Force exerted by a stick on its axle?

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SUMMARY

The discussion focuses on calculating the vertical component of the force exerted by a stick on its axle when the stick passes through the vertical position. The stick has a mass of 8.6 kg and a length of 0.7 m, with the axle located 0.24 m from one end. The angular velocity at the vertical position is determined to be 6.38 rad/s, and the center of mass is positioned 0.11 m below the axis of rotation. The participant initially calculated the centripetal force but encountered an error, likely due to not squaring the speed in their calculations.

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Homework Statement



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?

Homework Equations





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!
 
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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.
 

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