Circular motion of hoop and mass

[HardestPart]

Homework Statement

A mass M of 6.00E-1 kg slides inside a hoop of radius R=1.40 m with negligible friction. When M is at the top, it has a speed of 5.23 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 31.0°.

I have no idea from where to begin

I don't want you to do my homework,Can someone give me a hint?

 

[rock.freak667]

Start by drawing a free body diagram. The use conservation of energy.

 

[HardestPart]

When I do the diagram
the total forces work on the mass is :
mg=mv^2/r

the low of energy is:
Mgh=1/2mv^2r

But I can't see how the angle is related to the sloution?

 

[ideasrule]

Can you use the conservation of energy to figure out the speed of the mass at 31 degrees? If so, do that first, and make sure you can get a numerical value.

Also, mg is not the only force on the block. N, the normal force, provides part of the centripetal acceleration.

 

[HardestPart]

But I don't know which force I have to do to components
I need cos ans sin right?
How can I do that?

 

[HardestPart]

I tried to find the speed of the mass at 31 degrees by:
N+mgcosa=mv^2/r
but i don't know what to but instead of N?

 

[HardestPart]

I tried to slove the question by :
N+mgcosa=mv^2/r
but i don't know what to put instead of N!?

 

[rock.freak667]

I tried to slove the question by :
N+mgcosa=mv^2/r
but i don't know what to put instead of N!?

N is what you want to find.

 

[HardestPart]

I tried to fins the speed at angle 31 by:
I/2mv^2+mg2R=1/2mv^2+mgRsina
all m goes together
v^2+4gR=v^2+2gRsina
the answer i get from solving it i plugg into this:
N+mgcosa=mv^2/R
N=mv^2/Rsina-mgcosa
I get N=18.27N
but it is a wrong answer
Can you tell me where i went wrong?