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 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 time t = 0. Assuming that m = 2.0 kg and d = 2.0 cm,

A.)Find the moment of inertia, torque, and angular acceleration of the system at t = 0.

B.)Find the maximum kinetic energy of the system and the maximum angular speed reached

by the rod.

C.)Find the maximum angular momentum of the system

Explanation / Answer

m =2 kg , d = 2cm

(a) Moment of inertia

I = mr² = m*(d/3)² + m*(2d/3)² + m*(4d/3)² = 21md² / 9

I= 7md² / 3 =(7*2*0.02*0.02)/3 = 0.00187kg.m^2

Torque : = (m*4d/3 + m*d/3 - m*2d/3)*g = mdg

= (2*0.02*9.8)= 0.392 N.m

Angular acceleration = / I = mdg / (7md²/3) = 3g / 7d

= (3*9.8)/(7*0.02) = 210 rad/s^2

(B) Relative to a vertical position, the system has
PE = m*(4d/3 + d/3 - 2d/3)g = mdg

so that's the maximum KE (achieved when the rod is vertical).

K.E =mgd = (2*9.8*0.02) = 0.392 J

(f) KE = ½*I*²

0.392 = (1/2) (0.00187)w2

w= 20.48 rad/s

(C) Maximum angular momentum = I * = (0.00187)(20.48)

=0.0383 kg.m2/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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