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Unformatted text preview: ME163 Mechanical Vibrations (Winter 2009) HW8 Due Wednesday 4.04.09 7pm (in the box or to a TA) Answers on web by 9pm, midterm the day after 1. 2DOF Modeling. Consider the system shown in Figure 1 which is a crude model of a coal cart with a scoop attached to its side. The cart has a mass M and is propelled along a horizontal track by a force f . The scoop hangs off the side and is actuated by a torque T (the scoop has a total mass m and moment of inertia I about the center of mass C ). Obtain the equations of motion using x 1 and as coordinates and f and T as the inputs. Consider gravity. T M f g m C a y x 1 x 2 Figure 1: Coal cart with attached scoop. One choice of coordinates is x 1 and . The free body diagrams are shown in Figure 2. F x and F y are the reaction forces in the x and y directions due to the physical connection between f F M M F R R g 1 2 x y T x F F y mg C Figure 2: Free body diagrams the pendulum and the mass M . R 1 and R 2 are the vertical reaction forces due to the wheels. Summing forces on M in the horizontal and vertical directions gives M x 1 = F x + f ME163 Mechanical Vibrations 1 HW8 M y 1 = F y + R 1 + R 2 Mg For mass m , summing forces in the horizontal and vertical directions gives m x 2 = F x m y 2 = F y mg Summing moments about the center of mass gives I = aF x cos + aF y sin + T. Using the previous equations to eliminate F x and F y we obtain M x 1 + m x 2 = f (1) I + m x 2 a cos + ma ( y 2 + g )sin = T (2) The displacements are related as follows: x 2 = x 1 + a sin y 2 = a cos Thus x 2 = x 1 a 2 sin + a cos y 2 = a 2 cos + a sin Substituting x 2 and y 2 in (1) and (2) we obtain ( I + ma 2 ) + ma x 1 cos + mga sin = T ( M + m ) x 1 ma 2 sin + ma cos = f....
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 Winter '08
 Mezic,I

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