The figure at the right shows a top view of a uniform rod of mass 430 g and leng

Question

The figure at the right shows a top view of a uniform rod of mass 430 g and length 88 cm sitting on a frictionless surface. The rod is going to be hit at the bottom by a bullet of mass 18 g with a speed of 110 m/s. The bullet embeds in the rod. What is the initial angular momentum of the system? Explain. What will the final linear speed of the system be? Explain. What will the angular speed of the system be after the collision? Explain. For the next two questions the uniform rod is replaced by one that has four point masses 200 g, 100 g, 80 g, and 50 g, at 0.25 L, 0.5 L, 0.75 L, and L respectively, connected by thin sticks of negligible mass. The same bullet hits and embeds in the 50 g mass. How will the final linear speed for this case compare to the case above? Increase Same Decrease Explain. How will the final angular speed for this case compare to the case above? Increase Same Decrease Explain.

Explanation / Answer

a) Initial angular momentum of the system = Initial angular momentum of bullet (since rod is at rest) = mass of bullet*velocity of bullet*perpendicular distance

=> Linitial = .018*110*0.88 = 1.7424 Kg m2/s

b) moment of inertia of rod about its end = (1/3)mL2, moment of inertia of bullet = m'L2, angular velocity, w = v/L

Hence Lfinal = ((1/3)mL2+m'L2)(v/L) = (0.43L2/3+0.018L2)v/L = 0.16*0.88v = 0.1408v Kg m2/s

By conservation of angular momentum, 0.1408v = 1.7424 => v = 12.375m/s

c) Angular speed, w = v/L = 12.375/0.88 = 14.0625 rad/s

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