A uniform rod of mass 3.20×102 kg and length 0.450 m rotates in a horizontal pla

Question

A uniform rod of mass 3.20×102 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.10×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 30.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

What is the angular speed of the system at the instant when the rings reach the ends of the rod?

What is the angular speed of the rod after the rings leave it?

Explanation / Answer

moment of inertia of a rod about an axis passing through its center is given by I=(1/12)*M*L^2

where M=mass of the rod

L=length of the rod

given that M=3.2*10^(-2) kg

L=0.45 m

then moment of inertia of the rod=Il=(1/12)*3.2*0.01*0.45^2=5.4*10^(-4) kg.m^2

moment of inertia of each ring about the axis passing through the center of the rod=mass*distance^2

=0.25*(5.1*0.01)^2=6.5025*10^(-4) kg.m^2

so total moment of inertia of the initial system

=moment of inertia of the rod+2*moment of inertia of each ring

=5.4*10^(-4)+2*6.5025*10^(-4)=1.8405*10^(-3) kg.m^2

part 1:moment of inertia of the system when rings are the end of the rod

=moment of inertia of the rod+2*moment of inertia of each ring

=5.4*10^(-4)+2*0.25*(0.45/2)^2=25.8525*10^(-3) kg.m^2

using the principle of conservation of angular momentum,

initial moment of inertia*initial angular veloicty=final moment of inertia*final angular veloicty

==>1.8405*10^(-3)*30=25.8525*10^(-3)*final angular veloicty

==>final angular veloicty=2.1357 rev/min

part 2:
after the rings leave the rod, the moment of inertia of the rod=5.4*10^(-4) kg.m^2

then using the principle of conservation of angular momentum,

initial moment of inertia*initial angular veloicty=final moment of inertia*final angular veloicty

==>1.8405*10^(-3)*30=5.4*10^(-4)*final angular veloicty

final angular veloicty=102.25 rev/min

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