Solution

1. Kinematic Setup Let's set up a coordinate system with the origin at the center of the ring O, the y-axis pointing vertically upwards, and the x-axis horizontally. The stick is assumed to be horizontal and moves downwards with a constant speed $v_0$. The intersection point M is on the right side of the ring.

The angle $\phi$ is defined as the angle between the position vector OM and the negative y-axis. The coordinates of point M can be expressed in terms of the radius $r$ and the angle $\phi$ as:

$$x_M = r\sin\phi$$ $$y_M = -r\cos\phi$$

The position vector of M is $\vec{R}_M = r\sin\phi \, \hat{i} - r\cos\phi \, \hat{j}$.

2. Velocity of Intersection Point M The stick moves vertically downwards with a constant velocity $\vec{v}_{stick} = -v_0 \hat{j}$. Since point M is always on the stick, its vertical velocity component must be equal to the stick's velocity.

$$v_{My} = -v_0$$

We can also find the vertical velocity component by differentiating $y_M$ with respect to time:

$$v_{My} = \frac{dy_M}{dt} = \frac{d}{dt}(-r\cos\phi) = r\sin\phi \frac{d\phi}{dt}$$

Equating the two expressions for $v_{My}$:

$$r\sin\phi \frac{d\phi}{dt} = -v_0 \implies \frac{d\phi}{dt} = -\frac{v_0}{r\sin\phi}$$

Now, we find the horizontal velocity component by differentiating $x_M$ with respect to time:

$$v_{Mx} = \frac{dx_M}{dt} = \frac{d}{dt}(r\sin\phi) = r\cos\phi \frac{d\phi}{dt}$$

Substitute the expression for $\frac{d\phi}{dt}$:

$$v_{Mx} = r\cos\phi \left(-\frac{v_0}{r\sin\phi}\right) = -v_0 \cot\phi$$

The velocity vector of point M is:

$$\vec{v}_M = v_{Mx} \hat{i} + v_{My} \hat{j} = -v_0 \cot\phi \, \hat{i} - v_0 \, \hat{j}$$

The speed of point M is the magnitude of its velocity vector:

$$v_M = |\vec{v}_M| = \sqrt{(-v_0\cot\phi)^2 + (-v_0)^2} = v_0\sqrt{\cot^2\phi + 1} = v_0\sqrt{\csc^2\phi}$$ $$v_M = \frac{v_0}{\sin\phi}$$

3. Acceleration of Intersection Point M The acceleration of M is the time derivative of its velocity vector $\vec{v}_M$. Since $v_0$ is constant, we have:

$$\vec{a}_M = \frac{d\vec{v}_M}{dt} = \frac{d}{dt}(-v_0 \cot\phi \, \hat{i} - v_0 \, \hat{j})$$ $$\vec{a}_M = -v_0 \frac{d}{dt}(\cot\phi) \, \hat{i}$$

Using the chain rule:

$$\frac{d}{dt}(\cot\phi) = (-\csc^2\phi)\frac{d\phi}{dt}$$

Substituting this and the expression for $\frac{d\phi}{dt}$:

$$\vec{a}_M = -v_0 (-\csc^2\phi) \left(-\frac{v_0}{r\sin\phi}\right) \hat{i}$$ $$\vec{a}_M = -v_0 \left(\frac{1}{\sin^2\phi}\right) \left(\frac{v_0}{r\sin\phi}\right) \hat{i}$$ $$\vec{a}_M = -\frac{v_0^2}{r\sin^3\phi} \hat{i}$$

The magnitude of the acceleration is:

$$a_M = |\vec{a}_M| = \frac{v_0^2}{r\sin^3\phi}$$

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