Two particles A and B with masses m and 2m respectively are

Two particles A and B with masses m and 2m, respectively, are linked by a massless rigid rod and slide along a frictionless circular path of radius r. If the particles are released from rest at theta = 0 degree, find the speed of each particle and the angular velocity of the rod when the rod achieves the horizontal position. State your answers in terms of g and r. Motion occurs in the vertical plane.

Solution

The velocity of the particle in circular motion is given by,

v = r * omega

The velocities of masses m and 2m is assumed to be,

v = r * omega

The radius of the circles can be seen as a string with a ball at its end and under tension, the force required to stay intact is given by,

T = mg cos theta (centripetal)

The tangential force is mg sin theta which creates motion theta,

mg sin theta = m * r * omega^2

replace omega = v / r,

mg sin theta = m * v^2 / r

When the masses are released, the eqn of motion is given by,

T = mg sin theta

mg cos theta = m * v^2 / r

solve for V in the relation,

v = sqrt (r cos theta * g)

Since the velocity is independent of the mass of the particle but the displaced angle is always shifted by 90 deg,

Velocity of mass, 2m = sqrt (r cos theta * g)

Velocity of mass, m = sqrt (r cos (90+theta) * g)

The angular velocity of the rod omega1 given by,

omega 1 = v / r

omega 1 = sqrt (r cos theta * g) / r