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ME240_CA1_W08_Solution

# ME240_CA1_W08_Solution - ME 240 Introduction to Dynamics...

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ME 240: Introduction to Dynamics and Vibrations Mechanical Engineering Department The University of Michigan Winter 2008 Computer Assignment #1 Solution February 8, 2007 Prepared by Joosup Lim ( [email protected] ) (i) Free body diagram is shown in Fig.( 1 ). m θ R e ^ n e ^ t θ F D mg N Figure 1: Free Body Diagram (ii) Using tangential-normal coordinates, and Newton’s second law, or F = m a , we can write the equations as, ( F - D - mg sin θ ) e t + ( N - mg cos θ ) e n = m ( R ¨ θ e t + R ˙ θ 2 e n ) (1) We can consider tangential components only because there are no friction between the ring and the mass. Hence, summationdisplay F t = F - D - mg sin θ = mR ¨ θ (2) As given in the handout, the drag force D can be expressed as, D = bv = bv t = bR ˙ θ (3) 1

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By substituting D and rearranging Eq.( 2 ), we can get the required equation of motion, mR ¨ θ + bR ˙ θ + mg sin θ = F (4) After getting the answer, we can simplify Eq.( 4 ) for numerical simulation, ¨ θ + b m ˙ θ + g R sin θ = F mR (5) (iii) Consider two cases of (a) b = 1 and (b) b = 10 for problem (iii) and (iv).
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ME240_CA1_W08_Solution - ME 240 Introduction to Dynamics...

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