A rigid massless rod is rotated about one end in a horizontal circle.?

Question

There is a mass m1 attached to the center of the rod and a mass m2 attached to the outer end of the rod. The inner section of the rod sustains three times as much tension as the outer section. Find the ratio m2/m1

Answer 1

Let's say F1 and F2 are the tensions (forces) in the inner and outer sections of the rod respectively, w is the angular velocity, and r the distance to the center (half the length of the rod). Equating the radial force to (mass x centripetal acceleration) for each mass gives:

r*w^2*m1 = F1 - F2 = 2F2
(2r)*w^2*m2 = F2

Dividing the second equation by the first all variables cancel except for the ratio of the masses:

m2/m1 = 1/4

Answer 2

You need to know the horizontal tension on the rod. It will equal the horizontal component of the string tension to the outer mass. This will be determined by how fast the mass is orbiting, which we don't know.