- #1

- 53

- 0

## Main Question or Discussion Point

I have in my problem, a ball sitting on a cylindrical rod that pivots at the bottom,

some guy in the explanation said parallel axis theorem and came up with moment of inertia of:

I = (mL^2)/3 + [2Mr^2/5 + M(L+r)^2]

where L is the length of the rod, m is the mass of the rod, M is the mass of the ball, r is the radius of the ball.

some guy in the explanation said parallel axis theorem and came up with moment of inertia of:

I = (mL^2)/3 + [2Mr^2/5 + M(L+r)^2]

where L is the length of the rod, m is the mass of the rod, M is the mass of the ball, r is the radius of the ball.