sample_final_ss

# sample_final_ss - ME 274 Spring 2008 Final Examination -...

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1 ME 274 – Spring 2008 SOLUTION Final Examination - Wednesday PROBLEM NO. 1 Given : Disk P (having a mass of m = 2 kg) is constrained to move on a smooth, horizontal surface. An inextensible cord is attached to P with the cord passing through a hole in the surface at point O. A particle having a mass of M = 8 kg is attached to the other end of the cord, as shown below. At position 1 where R = 2 meters, P has a velocity that is perpendicular to OP with v 1 = 3 meters/sec. At position 2 where R = 1.5 meters, the velocity of P has both e R and e ! components. Find : For position 2 of P, a) find the value of ! = d / dt . b) find the value of ! R = dR / dt . c) write down the velocity vector for P in terms of its polar coordinates. FBD of P – shown Angular momentum H O ( ) 1 = mr P / O ! v P 1 = m R 1 e R ( ) ! v 1 e " ( ) = mR 1 v 1 k H O ( ) 2 = mr P / O ! v P 2 = m R 2 e R ( ) ! ! R 2 e R + R 2 ! 2 e ( ) = mR 2 2 ! 2 k M O ! = 0 " H O ( ) 1 = H O ( ) 2 " mR 1 v 1 = mR 2 2 ! # 2 " ! 2 = R 1 v 1 R 2 2 = 2 ( ) 3 ( ) 1.5 ( ) 2 = 2.67 rad / sec Work-energy (for P and B together) T 1 + V 1 + U 1 ! 2 nc ( ) = T 2 + V 2 " 1 2 mv P 1 2 + Mg R 1 # R 2 ( ) = 1 2 mv P 2 2 + 1 2 Mv B 2 2 Kinematics v P 2 = ! R 2 e R + R 2 ! e " v P 2 2 = ! R 2 2 + R 2 ! 2 ( ) 2 v B 2 = ! R 2 Solving F B O P e R e ! R B

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1 2 mv 1 2 + Mg R 1 ! R 2 ( ) = 1 2 m ! R 2 2 + R 2 ! " 2 ( ) 2 # \$ % ' ( + 1 2 M ! R 2 2 = 1 2 m R 2 ! 2 ( ) 2 + 1 2 m + M ( ) ! R 2 2 ) ! R 2 = ± mv 1 2 + 2 Mg R 1 ! R 2 ( ) ! m R 2 ! 2 ( ) 2 m + M = ± v 1 2 + 2 M / m ( ) g R 1 ! R 2 ( ) ! R 2 ! 2 ( ) 2 1 + M / m = ± 3 2 + 2 ( ) 8 / 2 ( ) 9.806 ( ) 2 ! 1.5 ( ) ! 1.5 ( ) 2 2.67 ( ) 2 1 + 8 / 2 ( ) = ± 2.54 m / sec ! v 2 e ( ) m / sec
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## This document was uploaded on 12/25/2011.

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sample_final_ss - ME 274 Spring 2008 Final Examination -...

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