Simple fixed axis rotational problem

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A uniform stick of mass 2.1 kg and length 1.3 m is pivoted at one end and released from a horizontal position. The angular acceleration calculated is 11.3 rad/s², leading to a torque of 13.368 Nm due to gravity. To find the force exerted by the pivot, the moment of inertia is considered, and the net torque is calculated using an axis at the center of the stick. The correct force exerted by the pivot is determined to be 5.14 N by dividing the torque by the length of the stick. This approach effectively uses the principles of rotational dynamics to solve the problem.
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Homework Statement
A uniform stick of mass M = 2.1 kg and length L = 1.3 m is pivoted at one end. It is held horizontally and released. Assume the pivot is frictionless. Find the magnitude in Newtons of the force Fo exerted on the stick by the pivot immediately after it is released.

I calculated the angular acceleration immediately after the stick's release to be 11.3 rad/s^2 for another step of the problem (and that was the correct solution for that part of the problem), but other than that I don't really know where to begin this problem. Could someone just provide a hint as to the direction I should be going in?
 

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Running this through my head..., keeping in mind you have alpha, I would next think about what is causing that alpha.
 
Not entirely sure what you mean.

Gravity acts on the mass producing a torque of 13.368 Nm (from \alpha \frac{1}{3}ML^2 = \tau)...but I don't know where to go with that...
 
You specify a moment of inertia, this depends upon where the axis of rotation is taken. This can be used to calculate the net torque, again with a specific axis of rotation in mind.
 
Ah, thanks. I got 5.14, which was correct, by taking the net torque with an axis at the center of the stick. This allowed me to get the torque provided by the pivot by dividing the 13.368 Nm from my previous post by two. The result could then be divided by L=1.3 to get the force.

Thanks again for your help.
 
nice job !
 
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