newtonian mechanics - Mass inside cavity in block free to move

So this is a problem I've been trying to solve for a while now. I've tried conservation of energy with conservation of momentum, as well as setting up the Lagrangian, but I have no way of checking my work and would love an outside opinion.

Here is the question text: A small particle of mass m is held at rest at the edge of a hemispherical cavity made in a solid block of mass 3m. The block is placed on a horizontal surface. The particle is released and slides downward along the surface of the cavity. Treat the particle as a point and find the normal force acting on the block from the floor at the moment the point is at the bottom of the cavity. All surfaces are frictionless.

The normal force I have calculated is 11mg/2. The sticking point is I'm not sure if I should calculate the acceleration of the point mass using its velocity relative to the floor or relative to the block. Thoughts?

Answer

Using an inertial frame avoids the need for fictitious forces.

If you use the ground frame of reference, then you need to use the speed of the particle in this frame, and also the radius of curvature of the path of the particle in this frame. It is easy to calculate the speed of the particle in this frame, but it is not obvious what is the radius of curvature.

You cannot use the radius of the block : since the block is moving horizontally opposite to the direction of the particle, the particle falls and rises within a shorter horizontal distance than if the block were stationary. (See diagram below.) So the local radius of curvature of the path of the particle is smaller at the bottom of the cavity if the block is moving than if the block were stationary.

The alternative is to use the frame of reference in which the block is stationary.

When the particle reaches the bottom of the hemispherical cavity, the normal contact force between the two objects is vertical, so neither object is accelerating at this instant. That is, the block is an inertial frame of reference at this instant. So you can use the speed of the particle and the radius of curvature in this frame to find the centripetal acceleration of the particle. This method is easier than using the ground frame.

The acceleration, and therefore the normal force, is the same in both frames of reference, because there is no relative acceleration between them. So both methods will give the same answer.