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    Rigid Body Dynamics

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    Please do probs 6/148 and 6/136 only. Refer attachment for fig.

    Problem 6/148 : The fig. shows cross section of a garage door which is a uniform rectangular panel 8 x 8 ft ans weighing 200 lb. The door carries two spring assemblies, one on each side of the door, like the one shown. Each spring has a stiffness of 50 lb/ft, and is unstretched when the door is in the open position shown. If the door is released from rest in this position, calculate the velocity of the edge at A as it strikes the garage floor.

    Problem 6/136 : The centre of the 200 lb wheel with centroidal radius of gyration of 4 in. has a velocity of 2 ft/sec down the incline in the position shown. Calculate the normal reaction N under the wheel as it rolls past position A. Assume that no slipping occures.

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    Please refer to the attachment.

    Problem 6/148

    2' CM
    Q 8'



    From the fig. we can determine the angle θ made by the gate with horizontal just before being released as follows :

    sinθ = 2/8 = 0.25 or θ = 14.48O

    Centre of mass of the door lies at the centre of the door. Its initial height from the floor = 10' - PQ = 10 - 4sin14.48 = 10 - 4x0.25 = 9'

    When the door has hit the floor, centre of mass will be at a height of 4' [half of the height of the door].

    Net fall in the height of the centre of mass = 9 - 4 = 5'

    Decrease in potential energy of the door between its shown position and hitting the ground = mgh = 200x5 = 1000 lb ft

    Neglecting loss of energy due to friction etc., the decrease in PE of the door is equal to i) increase in the PE ...

    Solution Summary

    Step by step solutions are provided for problems on rigid body dynamics.