PROBLEM A: A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path. Find the velocity v of the cube as it leaves the block.
PROBLEM B: A small block slides from rest from the top of a frictionless sphere of radius R. How far below the top x does it lose contact with the sphere? The sphere does not move.
Problem A: Conservation of Energy, Conservation of Momentum
Problem B: Conservation of Energy, FBD on m.
The Attempt at a Solution
I've solved both problems with what is the correct answer. Problem A: v = [(2gR)/(1+m/M)]1/2; Problem B: R/3.
However, I'm looking back at how both of these problems are correctly solved and am second-guessing the thinking behind both of them.
PROBLEM A: Examining the M-m system, the net external force-x equals zero, so momentum-x is conserved in the system. I understand this. From the energy perspective though, we consider the m-M-earth system. I define h=0 to be the vertical height at which m is leaving M with a horizontal velocity at. So, our initial energy (ONLY the grav. potential energy of m) must equal our final energy (the kinetic energy of m AND M). As M is a rigid body, we may examine any one point on it and see that its change in grav. potential energy is zero, so we are not concerned with that.
But, what if I were for some reason analyze just the m-earth system with the same h=0 height? Now, we have the equation that the initial grav. potential of m is equal to the final kinetic energy of ONLY m, clearly disagreeing with our equation from before. But what would make this second, "wrong" equation wrong and our first equation right? In the m-earth system, no friction from M acts on m (we are told its frictionless) AND the only force on m by M would be a normal force that is always perpendicular to m's displacement. So wouldn't energy be conserved in our "incorrect" way of looking at things? What's going on here?
PROBLEM B: The problem in this case is solved correctly by ONLY analyzing the m-earth system. (I'll say that m is the mass of the sliding block.) The mass of the sphere isn't even assigned a variable (although neither was the sliding block) and we are told that the sphere does not move. Analyzing the m-earth system (with h=0 being the vertical position of the "flight" point), we get that our initial energy (grav. potential of m) must equal our final energy (kinetic energy of m at "flight" point or at the angle with respect to the vertical at which it takes flight). Taking this with a FBD equation for the centripetal acceleration of m solves the problem.
So, similar to my "incorrect" way of thinking about problem A, we are only considering the m-earth system. Because there's no friction AND because the normal force acting on m is perpendicular to m's displacement, energy should be conserved in this system, making our equations useful. ...But if it's conserved in the m-earth system in this case, why is it presumably not conserved in my "incorrect" way of thinking of problem A?
Additionally, what if we were to consider problem B from an m-M(sphere)-earth perspective. Is energy conserved in this system? My assumption is that energy would be conserved BECAUSE we are told that the sphere does not move. As the diagram shows the sphere resting on some sort of stand attached to the table, I assume that this stand/table must be exerting a force onto the sphere to keep it from moving, BUT (this just hit me as I was about the argue that energy should not be conserved in this system) that his force is balancing what every force little m is exerting downwards, at an angle, on the M(sphere), meaning that the net force on the sphere is zero, meaning that no work is done on the sphere and that energy is conserved in the entire m-M(sphere)-earth system. Then, from this perspective, you would write down your energy equations only to see that kinetic-initial, kinetic-final, potential-initial, and potential-final of the sphere all equal zero, returning us back to the same equation in the "correct" way of solving the problem.
So, I think I've convinced myself that the m-Earth system analysis works in Problem B (it came to me as I was about to type why I think it should not work), but why does it not work for Problem A?
Additionally, as a side question, to have gravitational potential energy due to the earth, it makes sense for us to include the earth in our system, right? (Otherwise energy would clearly not be conserved.) So my notation of m-earth system and such is correct, right? Additionally, we would then have to take into account the kinetic and potential energy of the earth, but this would be calculated from our reference frame, right? (meaning that we would find the velocity of the earth and potential energy of it to always be zero?)
Thank you so much.