= Solution
Let $\theta$ measure the spoke's displacement from downward equilibrium. The total <moment of inertia> about the axle is $I=MR_2^2+mR_1^2$: the rim is a thin <thin hoop>, and the attached body is a point <mass>. The rim's weight has no axle <torque>, while the attached <mass> gives restoring <torque> $-mgR_1\sin\theta$. Hence
$$
\boxed{(MR_2^2+mR_1^2)\ddot\theta+mgR_1\sin\theta=0,\quad \theta(0)=\theta_0,\quad\dot\theta(0)=0.}
$$
Equivalently <conservation of energy> gives $\tfrac12I\dot\theta^2=mgR_1(\cos\theta-\cos\theta_0)$. For small <angles>, $\sin\theta\simeq\theta$, so the motion is approximately $\theta=\theta_0\cos(\omega t)$ with $\omega^2=mgR_1/I$. The small-amplitude period is
$$
\boxed{T=2\pi\sqrt{\frac{MR_2^2+mR_1^2}{mgR_1}}.}
$$
At finite amplitude the nonlinear pendulum equation remains valid but this period approximation does not. The $M\to0$ limit recovers a simple pendulum of length $R_1$.
Back to article page
