The rotationalinertia of a rod rotating about its center is given in Table 10-2.

Question

The rotationalinertia of a rod rotating about its center is given in Table 10-2.How do you calculate the rotational inertia when the rotation axisis shifted to the end of the rod? What is the total rotationalinertia of the block-rod-bullet system just after impact? Afterimpact, what is the angular momentum of the rod, the bullet, andthe block? There is no external torque to change the angularmomentum of the system. So, the system's angular momentum must beconserved.
Section 11-11 Conservation of AngularMomentum

Explanation / Answer

Given :               Mass of the block Mb = 0.30 kg               Lenght ( l ) = 0.60 m               Mass of the rod ( mr ) = 0.50 kg               Moment of inertia is : I = 0.060 kg -m2 (a)         Is   = Ir + I               = 0.060 + ( 0.30 + 0.003 )(0.6)2               = 0.060 + 0.303*  (0.6)2               = 0.168 kg - m2 (b)             Li   = Lf        mb v l = Is          velocity v = (  Is ) / ( mb  * l )                               = ( 0.168 * 4.5 ) / ( 0.003 * 0.6 )                               =   420 m /s                               = 4.20 x 102 m /s Hope this helps u!        mb v l = Is          velocity v = (  Is ) / ( mb  * l )                               = ( 0.168 * 4.5 ) / ( 0.003 * 0.6 )                               =   420 m /s                               = 4.20 x 102 m /s Hope this helps u!

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