Kinetic Energy and Angular Momentum of General Rotation System

[slayerdeus]

Five identical particles of mass m = 0.30 kg are mounted at equal intervals on a thin rod of length l = 1.01 m and mass M = 2.0 kg, with one mass at each end of the rod. If the system is rotated with angular velocity = 60 rev/min about an axis perpendicular to the rod through one of the end masses, determine
(a) the kinetic energy and
(b) the angular momentum of the system.

I have KE = 1/2*Irod*w^2 + 1/2*Iparticle*w^2
= 1/2(1/3*Mass of rod + 1/2*Mass of particles)*L^2*w^2

What am I doing wrong? I haven't looked at part b yet.

 

[Janitor]

Are you using a separate value of L for each of the particles?

 

[HallsofIvy]

You can treat the rod as an addition particle with mass 2.0 kg and positioned at distance 0.55 m from the axis of rotation. Since one of the particles is at the axis of rotation, it isn't moving and contributes nothing to the problem. You can think of this as five particles, moving in circles of radius 0.55 m (for the 2.0 kg mass) and of radius
1.1/4= 0.275 m, 2.2/4= 0.55 m, 3.3/4= 0.825 m, and 4.4/4= 1.1 m. (for the 3 kg masses)
(Actually, you could combine the mass at 0.55 and the rod into one 5 kg mass at 0.55 m if that is easier.)