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Unformatted text preview: 2 v = √ (2gh) = 3.13 m/s During the collision, the angular momentum of the system is conserved. Since the rod and block rotate together after the collision, the total rotational inertia of the system is I tot = I ro + I bl = (1/3)ML 2 + mL 2 = 0.181 kg·m 2 The angular velocity just after the collision is l i = l f mvL = I tot ω ω = mvL/I tot = 2.08 rad/s b) What is θ ? While the rod and block are rotating together, only the gravitational force does work on the system. Therefore, the mechanical energy of the system is conserved. When the rod makes angle θ to the horizontal, the change in height of the block is L(1 – cos θ ) and that of the center of mass of the rod is (L/2)(1 – cos θ ). We have E i = E f (1/2)I tot ω 2 = mgL(1 – cos θ ) + Mg(L/2)(1 – cos θ ) cos θ = 1 - I tot ω 2 /(2Lg(m + M/2)) = 0.857 θ = 31.0º...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
- Summer '08