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 equal-and-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. 10-18, their angular velocity (counterclockwise as seen in Fig. 11-48) is 1.4 m/s0.93 rad/s.1.5 mvrω===(c) Their rotational inertia is that of two particles in circular motion at r= 1.5 m, so Eq. 10-33 yields ()2222 50 kg 1.5 m225 kg m .Imr=⋅¦Therefore, Eq. 10-34 leads to 211225 kg m0.93rad/s98 J.KI⋅=(d) Angular momentum is conserved in this process. If we label the angular velocity
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.