This preview shows page 1. Sign up to view the full content.
51. We assume that from the moment of grabbing the stick onward, they maintain rigid
postures so that the system can be analyzed as a symmetrical rigid body with center of
mass midway between the skaters.
(a) The total linear momentum is zero (the skaters have the same mass and equaland
opposite velocities). Thus, their center of mass (the middle of the 3.0 m long stick)
remains fixed and they execute circular motion (of radius
r
= 1.5 m) about it.
(b) Using Eq. 1018, their angular velocity (counterclockwise as seen in Fig. 1148) is
1.4 m/s
0.93 rad/s.
1.5 m
v
r
ω
==
=
(c) Their rotational inertia is that of two particles in circular motion at
r
= 1.5 m, so Eq.
1033 yields
()
2
22
2 50 kg 1.5 m
225 kg m .
Im
r
=⋅
¦
Therefore, Eq. 1034 leads to
2
11
225 kg m
0.93rad/s
98 J.
KI
⋅
=
(d) Angular momentum is conserved in this process. If we label the angular velocity
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Center Of Mass, Mass, Momentum

Click to edit the document details