Question

A weightless rod of length l with a small load of mass m at the end is hinged at point A (Fig. 20) and occupies a strictly vertical position, touching a body of mass M. A light jerk sets the system in motion.

For what mass ratio M/m will the rod form an angle α = pi/6 with the horizontal at the moment of the separation from the body? What will be the Velocity u of the body at this moment? Friction should be neglected.

Answer 1

As long as the load touches the body, the velocity of the latter is equal to the horizontal component of the velocity of the load, and the acceleration of the body is equal to the horizontal component of the acceleration of the load.

Let a be the total acceleration of the load. Then we can write a = a_t + a_c, where a_c is the centripetal acceleration of the load moving in the circle of radius l, i.e., a_c = v²/l, where v is the velocity of the load (Fig. 154). The horizontal component of the acceleration is

The body also has the same acceleration. We can write the equation of motion for the body:

where N is the force of normal pressure exerted

by the load on the body. At the moment of separation of the load, N = 0 and 

The acceleration component at at the moment of separation of the load is only due to the force of gravity:

Thus, the velocity of the load at the moment of separation is

and the velocity of the body at the same moment is

According to the energy conservation law, we have

Substituting the obtained expression for v at the moment of separation and the value of sin α = sin pi/6 = 1/2 into this equation, we obtain the ratio:

The velocity of the body at the moment of separation is