Question

The figure is an overhead view of a thin uniform rod of length 0.467 m and mass M rotating horizontally at angular speed 15.7 rad/s about an axis through its center. A particle of mass M/3 initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed v_p is 3.32 m/s greater than the speed of the rod end just after ejection, what is the value of v_p?

Answer 1

Hello!

Well I can't see the figure, I don't know why, but from what I read I think I have an answer

As I read, this problem give us the data of the final state at the rod, because when the rod and the particle were attached to each other, they had a different amount of momentum, by conservation of energy we can determine what was the initial angular velocity, which will guide us to the final velocity of the particle

Lets reduce all the terms to continue

The initial mass is the sum of the rod plus the particle

The inertia of the rod after the detachment is reduced to only the mass of the rod

and finally the kinetic energy of the particle

Once joined we have

Doing some repetitive algebraic arragement we got this

From this we obtain two values:

From which we use the first value, because if we use the second, that would mean that the rod was moving in the oposite direction which is not posible, so the velocity of the particle is