A 39-kg girl sits near the rim of a 330-kg playground \"merry-go-round\" (large
ID: 1608925 • Letter: A
Question
A 39-kg girl sits near the rim of a 330-kg playground "merry-go-round" (large rotating disk) that spins at a rate of one turn in 2.2 s. Consider what happens when she carefully moves to the center of the disk.
Part A
If the radius of the disk is 1.5 m, compute its angular speed once the girl has made it all the way to the center. Approximate the "merry-go-round" as an ideal uniform disk, and treat the girl as a "point mass" (a good approximation as long as her diameter is much smaller than that of the disk!)
Explanation / Answer
initial angular momentum Li = (Idisk + m*R^2)*wi
wi = 2pi/2.2 = 2.86 rad/s
Li = ((1/2)*330*1.5^2 + (39*1.5^2))*2.86
Li = 1312.74 kg m^2/s^2
as the child moves to the center the inertia of child is zero
final angular momentum Lf = Idisk*wf
Lf = (1/2)*330*1.5^2*wf = 371.25 *wf
from conservation of angular momentum
Lf = Li
371.25 *wf = 1312.74
wf = 3.54 rad/s <<<==========ANSWER