Tricky physics problem about rod touching a box

Question

A light rod of lenght l is attached to the ground A with a movable joint, which is in a perpendicular equilibrium position. On the top of the rod, a sphere of mass m is touching a box of mass M next to it. After bringing the rod out of equilibrium, the rod is moving to the right pushing the box on a horizontal frictionless surface.

Find:

a) what is the necessary ratio of masses M/m when both objects move away from one another, and the rod and the surface make an agle of π/6 rad.

b) What will be the velocity of the box u during the moment a)

Answer 1

Great question. Before diving into the math, here’s a simpler way to think about what’s going on.

Imagine the rod like a bat swinging down and the box like a hockey puck sitting on frictionless ice. As the rod swings, it pushes the box. Since the surface is frictionless, the only force on the box is from the rod tip.

Now here’s the key insight: the moment the rod stops pushing the box (they lose contact), it must be because the tip of the rod and the box are moving at the exact same horizontal speed. If the rod tip were faster, it would still be pushing. If it were slower, it would fall behind. So we analyze the system at that exact moment — when both are moving together horizontally but no longer touching.

a) To find the mass ratio M/m at that moment:

The horizontal speed of the tip of the rod is v_tip = l * omega * cos(theta). At theta = pi/6, cos(pi/6) = sqrt(3)/2, so v_tip = l * omega * (sqrt(3)/2). The box must have this exact speed at the moment of separation.

Now apply conservation of momentum. The rod has mass m (concentrated at the top), and the box has mass M. The horizontal speed of the rod’s center of mass is v_cm = (l/2) * omega * (sqrt(3)/2). Since there’s no external force in the horizontal direction, total momentum must stay zero.

So:

M * u = m * v_cm

M * (l * omega * sqrt(3)/2) = m * (l/2 * omega * sqrt(3)/2)

Cancel common terms, solve, and you get:

M = m / 2

So the ratio M/m = 1/2

b) The velocity of the box at that moment is:

u = l * omega * (sqrt(3)/2)

Hope that clears it up. Let me know if you’d like help visualizing the torque and motion in more detail. This kind of question shows up a lot in mechanics problems that combine rotation and linear motion... somewhat common on the MCAT just a little bit more simplified!