ch15-p048 - 48. (a) The rotational inertia of a uniform rod...

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48. (a) The rotational inertia of a uniform rod with pivot point at its end is I = mL 2 /12 + mL 2 = 1/3 ML 2 . Therefore, Eq. 15-29 leads to () 2 2 1 3 2 3 2 28 ML gT T Mg L π = ¡ so that L = 0.84 m. (b) By energy conservation bottom of swing end of swing mm E EK U = ¡ = where U Mg =− A (c o s ) 1 θ with A being the distance from the axis of rotation to the center of mass. If we use the small angle approximation ( cos ≈−
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