qz5sol_6942s11

# qz5sol_6942s11 - From part b velocity of the block is v =...

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TA: Tomoyuki Nakayama Monday, February 28, 2011 PHY 2048: Physic 1, Discussion Section 6942 Quiz 5 (Homework Set #7) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A uniform solid sphere of mass M = 15.0 kg and radius R = 0.50 m can rotate about a vertical axis on frictionless bearings. A massless cord passes around the equator of the sphere, over a pulley of rotational inertia I P = 0.400 kg·m 2 and radius r = 0.20 m, and is attached to a small object of mass m = 5.00 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. The solid sphere has a rotational inertial of I S = (2/5)MR 2 a) What is the acceleration of the block? (If you want, you can solve this part after part b).)

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Unformatted text preview: From part b), velocity of the block is v = 3.06 m/s after it falls a distance d = 2.00 m. Applying one of the kinematics equations, we have v 2 – 0 2 = 2ad ⇒ a = v 2 /(2d) = 2.34 m/s 2 b) What is the speed of the object when it has fallen a distance 2.00 m after being released from rest? Only the gravitational force exerted on the block does work on the system. Hence the mechanical energy of the system is conserved. The energy conservation equation yields mgd = (1/2)mv 2 + (1/2)I S ω S 2 + (1/2)I P ω P 2 where d = 2 m. Since the cord does not slip on the sphere or pulley, the angular velocity of the sphere and pulley are related to the translational velocity of the block as v = R ω S = r ω P Expressing ω S and ω P in terms of v and plugging them into the energy conservation equation, the speed of the block after the block falls distance d is mgd = (1/2)mv 2 + (1/2)(2/5)MR 2 (v/R) 2 + (1/2)I P (v/r) 2 v = √ [2mgd/((2/5)M+ I P /r 2 + m )] = 3.06 m/s...
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## This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.

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qz5sol_6942s11 - From part b velocity of the block is v =...

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