A pole P (300kg, length = 12m) is sliding on a frictionless surface at 8.28m/s. The pole's velocity is perpendicular to the poles length. A mass N (.5kg) collides with one end of the pole at 1043m/s and sticks.
How fast does the pole rotate after the collision?
I'm assuming conservation of energy.
Kinetic energy: .5 x M x V2
Rotational Energy: .5 x I x w2
Where I = moment of inertia = 1/12M x L2 (radius was not given in the problem, I'm assuming the moment of inertia for a thin rod)
w = angular speed of rotation
The Attempt at a Solution
Pole's kinetic energy before impact:
.5 x 300kg x 8.28m/s2
Pole's kinetic + rotational energy after impact:
.5 x 300.5kg x v2 + .5 x 1/12*(300.5kg x (12m)2)w2
Due to the conservation of energy:
.5 x 300kg x 8.28m/s2 = .5 x 300.5kg x v2 + .5 x 1/12*(300.5kg x (12m)2)w2
I'm assuming the pole's center of mass is not significantly changed after the collision.
I also know that w(angular speed) = VCenter Mass/R
However, using this strategy, I result with two unknowns within my equation.
Any help with this is greatly appreciated.