please do all of them and specify which is which A rigid, massless rod has three

Question

please do all of them and specify which is which

A rigid, massless rod has three particles with equal masses attached to it as shown in the figure above. 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, show that the moment of inertial of the system (rod plus particles) about the pivot is 7md2 /3, the torque acting on the system at t = 0 is mgd (in which direction?), the angular acceleration of the system at t = 0 is 3g / 7d (in which direction?), the linear acceleration of the particle labeled 3 at t = 0 is 2g/7 (in which direction?), the maximum kinetic energy reached by the rod is mgd, the maximum angular speed reached by the rod is the maximum angular momentum of the system is m and the maximum speed reached by the particle labeled 2 is

Explanation / Answer

A) we need moment of inertia about the point p, we calculate the indivisual moment of inertia of the bodies and add them.

moment of intertia of body 1= m(d+d/3)^2=16md^2/9

of body 2= md^2/9

of body 3=4md^2/9

(NOTE that as the dimensions of the bodies not given , we assumed them as a point mass)

now total moment of inertia=21md^2/9=7md^2/3

B) CONVENTION TAKEN (ccw=+ve)

total torque about p=cw torque + ccw torque

=torque due to body 3+(torque due to bodies 1 and 3)

=-mg(2d/3)+(mg(4d/3)+mg(2d/3))

=mgd

as the answer is positive , the torque will be ccw direction

C) NOW angular acceleration=net torque/moment of inetia about p

=mgd/7md^2/3

=3g/7d(direction is ccw)

D)LINEAR ACCELERATION here is the tangential acceleration of particle 3.

so tangential acceleration=(cross product of angular acceleration and radius vector taken in thesame order)

=(3g/7d)(2d/3)[cross product of k and i)

=2g/7j.

j is direction of +ve y axis that is upward direction(z axis i.e k is in direction perpendicular and above the paper)

E)TOTAL ENERGY OF THE ROD IS CONSTANT IN ANY OF ITS CONFIGURATION.

so if the given time t=0 is taken as reference, we have our total energy=kinetic + potential energy =0

now consider another configuration i.e the rod making an angle v with the vertical axis and the two bodies 1 and 2 are below the reference.

so now our total energy will be k.e+p.e

where we know that p.e = -(4mgd/3+mgd/3-2mgd/3)cos(v)=-mgdcos(v)

now as we know that total energy is constant and is equal to 0

=> k.e-mgdcos(v)=0

=>k.e=mgdcos(v)

so the maximum kinetic energy =mgd [since maximum value o cos is 1]

F) we can also expess our kinetic energy in terms of angular velocity=Iw^2/2

=> w^2=2mgdcos(v)/I=6gcosv/7d=6g/7d(for maximum value cos(v)=1)

=>maximum angular velocity=square root of(6g/7d)

G)angular momentum=I.w

for max ang momentum w must be max

=>max angular momentum=7md^2/3(sq root of(6g/7d))=m.sq root of(14gd^3/3)

H)maximum speed reached by body 2=r.w=d/3(max value of w)=square rootof(6gd/63)

=sq root of(2gd/21)

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