hw11 - x =--16 . 1 1 / 3 . 11 + A cos0 . 567 t + B sin0 ....

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 11 (2.53, 2.81) 2.53 N C mg 8 15 8 15 N 2 F rx = m ¨ x c 8 17 N = m ¨ x c Also, F ry = m ¨ y c 15 17 N - mg = 0 N = 17 15 mg m ¨ x c = 8 17 17 15 mg ¨ x c = 8 15 g This the minimum acceleratio, a MIN , needed to prevent the ball from rolling down. Now, we are given that the acceleration is 2 a MIN . So an additional force, N 2 , is exerted between the vertical surface and the ball. F rx = m ¨ x c N 2 + 8 17 N = m 2 a MIN N + 2 + 8 17 17 15 mg = m 16 15 g 1
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N 2 = 8 15 mg = 8 15 0 . 06 (since weight is given to be 0.06 lb = 0 . 032 lb 2.81 x 100 150 - 1x F rx = m ¨ x c 150 - 1 x - 100 = 100 32 . 2 ¨ x c ¨ x + 1 3 . 11 x = 16 . 1 ¨ x + (0 . 5667) 2 x = 16 . 1 Solution of a general second order equation of this kind is given as ¨ y ( t ) + Py ( t ) + Q = 0 y ( t ) = - Q P + A cos Pt + B sin Pt Using the above to solve our equation, we get
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Unformatted text preview: x =--16 . 1 1 / 3 . 11 + A cos0 . 567 t + B sin0 . 567 t x = 50 + A cos0 . 567 t + B sin0 . 567 t (1) The initial conditions for the problem are 2 @ t = , x = @ t = , x = Using these on equation (1) we get A =-50 B = x = 50(1-cos0 . 567 t ) When x = 25 f t , the weight is at the top, 25 = 50(1-cos0 . 567 t TOP ) t TOP = 1 . 567 cos-1 1 2 = 1 . 85 sec At top, the velocity is determined as x = 50(0 . 567)sin(0 . 567)1 . 85 = 24 . 6 f t / sec 3...
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This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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hw11 - x =--16 . 1 1 / 3 . 11 + A cos0 . 567 t + B sin0 ....

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