How Does Angular Velocity Change in a Swinging Rod with Masses?

[dkgojackets]

Homework Statement

A massless rod length L has a small mass m attached to the center and another mass m attached at one end. On the opposite end, the rod is hinged to a frictionless hinge. The rod is released from rest at a horizontal position and swings down. What is the angular velocity as it swings through its lowest (vertical) point? Solve in terms of g and L.

Homework Equations

use moment of inertia, energy conservation

The Attempt at a Solution

I determined the total moment of inertia of the two masses to be (5/4)mL^2. I know that initial potential energy is 2mgL (setting bottom point of swing as zero). At the bottom, the mass in the middle has PE mgL, the system has KEr of (.5)Iw^2, and both masses have a translational kinetic energy. All of this must sum to the intial 2mgL? I think I am just missing a step in the algebra.

 

[Doc Al]

The position of the mass in the middle is L/2 from one end, not L (redo your PE terms). Also, you can treat the system as being in pure rotation about the hinge.

 

[m2003]

Your approach is correct. To find the angular velocity, you can use the principle of conservation of energy. At the top of the swing, the only form of energy present is potential energy, which is equal to 2mgL. At the bottom of the swing, the potential energy is converted into kinetic energy, which is equal to the sum of the rotational kinetic energy and the translational kinetic energy. This can be expressed as:

2mgL = (1/2)(5/4)mL^2w^2 + 2(1/2)mw^2

Solving for w, we get:

w = √(8g/5L)

Therefore, the angular velocity at the bottom of the swing is √(8g/5L).