A bead rests at the top of a fixed, frictionless hoop.

[haruspex]

So the accelerations at these points is g downwards. Correct?

Only for those where the normal force is zero.

 

[physicsdude101]

Only for those where the normal force is zero.

For Φ=0, a=0 and for Φ=π, a=-6g. Correct?

 

[haruspex]

For Φ=0, a=0 and for Φ=π, a=-6g. Correct?

Right for φ=0, but how do you get -6g?

 

[physicsdude101]

Right for φ=0, but how do you get -6g?

In my equation N=mg(3cosφ-2) I put φ=π and get N=-5mg. Then the net force is -5mg-mg=-6mg.

 

[TomHart]

Then the net force is -5mg-mg=-6mg.

The normal force has to be acting vertically upward at the bottom of the loop and the weight vertically downward. So those 2 terms cannot be the same sign.

 

[physicsdude101]

The normal force has to be acting vertically upward at the bottom of the loop and the weight vertically downward. So those 2 terms cannot be the same sign.

Ok so would it be 4mg then? How do I get -5mg using the formula though? Isn't that pointing downwards?

 

[haruspex]

Ok so would it be 4mg then? How do I get -5mg using the formula though? Isn't that pointing downwards?

If you look at how you obtained your equation for N, I think you will find that you have effectively defined it as positive radially outwards.

 

[physicsdude101]

If you look at how you obtained your equation for N, I think you will find that you have effectively defined it as positive radially outwards.

Ah I think I get it. So my value of -5mg is correct the way I defined N, and it just means that at φ=π it has magnitude 5mg in the positive y direction? Thus the net force would have a magnitude 5mg-mg=4mg in the positive y direction?

 

[haruspex]

Ah I think I get it. So my value of -5mg is correct the way I defined N, and it just means that at φ=π it has magnitude 5mg in the positive y direction? Thus the net force would have a magnitude 5mg-mg=4mg in the positive y direction?

Yes.

 

[physicsdude101]

Thank you. I think all my questions have been answered then. :)