I don’t even know where to begin on this one!

A light rod 1.00 m in length rotates about an axis perpendicular to its length and passing through its center as in the figure below. Two particles of masses 4.00 kg and 3.00 kg are connected to the ends of the rod.

(a) Neglecting the mass of the rod, what is the system’s kinetic energy when its angular speed is 2.50 rad/s?

(b) Repeat the problem, assuming the mass of the rod is taken to be 2.10 kg.

### 1 Answer

The kinetic energy of a mass M rotating at distance R from an axis with angular velocity w , is

K = (1/2)Iw^2

where I is the mass’s moment of inertia, which is just MR^2 if its concentrated at distance R.

So for part (a)

K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2

=(1/2)[M1 + M2]R^2w^2

= (1/2)4 + 3^2(2.5)^2

= 5.47 J

In part (b) you have to add the moment of inertia of the rod which is I = (1/12)ML^2 = (1/12)(2.1)(1)^2 = .175

K = (1/2)M1R^2w^2 + (1/2)M2R^2w^2 + (1/2)Iw^2

= 5.47 + (1/2)(.175)(2.5)^2

= 6 J