# Conservation of energy and centripital force

The Question is:

A mass M of 5.00×10-1 kg slides inside a hoop of radius R=1.50 m with negligible friction. When M is at the top, it has a speed of 5.65 m/s. Calculate size of the force with which the M pushes on the hoop when M is at an angle of 27.0 degrees.

My workings thus far:

1st i know i have to use the radius and theta to get the distance of the mass below the hoop
then i have to use the speed at the top and the conservation of mechanical energy to get the speed of the mass at theta
then i make a fbd and get the normal force

so
using trig
1.5 cos 27= x
x= 1.3365m (distance under center of hoop)
distance from bottem then equals 0.1635 m as the hoop is 1.5 m in radius

so using conservation of energy i get
total energy top= total energy bottem
potentical energy top + kE top = pot Energy bottem + Ke bottem
mgh + 1/2 mv^2 = mgh+1/2mv^2 (all masses cancel out
(9.81)(3) +1/2(5.65)^2 = (9.81)(0.1635) + 1/2 v^2
solve for v = 5.88223 m/s

so the speed at theata is 5.88223 m/s, then using centripical force =mv^2/r we can get 23.067 N for the force.

Then we draw a free body diagram and we get that Fc=Fn (what were looking for)+ Fg
This is were i get stuck..
Help from here??

tiny-tim
Homework Helper
Welcome to PF!

… when M is at an angle of 27.0 degrees

Then we draw a free body diagram and we get that Fc=Fn (what were looking for)+ Fg …

Hi Megz27! Welcome to PF! i] 27.0 degrees from what? ii] Fc = Fn + Fg is not correct.

27 degress from the bottem of the circle..

how would you approach the FC is that is not correct?

tiny-tim
What equation does it give you? 