a 8.11 g bullet is fired into a 0.524 kg block attached to the end of a 0.694 m

Question

a 8.11 g bullet is fired into a 0.524 kg block attached to the end of a 0.694 m nonuniform rod of mass 0.424 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 is 0.0970 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) If the angular speed of the system about A just after impact is 7.06 rad/s, what is the bullet's speed just before impact?

Explanation / Answer

r = 0.694 m

Moment of inertia of the system I = 0.0970 + (0.524 + 0.00811)r^2

                                                 = 0.0970 + 0.5321(0.694)^2

                                                 = 0.0970 + 0.2562

                                                 = 0.3532 Kg m^2

According to law of conservation of angular momentum

                                            lo = L

                                        mvr = I

                             (0.00811)v(0.694) = 0.3532 * 7.06

                                            v = 443.14 m/s        

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