A rigid mass-less rod has three particles with equal masses m = 1 kg attached to

Question

A rigid mass-less rod has three particles with equal masses m = 1 kg attached to it as show in the Figure 5 below. The rod is free to rotate in a vertical plane with a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at t = 0. The distance between each mass is d = 2 m.

(a) Determine the moment of inertia of the system (rod plus masses) about the pivot point.

(b) The torque acting on the system at t = 0.

(c) The angular acceleration of the system at t = 0.

(d) The maximum angular momentum of the system.

Explanation / Answer

I have found moment of inertia, I = 7md ^2/3 torque = m gd k, where k is unit vector angular accel = 3 g/7d accel = 2 g/7 The KE will be maximum when the PE is minimum. So what position will have the masses as low as they can get? Use PE = mgy. If you choose y = 0 at the height of point P, then the initial PE will be 0.

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