A bullet with a mass of 12.0 g is fired at a wooden rod that hangs vertically do

Question

A bullet with a mass of 12.0 g is fired at a wooden rod that hangs vertically down from a pivot point that passes through the upper end of the rod. The bullet embeds itself in the exact center of the rod and the rod/bullet system swings up, reaching a maximum angular displacement of 56.0 degrees from the vertical. The rod has a mass of 375. g, a length of 1.50 m, and we can assume the rod rotates without friction about the pivot point. Assume the bullet is traveling horizontally when it hits the rod, and use g = 10.0 m/s2. (a) What is the rotational inertia of the rod/bullet system after the collision, rotating about the pivot point at the upper end of the rod? kg m2 (b) What is the angular speed of the rod/bullet system immediately after the collision? rad/s (c) What is the bullet?s speed just before it hits the rod? m/s

Explanation / Answer

let m1=mass of bullet m2=mass of rod v=velocity of rod before hitting

so, (m1+m2)g(1-cos56)l/2=(1/2)m1*v^2 i.e potential energy=kinetic energy

calculating,v=14.59m/s

v=(l/2) ,so =19.45

I^2=mv^2 so,I=0.00675 Kg-m^2

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