KCET · Physics · Laws of Motion
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)
- A \(\frac{\pi}{4}\)
- B zero
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
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
\(\frac{T \sin \theta}{T \cos \theta}=\frac{m v^2 / r}{m g} \Rightarrow \tan \theta=\frac{v^2}{r g}=\frac{10^2}{10 \times 10}=1 \)
\(\Rightarrow \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad}\)
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
\(\frac{T \sin \theta}{T \cos \theta}=\frac{m v^2 / r}{m g} \Rightarrow \tan \theta=\frac{v^2}{r g}=\frac{10^2}{10 \times 10}=1 \)
\(\Rightarrow \tan \theta=\tan 45^{\circ} \Rightarrow \theta=45^{\circ}=\frac{\pi}{4} \mathrm{rad}\)
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