- #1
Megz27
- 5
- 0
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??
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??