a rigid, massless rod has three particles with equal masses attached to it. the

Question

a rigid, massless rod has three particles with equal masses attached to it. the rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through point p and is released from rest in the horizontal position t=0. assuming m and d are known, show that:

(a) the moment of inertia of the system (rod plus particles) about the pivot is 7md^2/3

(b) the torque acting on the system at t=0 is mgd (in which direction?)

(c) the angular acceleration of the system at t=0 is 3g/7d (in which direction?)

(d) the linear acceleration of the particle labeled 3 at t=0 is 2g/7 (in which direction?)

(e) the maximum kinetic energy reached by the rod is mgd

(f) the maximum angular speed reached by the rod is Sqrt(6g/7d)

(g) the maximum angular momentum of the system is m(sqrt(14gd^3/3))

(h) the maximum speed reached by the particle labeled 2 is sqrt(2gd/21)

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 θ, we get

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

T = Iα

Angular acceleration α = T/I = mgd*Cosθ/(7/3*md^2) = 3gCosθ/(7d)

This is max. when Cosθ = 1.

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

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

α = dω/dt = (dω/dθ)(dθ/dt) = ω(dω/dθ)

So, α dθ = ω dω

ω dω = 3gCosθ/(7d) dθ

Integrating both the sides,

ω^2/2 = 3gSinθ/(7d) + C

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

Thus, ω^2/2 = 3gSinθ/(7d)

ω = √[6gSinθ/(7d)]

Max. ω will be when Sinθ = 1.

ω_max = √[6g/(7d)]

Max. KE = 1/2*I*ω_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

not sure how to findout f, g and h ....... hope this helps atleast

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