A smooth metal ring of radius R = 0.8 m lies in a horizontal…
A smooth metal ring of radius $R = 0.8\ \mathrm{m}$ lies in a horizontal plane. A rod AB rotates about the point A on the ring with constant angular velocity $\omega = 8\ \mathrm{rad/s}$. A small bead C is threaded on both the ring and the rod. Find the velocity and acceleration of the bead.
Explanation of Method
A bead on a ring pushed by a rotating rod.
Step 1: Where is the bead?
The bead sits where the rod crosses the ring. Let the rod make angle $\varphi$ with the diameter through A. The angle $\varphi$ at A is an inscribed angle, and the angle at the centre O on the same arc is twice as large:
\[ \angle(\text{diameter}, OC) = 2\varphi. \]
Step 2: Angular velocity of the bead about O
The rod turns uniformly, $\varphi = \omega t$, so the bead's angle about the centre is $2\omega t$. The bead moves uniformly around the ring with angular velocity
\[ \Omega = 2\omega = 16\ \mathrm{rad/s}. \]
Step 3: Velocity
\[ v = \Omega R = 2\omega R = 2 \cdot 8 \cdot 0.8. \]
\[v = 12.8\ \mathrm{m/s} \ \text{(tangent to the ring)}\]
Step 4: Acceleration
Uniform circular motion, so only a normal acceleration toward the centre O: \[a = \Omega^2 R = 4\omega^2 R = 4 \cdot 64 \cdot 0.8.\]
\[a = 204.8\ \mathrm{m/s^2} \ \text{(toward the centre of the ring)}\]
Result
\[\boxed{\,v = 12.8\ \mathrm{m/s} \ \text{(tangent to the ring)}\,}\]
\[\boxed{\,a = 204.8\ \mathrm{m/s^2} \ \text{(toward the centre of the ring)}\,}\]
Remark
About the original solution. It assumed the bead moves like the end of a rod of length $R$ turning at $\omega$, which gave 6.4 m/s and 51.2 m/s². But the rod turns about A on the ring, not about the centre, and the inscribed-angle theorem doubles the bead's angular velocity about O. That makes the speed twice and the acceleration four times as large.