An object is made from 2 solid disks: m1 = 23 kg, m2 = 39kg, and r1 = 17cm, and r2 = 33cm. The disks are connected by a thin rod of M = 11kg, L = 67cm. The object is held in a horizontal position and then let go. The object is free to rotate about a pivot through centre of mass of of larger disk. Find the speed of the CM of smaller disk when it has swung 90degrees from original horizontal position.
I = ML^2 / 3 (rod - axis running through edge perpendicular to bar) or ML^2/12 for CM axis?
I = mr^2 / 2 (disks)
The Attempt at a Solution
To compute the net rotational inertia:
I (total) = I(rod) + I(small disk) + I(large disk)
Do I also have to incorporate the rotational inertia of both the small disk and rod from the distance from CM of large disk (which acts as pivot)?
(ie. would I also include mr^2 (small disk, where r is length from CM of large disk), mr^2 (for rod))?
And for the height that the CM of small disk falls (which is on the far end), would it be the length of rod + radii of both small and large disk?