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qz8sol_3885f10

# qz8sol_3885f10 - 2 v = √(2gh = 3.13 m/s During the...

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TA: Tomoyuki Nakayama Monday, October 25, 2010 PHY 2048: Physic 1, Discussion Section 3885 Quiz 8 (Homework Set #9) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ In the figure below right, a small 150 g block slides down a frictionless surface through height h = 50.0 cm and then sticks to a uniform rod of mass 400 g and length 80.0 cm. The rods pivots about point O though angle θ before stopping. a) What is the angular velocity of the rod-block system about point O just after the block sticks to the rod? The surface is frictionless. Thus the mechanical energy of the block is conserved while it is sliding down the surface. The velocity of the block just before hitting the rod is E i = E f mgh = (1/2)mv
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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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