Two bodies of equal mass m are connected by a rope passing…

Two bodies of equal mass $m$ are connected by a rope passing through a hole in a horizontal table. One body moves on the table in a circle of radius $R$ around the hole; the other hangs below. What angular velocity must the circling body have for the hanging body to stay at the same height? Neglect all friction.

Explanation of Method

A circling body holding up a hanging body.

Step 1: Hanging body at rest

\[ T = mg. \]

Step 2: Circling body

The same tension is the only horizontal force on it, so it supplies the centripetal force: $T = m\omega^2 R$.

Step 3: Condition

\[ m\omega^2 R = mg. \]

\[\omega = \sqrt{\frac{g}{R}}\]

$T$ $T$ $mg$ $R$ $\omega$ $T = mg = m\omega^2 R$

The same rope tension holds the hanging body up and pulls the circling body toward the hole. It balances the weight exactly when it equals the needed centripetal force.

Result

\[\boxed{\,\omega = \sqrt{\frac{g}{R}}\,}\]

Remark

Check. The original result is correct. For example, with $R = 0.5\ \mathrm{m}$: $\omega \approx 4.43\ \mathrm{rad/s}$, about 0.7 revolutions per second. If it spins faster, the hanging body rises; slower, and it sinks.