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Homework Statement
A mass M of 5.20E1 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 5.27 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.
Picture attached at the bottom.
The attempt at a solution
E at the top of the circle should equal the energy at that particular part of the circle. So,
mg(2r) + (1/2)mvtop2 = mg(rrcosθ) + (1/2)mv^2
m's cancel, so
g(2r) + (1/2)vtop2 = g(rrcosθ) + (1/2)v^2
2(g(2r) + (1/2)vtop2  g(rrcosθ) = v^2
a = v^2/r
F = ma
When I plug everything in, I get
2(39.831) = v^2
v^2 = 79.66
a = 79.66/1.4
a = 56.90
F = (.52)(56.90)
F = 29.59 N
This answer is incorrect.
Please, can someone let me know where I'm going wrong?
A mass M of 5.20E1 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 5.27 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.
Picture attached at the bottom.
The attempt at a solution
E at the top of the circle should equal the energy at that particular part of the circle. So,
mg(2r) + (1/2)mvtop2 = mg(rrcosθ) + (1/2)mv^2
m's cancel, so
g(2r) + (1/2)vtop2 = g(rrcosθ) + (1/2)v^2
2(g(2r) + (1/2)vtop2  g(rrcosθ) = v^2
a = v^2/r
F = ma
When I plug everything in, I get
2(39.831) = v^2
v^2 = 79.66
a = 79.66/1.4
a = 56.90
F = (.52)(56.90)
F = 29.59 N
This answer is incorrect.
Please, can someone let me know where I'm going wrong?
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