A 0.21-kg bullet is fired with a velocity of 72.3 m/s into a solid cylinder of m

Question

A 0.21-kg bullet is fired with a velocity of 72.3 m/s into a solid cylinder of mass 7.1 kg and radius 0.49 m. The cylinder is initially at rest and is mounted on a fixed vertical axis that runs through its center of mass. The line of motion of the bullet is perpendicular to the axis and at a distance d = 0.0015 m from the center. Find the angular speed of the system after the bullet strikes and adheres to the surface of the cylinder.

Hint:The angular momentum before and after the collision is the same with respect to the axis.

Explanation / Answer

conserving angular momentum of the system about the axis shown;
0.21*72.3*0.0015 = (Icylndr + Ibult)*

So, = 0.21*72.3*0.0015/(Icylndr + Ibult) = 0.21*72.3*0.0015/[0.5*71*0.49^2 + 0.21*0.49^2] = 2.656* 10-3 rad/sec

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