- #1

Ebola_V1rus

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- 0

## Homework Statement

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?

## Homework Equations

I'm assuming conservation of energy.

Kinetic energy: .5 x M x V

^{2}

Rotational Energy: .5 x I x w

^{2}

Where I = moment of inertia = 1/12M x L

^{2}(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/s

^{2}

Pole's kinetic + rotational energy after impact:

.5 x 300.5kg x v

^{2}+ .5 x 1/12*(300.5kg x (12m)

^{2})w

^{2}

Due to the conservation of energy:

.5 x 300kg x 8.28m/s

^{2}= .5 x 300.5kg x v

^{2}+ .5 x 1/12*(300.5kg x (12m)

^{2})w

^{2}

I'm assuming the pole's center of mass is not significantly changed after the collision.

I also know that w(angular speed) = V

^{Center Mass}/R

However, using this strategy, I result with two unknowns within my equation.

Any help with this is greatly appreciated.