{x}}{m}=\frac{5 \sqrt{3}}{8} g \Rightarrow a_{T}=\frac{5 \sqrt{3}}{8} g $$

Normal reaction, $N$ Net force along $y$-axis on the ring just after the release will be zero.

$$ \begin{aligned} & F_{y}=0 \\ \therefore \quad & N+F \cos 60^{\circ}=m g \cos 60^{\circ} \\ \therefore \quad & =m g \cos 60^{\circ}-F \cos 60^{\circ} \\ & =\frac{m g}{2}-\frac{m g}{4} \frac{1}{2}=\frac{m g}{2}-\frac{m g}{8} \\ & N=\frac{3 m g}{8} \end{aligned} $$



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