Hw 4 Solutions

# Rad s rad s s 2 4 p1037 we use the parallel axis

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. rad / s rad / s s. 2 4. P.10.37. We use the parallel axis theorem: I = I com + Mh 2 , where I com is the rotational inertia about the center of mass (see Table 10-2(d)), M is the mass, and h is the distance between the center of mass and the chosen rotation axis. The center of mass is at the center of the meter stick, which implies h = 0.50 m – 0.20 m = 0.30 m. We find ( )( ) 2 2 2 com 1 1 0.56 kg 1.0 m 4.67 10 kg m . 12 12 I ML = = = × 2 Consequently, the parallel axis theorem yields ( )( ) 2 2 2 2 2 4.67 10 kg m 0.56 kg 0.30 m 9.7 10 kg m . I = × + = × 5. P.10.39. The particles are treated “point-like” in the sense that Eq. 10-33 yields their rotational inertia, and the rotational inertia for the rods is figured using Table 10-2(e) and the parallel-axis theorem (Eq. 10-36). (a) With subscript 1 standing for the rod nearest the axis and 4 for the particle farthest from it, we have

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