In the figure, a 6.76 g bullet is fired into a 0.656 kg block attached to the en

Question

In the figure, a 6.76 g bullet is fired into a 0.656 kg block attached to the end of a 0.829 m nonuniform rod of mass 0.545 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 o.0988 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 2.13 rad/s, what is the bullet's speed just before impact? Rod Block Bullet Units (a) Number Units (b) Number

Explanation / Answer

(A) I = (0.0988) + (0.656 x 0.829^2) + (6.76 x 10^-3 x 0.829^2)

I = 0.554 kg m^2

(B) Applying angular momentum conservation,

Li = Lf { L = m v r Or L = I w }

(0.00676 x 0.829 x v0 ) + 0 = (0.554) (2.13)

v0 = 210.6 m/s ........Ans

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