A uniform disk of mass 10M and radius 3R can rotate freely around its fixed cent

Question

A uniform disk of mass 10M and radius 3R can rotate freely around its fixed center like a merry-go-round. A smaller uniform disk of mass M and radius R lies on top of the larger disk, concentric with it. Initially both disks rotate together with angular velocity . Then a small disturbance causes the smaller disk to slide outward until its edge rests against the edge of the larger disk. The disks rotate together without further sliding. Derive an expression for the final angular speed of the disks, in terms of system parameters.

http://web.mst.edu/~vojtaa/engphys1/handouts/HW23.pdf (see question 4 for diagram)

Explanation / Answer

When stuck together, the disks have equal angular velocities,
regardless of whether they share the same central axis.
The initial rotational inertia of the system is
Ii = Ibigdisk + Ismalldisk where,
Ibigdisk = 1/2 MR2
Since the small disk is initially concentric with the big disk,
Ismalldisk = 1/2 mr2

After it slides, the rotational inertia of the small disk is found from the PARALLEL AXIS THEOREM :
I = ICM + MD2
I = 1/2 MR2 + 1/2 mr2 + m(R-r)2

Using the conservation of angular momentum,
Iii = Iff
we solve for the final angular velocity,
f = i[(1/2 MR2 + 1/2 mr2) / (1/2 MR2 + 1/2 mr2 + m(R-r)2)]
Substituting M = 10M and R = 3R, this becomes
f = i(91/99)
Since i = 20 rad/s,
f = 18 rad/s

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