A thin circular ring of mass $M$ and radius $R$ is rotating about its central axis with a constant angular velocity $\omega$. Two objects, each of mass $m$, are gently attached to the opposite ends of a diameter of the ring. The new angular velocity of the ring is:
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Since angular momentum is conserved, increasing the moment of inertia must decrease the angular velocity. The ratio of new velocity to old is the inverse ratio of their moments of inertia: $\frac{M}{M+2m}$.
Step 1: Understand the setup. A ring of mass $M$ and radius $R$ spins about its center at angular speed $\omega$. Two small objects, each of mass $m$, are gently stuck at the two ends of a diameter. We need the new spin speed.
Step 2: Use conservation of angular momentum. No outside twist acts on the ring, so its angular momentum stays the same. \[ I_1\omega_1 = I_2\omega_2 \] where $I$ is the moment of inertia, the spinning version of mass.
Step 3: Find the starting moment of inertia. For a ring spinning about its center, all the mass sits at distance $R$. \[ I_1 = MR^2 \]
Step 4: Find the new moment of inertia. The two added masses also sit at distance $R$ from the center. Each adds $mR^2$. \[ I_2 = MR^2 + mR^2 + mR^2 = (M+2m)R^2 \]
Step 5: Plug into the conservation rule. \[ (MR^2)\,\omega = (M+2m)R^2\,\omega_2 \] The $R^2$ cancels from both sides.
Step 6: Solve for the new speed. \[ \omega_2 = \frac{M\omega}{M+2m} \] Adding mass makes the ring harder to spin, so the speed drops. \[ \boxed{\omega_2 = \frac{M\omega}{M+2m}} \]
