Why Is Normal Force Not Equal to mg in Rotating Collar Problems?

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Homework Statement


Handwritten notes on graph paper: Given m_D = 300 g, omega = 5 rad/s, mu = 0 (frictionless); Find: r for no slip; Solution: If there is no slip, D is moving in a circular path of radius r about the axis AC; Use e_t and e_n.


Homework Equations


[/B]
Radial and Transverse coordinates to relate acceleration

Balancing forces

The Attempt at a Solution


Handwritten free-body diagram of a collar on a rotating rod at 40 degrees, with normal forces Nt and Nl, weight mg, and an acceleration diagram showing mat and man, followed by equations for force balance and the resulting expressions Nl = mg/sin 40 and Nl = mrw^2/cos 40.


I know that there is more to this problem, but my question is simple: Why don't we know what the normal force is? When I solved it myself I just set the Normal force to mass * gravity... why isn't this the case for this problem? All of the solutions I've found have them creating a relation between the X and Y force eqns. and setting them equal to solve. Why can't we just set N = mg as I was so used to doing?
 
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You haven't told us what problem you are trying to solve . Please post the original statement of the problem as given to you .
 
Nidum said:
You haven't told us what problem you are trying to solve . Please post the original statement of the problem as given to you .
My apologies, I completely forgot to include that part. Here it is:

Diagram of a bent rod AC rotating about a vertical axis, with collar D sliding on a segment marked B to D, an angle alpha at the lower part and a radius r marked from the axis to the collar.
 
N can only work in a normal direction (as the word normal force suggests :smile:) . Without friction, the only direction in which the rod can exercise force on the collar is perpendicular to the rod.
You have figured out the components of N, so you can add them up (vectorially) to get ##|\vec F_N|##.

[edit] no, sorry, you needed to calculate that |F| from its vertical component to get r.