1. The problem statement, all variables and given/known data 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. Find the interaction force 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. 2. Relevant equations Inertia about CoM of cylinder I0 = MR2 Force of friction ##\equiv f## I use capitals for the cylinder and lower case for the point-mass. For example: Acceleration of point-mass ##\equiv <a_x,a_y>## Acceleration of cylinder's CoM ##\equiv A## Velocity of point-mass ##\equiv <v_x,v_y>## Velocity of cylinder's CoM ##\equiv V## 3. The attempt at a solution Net force in the horizontal (x) direction is friction: f = MA+max The only torque on the cylinder is friction: fR = IA/R = MRA f = MA This implies that the point mass just falls straight downwards... not what I expected but I'll roll with it. vy is zero at the lowest point, so conservation of energy gives: mgR = 0.5(MV2+I0ω2) = MV2 V2 = mgR/M If we move to the frame of the cylinder's CoM when the point mass is at the lowest point, then the point mass will be traveling in a circular arc (of radius R) at a speed which is the relative speed of the two objects. Since the point-mass is stationary, the relative speed is just the speed V of the cylinder's CoM. (No need to consider fictitious forces because the x-acceleration is momentarily zero.) F-mg = mV2/R = m2g/M F = mg(1+m/M) This is not right.