Example 8-16 A Girl on a Merry-Go-Round
The girl in Figure 8-27 has a mass of 30 kg. She runs toward the merry-go-round at 3.0 m>s, then jumps on. The
merry-go-round is initially at rest, has a mass of 100 kg and a radius of 2.0 m, and can be treated as a uniform disk.
Find the rotation speed of the merry-go-round after the girl jumps on.
Set Up
As we described above, angular momentum
is conserved in this process. Before she jumps
on, the girl has all of the angular momentum,
with a magnitude given by Equation 8-23.
After she jumps on, she and the merry-goround rotate together with the same angular
speed v. Together they behave like a single
object made up of a rotating disk with an
extra mass (the girl) at its rim.
Angular momentum of the girl
before jumping on:
(8-23)
L = rp› = rp sin f
before girl jumps on
merry-go-round (top view)
=0
rotation axis
Angular momentum of the merrygo-round with the girl riding on it:
(8-22)
Lz = Ivz
Moment of inertia of the merrygo-round plus girl:
(8-6)
I = Imerry-go-round + Igirl
R
Figure 8-27 shows that the distance
r sin f is just equal to the radius of the
merry-go-round, R = 2.0 m. Use this to find
the magnitude of the angular momentum
before she jumps on.
After she jumps on, the magnitude of the
angular momentum is L = Iv, where v is what
we are trying to find. Calculate the value of I
for the system of merry-go-round plus girl.
girl
mgirl
after girl jumps on (top view)
rotation axis
Solve
vgirl
merry-go-round
with girl
Before the girl jumps on, her linear momentum has magnitude
p = mgirl vgirl = 130 kg2 13.0 m>s2 = 90 kg # m>s
Her angular momentum is
L = rp sin f = (r sin f)p
= Rp = 12.0 m2 190 kg # m>s2 = 180 kg # m2 >s
Moment of inertia of the merry-go-round alone, considered as a
uniform cylinder: From Table 8-1,
Imerry@go@round =
1
1
Mmerry@go@roundR2 = 1100 kg2 12.0 m2 2 = 200 kg # m2
2
2
Moment of inertia of the girl alone, considered as an object of mass
mgirl rotating a distance R from the axis:
Igirl = mgirlR2 = (30 kg)(2.0 m)2 = 120 kg # m2
Moment of inertia of the merry-go-round plus girl:
I = Imerry-go-round + Igirl = 200 kg # m2 + 120 kg # m2 = 320 kg # m2
Angular momentum is conserved, so set the
magnitude of angular momentum before the
girl jumps on equal to the ­magnitude after
she jumps on. Then solve for the final angular speed v. Note that if L is in kg # m2 >s
and I is in kg # m2, then v is in rad>s.
Reflect
To get a better sense of our result, we convert
it to rev>min. Our answer says that the
merry-go-round spins 5.4 times per minute.
In other words, it makes one revolution
every (1>5.4) minute, or about once every 11
s—a reasonable pace for a merry-go-round.
Before the girl jumps on, L = 180 kg # m2 >s
After she jumps on, angular momentum is Lz = Ivz
with magnitude L = Iv = (320 kg # m2)v
Angular momentum is conserved, so the magnitude L is the same before
and after:
180 kg # m2 >s = 1320 kg # m2 2v
180 kg # m2 >s
= 0.56 rad>s
v =
320 kg # m2
Convert v to revolutions per minute:
v = 10.56 rad>s2 a
= 5.4 rev>min
10.562 1602 rev
1 rev
60 s
ba
b =
2p rad 1 min
2p
min