Question

a uniform rod of mass1 = 4.7 kg and legth 80 cm lies horizontal frictionless table, with a vertical frictionless axle passing through a point P, located 25cm from center of mass of the rod. Also a particle of mass 2 = 1.2 kg is attached to the rod 5 cm from the far end.

A. Find the moment of inertia of the rod only about point P.

The system is initially rotating counter clock wise about P at 6.40 rad/s, at the instant rod is aligned with the y axis, an explosive charge fires m2 at speed 8.2 m/s at angle 35 with respect to x axis. Find angular momentum of particle at moment just after explosion as measured with respect to the axle at P

C, find final angular velocity of rod after explosion

Answer 1

A) Use parallel axis theorem,

I = Icm + m*d^2

= m*L^2/12 + m*d^2

= 4.7*0.8^2/12 + 4.7*0.25^2

= 0.54 kg.m^2

B) angular momentum of the particle = m*v*R*sin(35)

= 1.2*8.2*0.5*sin(55)

= 4.03 kg.m^2/s

C) Apply coonservation of momentum

Li = Lf

( 0.54 + 1.2*0.5^2)*6.4 = 4.03 + 0.54*w

==> w = ( ( 0.54 + 1.2*0.5^2)*6.4 - 4.03)/0.54

= 2.5 rad/s