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.

A) Find the moment of inertia of the system about the pivot point P.

B) Find the torque acting on the system at t=0. (Magnitude and direction)

C) Find the angular acceleraiton of he system at t=0. (Magnitude and direction)

D) Find the linear acceleration of the particle labeled 3 at t=0.

E) Find the maximum kinetic energy of the system.

F) Find the maximum angular speed reached by the rod.

http://s3.postimg.org/sjbfdl7yr/physics.png

Explanation / Answer

moment of inertia about P ,

I = ( m x (2d/3)^2) + (m x (d/3)^2) + ( m x (4d/3)^2)

I = 21md^2/9 = 7md^2 / 3

b) torque due to a force = r xF

net torque = (- 2d/3 x mg) + ( d/3 x mg) + ( 4d/3 x mg)

= mgd

c) torque = I xalpha

mgd = ( 7m d^2/3) alpha

alpha = 3g / 7d

d) a = alpha * r

r for 3rd particle = 2d/3

a = (3g/7d)(2d/3) = 2g/7 = 2.8 m/s^2

e) as this rod comes vertical.

now centre of mass position.

ycm = ( (-4md/3 - md/3 + 2md/3) / 3m = -d/3

so centre of mass comes -d/3 distance down .

so   3mg (d/3) = Iw^2 /2   = max. KE

KE = mgd

f) mgd = (7 m d^2 /3)w^2 /2

6g/7d = w^2

w = sqrt(6g/7d)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.