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. Assume that m and d are known. (Use the following as necessary: m, d, and g.) http://www.webassign.net/serpse8/11-p-049.gif K so i figured out most of it. how do i find the maximum angular momentum of the system. how do i find the linear acceleration of the particle labeled 3 at t = 0. magnitude ? Also how do i find the max kinetic energy of the system?

Explanation / Answer

Inertia about P is I = m*(2d/3)^2 + m*(d - 2d/3)^2 + m*(2d - 2d/3)^2 = 7/3*md^2

At any instant when angle between rod and horizontal is , we get

Torque about P is T = mg*(2d - 2d/3)Cos + mg*(d - 2d/3)Cos - mg*(2d/3)Cos = mgd*Cos

T = I

Angular acceleration = T/I = mgd*Cos/(7/3*md^2) = 3gCos/(7d)

This is max. when Cos = 1.

So, max acc = 3g/(7d)

Linear acc = r = (2d/3)*3g/(7d) = 2g/7

= d/dt = (d/d)(d/dt) = (d/d)

So, d = d

d = 3gCos/(7d) d

Integrating both the sides,

^2/2 = 3gSin/(7d) + C

At t = 0 when = 0, we have = 0. Hence, C = 0

Thus, ^2/2 = 3gSin/(7d)

= [6gSin/(7d)]

Max. will be when Sin = 1.

_max = [6g/(7d)]

Max. KE = 1/2*I*_max^2

Max KE = 1/2*(7/3*md^2) * ([6g/(7d)])^2

Max KE = 1/2*(7/3*md^2) * 6g/(7d)

Max KE = mgd

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