Source: Physics Competition - Mechanics
Problem Sets:
Problem
As shown in Figure, rod OA of length R rotates in a vertical plane about a horizontal axis through point O. Its endpoint A is attached to a light, inextensible string that passes over fixed pulleys B and C to a block M. Point B is directly above O at a distance H. At a certain instant, the angular velocity of the rod is $\omega$ and the angle between the string segment BA and the vertical line OB is $\alpha$.
P0726-problem-1
The speed of block M, $v_M$, is equal to the rate of change of the length of the string segment AB. This is the component of the velocity of point A, $\vec{v}_A$, along the direction of the string AB.
The speed of point A is $v_A = \omega R$, and its direction is perpendicular to the rod OA.
$$v_M = \vec{v}_A \cdot \hat{u}_{AB} = v_A \cos\psi$$where $\psi$ is the angle between the vector $\vec{v}_A$ and the string AB.
In triangle OAB, let $\angle OAB = \gamma$. Since $\vec{v}_A$ is perpendicular to OA, the angle between $\vec{v}_A$ and AB is $\psi = 90^\circ - \gamma$.
$$v_M = \omega R \cos(90^\circ - \gamma) = \omega R \sin\gamma$$Applying the law of sines to triangle OAB:
$$\frac{H}{\sin\gamma} = \frac{R}{\sin\alpha}$$ $$\sin\gamma = \frac{H\sin\alpha}{R}$$Substituting $\sin\gamma$ into the expression for $v_M$:
$$v_M = \omega R \left( \frac{H\sin\alpha}{R} \right) = \omega H \sin\alpha$$