# Rigid Body Dynamics

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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#### Solution Preview

Please refer to the attachment.

Problem 6/148

P

θ

2' CM

Q 8'

10'

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.