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A car is moving in a circular horizontal track of radius $10 \mathrm{~m}$ with a constant speed of $10 \mathrm{~ms}^{-1}$. A bob is suspended from the roof of the car by a light wire of length $1.0 \mathrm{~m}$. The angle made by the wire with the vertical is (in rad)
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Radius of circular track, $r=10 \mathrm{~m}$ speed of car, $v=10 \mathrm{~m} / \mathrm{s}$
The given situation is shown below
$$
\begin{aligned}
& T \sin \theta=\frac{m v^2}{r} \\
& T \cos \theta=m g
\end{aligned}
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
& \frac{T \sin \theta}{T \cos \theta}=\frac{m v^2 / r}{m g} \Rightarrow \tan \theta=\frac{v^2}{r g}=\frac{10^2}{10 \times 10}=1 \\
& \Rightarrow \quad \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad}
\end{aligned}
$$
The given situation is shown below
$$
\begin{aligned}
& T \sin \theta=\frac{m v^2}{r} \\
& T \cos \theta=m g
\end{aligned}
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
& \frac{T \sin \theta}{T \cos \theta}=\frac{m v^2 / r}{m g} \Rightarrow \tan \theta=\frac{v^2}{r g}=\frac{10^2}{10 \times 10}=1 \\
& \Rightarrow \quad \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad}
\end{aligned}
$$
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