Problem: A small puck of mass $m$ is carefully placed onto the inner surface of the thin hollow thin cylinder of mass $M$ and of radius $R$. Initially, the cylinder rests on the horizontal plane and the puck is located at the height $R$ above the plane as shown in the figure on the left. Find the interaction force $F$ between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is $g$.

Solution: Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force $m g$ and the reaction force from the cylinder $N$. The cylinder is subject to the gravity force $M g$, the reaction force from the plane $N_{1}$, the friction force $F_{f r}$ and the pressure force from the puck $N^{\prime}=-N$. The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows

$$
m a_{x}=N \sin \alpha,
\tag{A.1}
$$

where $a_{x}$ is the horizontal projection of the puck acceleration.

For the cylinder the equation of motion with the acceleration $w$ is found as

$$
M w=N \sin \alpha-F_{f r} .
\tag{A.2}
$$

Since the cylinder moves along the plane without sliding its

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angular acceleration is obtained as

$$
\varepsilon=w / R
\tag{A.3}
$$

Then the equation of rotational motion around the center of mass of the cylinder takes the form

$$
I \varepsilon=F_{f r} R
\tag{A.4}
$$

where the inertia moment of the hollow cylinder is given by

$$
I=M R^{2}
\tag{A.5}
$$

Solving (A.2)-(A.5) yields

$$
2 M w=N \sin \alpha
\tag{A.6}
$$

From equations (A.1) and (A.6) it is easily concluded that

$$
m a_{x}=2 M w .
\tag{A.7}
$$

Since the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from (A.7) after integrating that

$$
m u=2 M v .
\tag{A.8}
$$

It is obvious that the conservation law for the system is written as

$$
m g R=\frac{m u^{2}}{2}+\frac{M v^{2}}{2}+\frac{I \omega^{2}}{2},
\tag{A.9}
$$

where the angular velocity of the cylinder is found to be

$$
\omega=\frac{v}{R}
\tag{A.10}
$$

since it does not slide over the plane.

Solving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as

$$
u=2 \sqrt{\frac{M g R}{(2 M+m)}}
\tag{A.12}
$$
$$
v=\frac{m}{M} \sqrt{\frac{M g R}{(2 M+m)}}
\tag{A.13}
$$

In the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius $R$ and, at the lowest point of its trajectory, have the velocity

and the acceleration

$$
v_{r e l}=u+v
\tag{A.14}
$$

$$
a_{\text {rel }}=\frac{v_{\text {rel }}^{2}}{R}
\tag{A.15}
$$

At the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15).

$$
F-m g=\frac{m v_{r e l}^{2}}{R}
\tag{A.16}
$$

then the interaction force between the puck and the cylinder is finally found as

$$
F=3 m g\left(1+\frac{m}{3 M}\right) \text {. }
\tag{A.17}
$$

Answer: $F=3 m g(1+\frac{m}{3 M})$