Inertia of a pendulum with disc

1. Aug 15, 2012

A pendulum consists of a uniform thin rod of mass 5 kg and length 2 m to which is fixed a circular disc of mass 8 kg and radius 0.4 m. There is a pivot at one end.

(a) Find the CoM and Moment of Inertia when the disc is:
(i) half way along the rod;
(ii) at the opposite end of the rod to the pivot.

I cannot do part i. I found the moment of inertia of the rod to be
I = (1/3)ML^2
I = 1/3*5*2^2
I = 6.667 kgm^2

but I don't know how to calculate for the disc.

For part ii. I found the CoM using
M = ((5*1)+(8*2))/13
M = 1.62m from pivot

I found Inertia using:
I = (1/12)ML^2 +(1/2)MR^2+ML^2
I = 5.507 kgm^2

Have I got the correct answer for part ii? How would I calculate for part i? Also, how would I find angular velocity given angular impulse? Any help greatly appreciated.

2. Aug 15, 2012

Simon Bridge

Hint: parallel-axis theorem.
You seem to have assumed that the CoM of the disk is at the end of the rod ... is that actually the case or is the disc attached to the rod by it's rim?
You seem to have the mass of each object to be the same in the equation for I (same M in each term).
Where do you get the (1/12) part from?

Last edited: Aug 15, 2012