Consider the following situation: A uniform disk turns at 3.7 rev/s about a fric
ID: 1416815 • Letter: C
Question
Consider the following situation: A uniform disk turns at 3.7 rev/s about a frictionless spindle. A non-rotating rod, of the same mass as the disk and length equal to the diameter of the disk, is dropped onto the freely spinning disk. They turn together around the spindle, without slipping, with their centers right above one another. What is the moment of inertia for the spinning object(s) before the rod is dropped onto the disk? What is the moment of inertia for the spinning object(s) after the rod is dropped onto the disk? What is the angular velocity in rad/s of the system after the rod is dropped onto the disk?Explanation / Answer
The formula for angular momentum is:
L = Iw
Conservation of angular momentum says that L before = L after:
I1w1 = I2w2
1.
The moments of inertia of a disk or cylinder through the center is:
I = 1/2mr2
2.
Moment of inertia of a thin rod through the center is
I = 1/12ml2
and the rod is 2r long, so the moment of inertia in terms of r is:
I = 1/12m(2r)2
I = 4/12mr2 = 1/3mr2
So the moment of inertia before the rod is dropped on is just the disk:
I1 = 1/2mr2
And after it the rod is dropped on, it is both moments:
I2 = 1/2mr2 + 1/3mr2 = 5/6mr2
3.
since ,
I1w1 = I2w2
(1/2mr2)(7.0 rev/s) = (5/6mr2)w2
(1/2)(7.0 rev/s) = (5/6)w2
w2 = 4.2 rev/s