# Force exerted by a stick on its axle?

1. Oct 25, 2009

### EstimatedEyes

1. The problem statement, all variables and given/known data

A stick of uniform density with mass M = 8.6 kg and length L = 0.7 m is pivoted about an axle which is perpendicular to its length and located 0.24 m from one end. Ignore any friction between the stick and the axle. What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical?

2. Relevant equations

3. The attempt at a solution

I already correctly calculated that the angular velocity when it is in the vertical position is 6.38 rad/s and that the center of mass is .11m below the axis of rotation when it is in the vertical position. From this I calculated v of the center of mass by multiplying the angular velocity to the radius. Once I got this answer I calculated the centripetal force by v^2/r and then added it to mg. This answer is not right, but I don't know where I went wrong. I'm sure it's something really simple. Thanks in advance for your help!

2. Oct 26, 2009

### kuruman

The (net) centripetal force is mv2/r. Assuming that you did not forget to multiply by the mass, your method is correct. Redo the numbers - you may have forgotten to, say, square the speed.