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A light rigid wire of length $1 \mathrm{~m}$ is attached to a ball $A$ of mass $500 \mathrm{~g}$ to one end. The other end of the wire is fixed, so that the wire can rotate freely in the vertical plane about its fixed end. At the lowest point of the circular motion, the ball is given a horizontal velocity $6 \mathrm{~m} / \mathrm{s}$. Determined the radial component of the acceleration of the ball, when this rigid wire makes an angle $60^{\circ}$ with the upward vertical.
(Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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(Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$6 \mathrm{~m} / \mathrm{s}^2$
Given, length of a rigid wire, $l=1 \mathrm{~m}$ mass of the ball, $m=500 \mathrm{~g}=0.5 \mathrm{~kg}$ velocity of ball at a point, $C=v \mathrm{~m} / \mathrm{s}$
Motion of body in vertical plane is shown in the figure.
In $\triangle B O C$,
$$
\begin{aligned}
\cos 60^{\circ} & =\frac{O B}{O C} \\
\frac{1}{2} & =\frac{h}{1} \\
h & =\frac{1}{2} m \\
\therefore A B=A O+O B \quad & =l+h=1+\frac{1}{2} \\
A B & =\frac{3}{2} m
\end{aligned}
$$
According to the laws of conservation of energy, $(\mathrm{KE}+\mathrm{PE})$ at a point $A=(\mathrm{KE}+\mathrm{PE})$ at point $C$
$$
\begin{aligned}
& \frac{1}{2} m u^2+0=\frac{1}{2} m v^2+m g(A B) \\
& u^2=v^2+2 g(A B) \\
& 6^2=v^2+2 g \cdot \frac{3}{2} \quad \text { [From Eq. (i)] } \\
& 6^2=v^2+2 \times 10 \times \frac{3}{2} \\
& v^2=6
\end{aligned}
$$
$\therefore$ Radial component of acceleration of the ball
$$
a_C=\frac{v^2}{l}=\frac{6}{l}=6 \mathrm{~m} / \mathrm{s}^2
$$
Motion of body in vertical plane is shown in the figure.
In $\triangle B O C$,
$$
\begin{aligned}
\cos 60^{\circ} & =\frac{O B}{O C} \\
\frac{1}{2} & =\frac{h}{1} \\
h & =\frac{1}{2} m \\
\therefore A B=A O+O B \quad & =l+h=1+\frac{1}{2} \\
A B & =\frac{3}{2} m
\end{aligned}
$$
According to the laws of conservation of energy, $(\mathrm{KE}+\mathrm{PE})$ at a point $A=(\mathrm{KE}+\mathrm{PE})$ at point $C$
$$
\begin{aligned}
& \frac{1}{2} m u^2+0=\frac{1}{2} m v^2+m g(A B) \\
& u^2=v^2+2 g(A B) \\
& 6^2=v^2+2 g \cdot \frac{3}{2} \quad \text { [From Eq. (i)] } \\
& 6^2=v^2+2 \times 10 \times \frac{3}{2} \\
& v^2=6
\end{aligned}
$$
$\therefore$ Radial component of acceleration of the ball
$$
a_C=\frac{v^2}{l}=\frac{6}{l}=6 \mathrm{~m} / \mathrm{s}^2
$$
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