48. (a) The rotational inertia of a uniform rod with pivot point at its end is I= mL2/12 + mL2= 1/3ML2. Therefore, Eq. 15-29 leads to ()2213232 28MLgTTMg Lπ=¡so that L= 0.84 m. (b) By energy conservation bottom of swingend of swingmmEEKU=¡=whereUMg=−A(cos)1θwith Abeing the distance from the axis of rotation to the center of mass. If we use the small angle approximation (cos≈−
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