Question

A small bead is placed on a thin circular loop of radius \(25\,\text{cm}\) which is rotating about its vertical diameter with a constant velocity of \(10\,\text{rad s}^{-1}\). The angle made by the radius vector joining the center of the loop to the bead with the vertically downward direction is (Neglect friction and take \(g=10\,\text{m s}^{-2}\)):

• A. \(\cos^{-1}(0.4)\)
• B. \(\cos^{-1}(0.6)\)
• C. \(\sin^{-1}(0.4)\)
• D. \(\sin^{-1}(0.6)\)

Hint:

For a bead on a rotating hoop: \[ \cos\theta=\frac{g}{\omega^2 r} \] This is a standard equilibrium condition frequently used in rotational dynamics problems.

Answer 1

The Correct Option is A

Concept: For equilibrium of bead on a rotating circular hoop: \[ mg\sin\theta = m\omega^2 r\sin\theta\cos\theta \] Simplifying: \[ g=\omega^2 r\cos\theta \] Hence, \[ \cos\theta=\frac{g}{\omega^2 r} \]

Step 1: Substitute the given values. Radius: \[ r=25\,\text{cm}=0.25\,\text{m} \] Angular velocity: \[ \omega=10\,\text{rad s}^{-1} \] Thus, \[ \cos\theta = \frac{10}{(10)^2(0.25)} \] \[ = \frac{10}{25} \] \[ =0.4 \] Therefore, \[ \boxed{\theta=\cos^{-1}(0.4)} \]