Block On Turntable problem: A 200 g, 40.0-cm-diameter solid cylindrical turntabl
ID: 1283922 • Letter: B
Question
Block On Turntable problem:
A 200 g, 40.0-cm-diameter solid cylindrical turntable rotates on frictionless bearings at 60.0 rpm. A 20.0 g block sits at the center of the turntable. The block then slides outward along a frictionless groove in the surface of the turntable and stops when it reaches the edge.
For this problem, you can assume that the turntable can be treated as a solid disk and the block can be treated as a point mass.
What is the turntable's angular speed when the block reaches the outer edge of the turntable?
Explanation / Answer
Solve by conservation of angular momentum
Iw = Iw
The initial I is only of the disk
I is .5mr2 for a disk, so that is the initial I
The final I is the disk plus the block
I = .5mr2 + m'r2
60 rpm converts to 6.28 rad/s, so...
(.5)(.2)(.2)2(6.28) = [.5(.2)(.22) + .02(.2)2](w)
w = 5.24 rad/s
If you need that converted back into rpm, that is 50 rpm