Problem 19.58 In Example 19.6, we neglected the
moments of inertia of the two masses mabout the axes
through their centers of mass in calculating the total
angular momentum of the person, platform, and masses.
Suppose that the moment of inertia of each mass about
the vertical axis through its center of mass is IM=
0.001 kg-m2. If the person’s angular velocity with her
arms extended to r1=0.6misω1=1 revolution per
second, what is her angular velocity ω2when she pulls
the masses inward to r2=0.2 m? Compare your result
r1
r1
r2
Solution: Using the numbers from Example 19.6, we conserve
angular momentum
HO1=(IP+2mr2
1+2IM)ω1
=(0.4 kg-m2+2[4 kg][0.6m]
2+2[0.001 kg-m2])1rev
s
HO2=(IP+2mr2
2+2IM)ω2
s.
Problem 19.59 Two gravity research satellites (mA=
250 kg, IA=350 kg-m2;mB=50 kg, IB=16 kg-m2)
are tethered by a cable. The satellites and cable
rotate with angular velocity ω0=0.25 rpm. Ground
controllers order satellite Ato slowly unreel 6 m of
additional cable. What is the angular velocity afterward?
A
B
0
ω