In the figure shown, a 1.75 g bullet is shot into a 0.545 kg block that is mount

Question

In the figure shown, a 1.75 g bullet is shot into a 0.545 kg block that is mounted on the end of a 0.620 m nonuniform rod of mass 0.475 kg. The block-rod-bullet system then rotates about a fixed axis at point A. The moment of inertia of the rod alone about A is 0.0570 kg*m^2. Assume the block is small enough to treat as a particle on the end of the rod. (a) What is the moment of inertia of the block-rod-bullet system about point A? kg*m^2 (b) If the angular speed of the system about A just after the bullet's impact is 4.05 rad/s, what is the speed of the bullet just before the impact? m/s

Explanation / Answer

a) MI of block-rod-bullet system about A= ((1.75/1000)+0.545) *0.62*0.62 + 0.057 = 0.2671707 Kgmm

b) Conserving angular momentum ,

mvr=Iw

0.00175v*.62 = 0.2671707 * 4.05

=> v = 997.27311 m/s

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