A rigid, massless rod has three particles with equal masses attached to it as sh

Question

A rigid, massless rod has three particles with equal masses attached to it as shown in the figure below. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P, and is released from rest in the horizontal position at t = 0. Assuming m and d are known.

Find:

a) the moment of inertia of the system (rod plus particles) about the pivot,
(b) the torque acting on the system at t = 0,
(c) the angular acceleration of the system at t = 0,
(d) the linear acceleration of the particle labeled 3 at t = 0,
(e) the maximum kinetic energy of the system,
(f) the maximum angular speed reached by the rod,
(g) the maximum angular momentum of the system, and
(h) the maximum speed reached by the particle labeled 2.

Explanation / Answer

a) treatinq the masses as point masses
I=m*d^2*(16/9+1/9+4/9)
simplify

I=m*d^2*7/3

f) using energy
m*g*d*(1/6+4/6-2/6)=.5*I*^2
solve for
m*g*d*(1/6+4/6-2/6)=.5*m*d^2*21*^2/9
do some algebra

sqrt(3*g/(7*d))=

g) I*

sqrt(7*g/3)*m*d

...

Ip = md²[(1/3)²+(4/3)²+(2/3)²] = (7/3)md²

From E = ½Ip*w² we get
1) w = [2E/I]
You already have E = mgd; substituting in (1),

w = [(6/7)(g/d)]

V2 = w*R2 = w*d/3 = (1/3)[6gd/7]

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