one Physic problem.?

Question

A 50.0 g projectile moves to the right with a speed of 2.00 m/s toward a 0.500 kg stationary and uniform 50.0 cm long thin rod pivoted at its center of mass wich is free to rotate about this frictionless pivot ( in the center). What is the linear speed of the ends of the rod if the projectile undergoes a totally inelastic collision with the very end of the rod? The rod is mounted horizantally and you may ignore any effects due to gravity.

Answer 1

The bullet's momentum before is m*v.

The spinning rod's momentum is I*w. Where w is in rad/s. But what is I? The rotational inertia of the rod, with the projectile stuck on the end, is I.

The rotational inertia of the rod is
Irod = M*l^2/12
where l = length (in this font, capital I (i) looks identical to small l (L) - at least on my computer). You can consider the projectile to be a particle of mass m, so you can add the rotational inertia of a partical a distance l/2 from the pivot to Irod. So
I = Irod + m*(l/2)^2

So the spinning rod's momentum is (Irod + m*(l/2)^2)*w. From w, I trust you can get the linear speed of the end of the rod.