In the figure, a 5.25 g bullet is fired into a 0.547 kg block attached to the end of a 0.718 m nonuniform rod of mass 0.117 kg. The block-rod-bullet system then rotates in the plane of the figure, about a fixed axis at A. The rotational inertia of the rod alone about A is0.0577 kg-m2. Treat the block as a particle. (a) What then is the rotational inertia of the block-rod-bullet system about point A? (b) Ifthe angular speed of the system about A just after impact is 4.31 rad/s, what is the bullet's speed just before impact?RodBlockBullet(a) NumberUnits(b) NumberiUnits

Concept: Here, In this Question we will use the concept of Conservation of angular momentum. The total angular momentum of a system that is not subjected to external torque remains constant regardless of transformation or interaction with the system. That is, In form of Expression, Linitial=Lfinal For the whole system angular momentum will be Linitial=mvL here, m= mass of bullet v= velocity L= Length of rod Angular momentum of rotating body will be, Lfinal=Iω Here, I= Rotational inertia ω= Angular speed For The moment of Inertia of an object I=mr2Solution: The Rotational inertia of the given system is Ir=Irod+m+ML2On substituting the valuesIr=0.0577 kg·m2+5.25×10-3 kg+0.547 kg0.7182Ir=0.3423 kg·m2 This is the rotational inertia of Block-rod bullet system. (b) According to the conservation of angular momentum Linitial=LfinalExpression for Initial Angular momentum isLinitial=mvLExpression for Final Angular momentum isLfinal=IωThusmvL=IωSolving for vv=IωmLSubstituting the valuesv=0.3423 kg·m24.31 rad/s5.25×10-3kg0.718 mv=391.38 m/s This is the Speed of bullet just before the impact.

Answered: In the figure, a 5.25 g bullet is fired…