23 23 -2 points HRW9 P.060 My Notes Ask Your Teacher In the figure below, a 4.0

Question

23

23 -2 points HRW9 P.060 My Notes Ask Your Teacher In the figure below, a 4.0 g bullet is fired into a 0.40 kg block attached to the end of a 0.60 m nonuniform rod of mass 0.50 kg. The block system then rotates in the plane of the figure about a fixed axis at A. The rotational inertia of the rod alone about that axis at A is 0.060 kg.m2. Treat the block as a particle (a) What then is the rotational inertia of the block-rod-bullet system about point A? kg m (b) If the angular speed of the system about A just after impact is 4.5 rad/s, what is the bullet's speed just before impact? m/s Rod Block Bullet

Explanation / Answer

(a) Inertia of the rod is given = 0.06 kg -m2
And the inertia of the block about point A
= ml2 where m is the mass of the block and l is the length of the rod.
Therefore total inertia of the rod = 0.06 + ml2
= 0.06 + (0.4*0.62) = 0.204 kg-m2
(b) angular speed = 4.5 rad/s
Applying conservation of angular momentum
MVl = Iw
where M is the mass of bullet , V is the linear velocity of bullet
and I is the total inerita about A and w is the angular velocity after impact.
(0.004)*V*0.6 = 0.204*4.5
V = 382.5 m/s

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