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

Step 1: Picture the setup.
A bead sits on a spinning vertical ring. As the ring turns, the bead climbs up to some angle $\theta$ measured from the lowest point and stays there. Two effects balance, gravity pulling it down and the need for a center pull to keep it on a circle.

Step 2: Forces along the ring.
The bead moves in a small horizontal circle of radius $r\sin\theta$. The push it needs toward the center comes from a part of the normal force. Balancing the gravity component and this center pull along the ring gives \[ mg\sin\theta = m\omega^2 (r\sin\theta)\cos\theta \]
Step 3: Cancel common terms.
Both sides have $m$ and $\sin\theta$ (the bead is not at the very bottom), so we cancel them. \[ g = \omega^2 r \cos\theta \]
Step 4: Make cosine the subject.
\[ \cos\theta = \frac{g}{\omega^2 r} \]
Step 5: Put in the numbers.
Use $r = 0.25$ m, $\omega = 10$ rad/s, $g = 10$. \[ \cos\theta = \frac{10}{(10)^2 (0.25)} = \frac{10}{25} = 0.4 \]
Step 6: State the angle.
So the angle from the downward vertical is \[ \boxed{\theta = \cos^{-1}(0.4)} \]