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A simple pendulwm of length $L$ carries a bob of mass $m$. When the bob is at its lowest position, it is given that the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is
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Verified Answer
The correct answer is:
$\sqrt{10} m g$
According to the question, the bob is moving in a vertical circle, so the velocity at lowest point $v=\sqrt{5 r g}$ Now, applying conservation of mechanical energy, we get
$$
\frac{1}{2} m \times 5 r g=\frac{1}{2} m v^2+m g r
$$
or,
$$
\frac{5}{2} r g=\frac{v^2}{2}+r g \text { or, } \quad \frac{3}{2} r g=\frac{v^2}{2}
$$
or,
$$
v^2=3 r g
$$
Now, Horizontal force on the bob, $F x=F_c=\frac{m v^2}{r}$
$$
=\frac{m \times 3 r g}{r}=3 m g
$$
[from Eq. (i)]
Vertical force on the bob, $F_y=m g$
$$
\begin{aligned}
\therefore \quad F_{\text {net }} & =\sqrt{F_x^2+F_y^2} \\
& =\sqrt{(3 m g)^2+(m g)^2}=\sqrt{10} m g
\end{aligned}
$$
$$
\frac{1}{2} m \times 5 r g=\frac{1}{2} m v^2+m g r
$$
or,
$$
\frac{5}{2} r g=\frac{v^2}{2}+r g \text { or, } \quad \frac{3}{2} r g=\frac{v^2}{2}
$$
or,
$$
v^2=3 r g
$$
Now, Horizontal force on the bob, $F x=F_c=\frac{m v^2}{r}$
$$
=\frac{m \times 3 r g}{r}=3 m g
$$
[from Eq. (i)]
Vertical force on the bob, $F_y=m g$
$$
\begin{aligned}
\therefore \quad F_{\text {net }} & =\sqrt{F_x^2+F_y^2} \\
& =\sqrt{(3 m g)^2+(m g)^2}=\sqrt{10} m g
\end{aligned}
$$
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