# Rigid massless rod with three point masses attached Inertia and potential energy

## Homework Statement

An object consists of three point masses (m1=0.2 kg, m2=0.3 kg and m3=0.5 kg) attached to a massless rigid rod. The masses are located at 0 m, 0.4 m, and 0.8 m respectively. (The rod is 0.8 m in length so the 0.5 kg mass is attached to one end and the 0.2 kg mass to the other).
The center of mass is located at 0.52 m.

a) The object is pivoted about an axis so that it rotates from the position of the 0.2 kg mass.

(0.2-------0.3-------0.8 with circular rotation originating at the first mass). What is the object's moment of inertia about that axis?

b) Starting from rest the object is allowed to swing from horizontal to vertical (so the rod is now pointing down). By how much does the potential energy of the object change?

c) from (a) and (b)'s answers, calculate rotational kinetic energy.

I=Icm+MD2
Icm=???
E=1/2Iw2

## The Attempt at a Solution

I attempted 0.522*1 (since one is the total weight) + 0.5*0.522 and got an answer of 0.4 something. I have a study guide that tells me this answer is incorrect. My main problem with this question is calculating the inertia and then figuring out how it relates to the potential energy! Any help would be appreciated.