The 2.3-kg sphere and 2.3-kg block (shown in section) are secured to the arm of

Question

The 2.3-kg sphere and 2.3-kg block (shown in section) are secured to the arm of negligible mass which rotates in the vertical plane about a horizontal axis at O. The 5.6-kg plug is released from rest at A and falls into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity ?0 = 2.1 rad/s. Determine the angular velocity ? (positive if counterclockwise, negative if clockwise) of the arm immediately after the plug has wedged itself in the block.

Explanation / Answer

consider the moment just before the collison happens.

the arm is horizontal.

moment of inertia of the system of ball an the block =I1=2.3*0.26^2+2.3*0.56^2=0.87676 kg.m^2

then its angular momentum=moment of inertia*angular veloicty=1.8412 kg.m^2/s

speed of the 5.6 kg block after faling through 0.78 m=sqrt(2*9.8*0.78)=3.91 m/s

then its angular momentum just before collision=mass*speed*radius=5.6*3.91*0.56=12.262 kg.m^2/s

then total angular momentum=1.8412+12.262=14.103 kg.m^2/s

just after collison, moment of inertia of the system=I2=2.3*0.26^2+(2.3+5.6)*0.56^2=2.6239 kg.m^2

then using the principle of conservation of angular momentum,

if new angular velocity is w,

then 2.6239*w=14.103

==>w=14.103/2.6239=5.3748 rad/sec

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