Question

An ant of mass $m$ travels tangentially along a ring with radius $r$, mass $M$ and negligible thickness. When it reaches the opposite side of the ring (you can mark a dot opposite to its initial position to indicate its destination), find the angle through which the ring has rotated. The ring can only rotate about its center and cannot translate.

Answer 1

$\pi$

Answer 2

Let $\phi$ be the angle of the ant's position on the ring, measured in the ring's own coordinate system. Let $\theta$ be the angle of rotation of the ring relative to the lab frame. The ant moves tangentially along the ring, which means its velocity relative to the ring is tangential.

Let $\vec{r}_{a/R}'(t) = (r\cos\phi(t), r\sin\phi(t))$ be the ant's position in the ring's frame. The velocity of the ant relative to the ring is $\vec{v}_{a/R}' = (-r\sin\phi \dot{\phi}, r\cos\phi \dot{\phi})$.

The problem states that the ring cannot translate, implying its center is fixed. However, for the system to be isolated and have conserved momentum, the center of mass must be stationary. This leads to a relationship between the ring's center velocity and the ant's velocity relative to the ring. If we assume the system is isolated (which is typical for such problems, and the constraint "cannot translate" might be misleading or imply external forces keeping the CM fixed), the center of mass of the ant-ring system is stationary.

Let the center of mass be the origin. The position of the ring's center $\vec{R}$ is related to the ant's position relative to the ring $\vec{r}_{a/R}'$ by: $\vec{R}(t) = -\frac{m}{m+M} \vec{r}_{a/R}'(t)$.

The velocity of the ring's center is $\vec{V}_R = \frac{d\vec{R}}{dt} = -\frac{m}{m+M} \frac{d\vec{r}_{a/R}'}{dt} = -\frac{m}{m+M} \vec{v}_{a/R}'$. Substituting the components of $\vec{v}_{a/R}'$: $\vec{V}_R = -\frac{m}{m+M} (-r\sin\phi \dot{\phi}, r\cos\phi \dot{\phi})$.

The ring rotates with angular velocity $\omega_R = \dot{\theta}$. The velocity of the ring's center can also be expressed as $\vec{V}_R = \vec{\omega}_R \times \vec{R}$. $\vec{R} = -\frac{m}{m+M}(r\cos\phi, r\sin\phi)$. $\vec{V}_R = \dot{\theta}\hat{k} \times (-\frac{m}{m+M})(r\cos\phi \hat{i} + r\sin\phi \hat{j}) = -\frac{m}{m+M} r \dot{\theta} (-\sin\phi \hat{i} + \cos\phi \hat{j})$.

Comparing the two expressions for $\vec{V}_R$: $-\frac{m}{m+M} (-r\sin\phi \dot{\phi}, r\cos\phi \dot{\phi}) = -\frac{m}{m+M} r \dot{\theta} (-\sin\phi \hat{i} + \cos\phi \hat{j})$. This equality implies $\dot{\phi} = \dot{\theta}$.

The ant travels tangentially along the ring and reaches the opposite side. This means the angle $\phi$ in the ring's frame changes by $\pi$ radians. So, $\Delta\phi = \pi$.

Since $\dot{\phi} = \dot{\theta}$, integrating both sides with respect to time gives $\Delta\phi = \Delta\theta$. Therefore, the angle through which the ring has rotated is $\Delta\theta = \pi$.