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A mass M of 5.20E-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 5.27 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(r-rcosθ) + (1/2)mv^2

m's cancel, so

g(2r) + (1/2)vtop2 = g(r-rcosθ) + (1/2)v^2

2(g(2r) + (1/2)vtop2 - g(r-rcosθ) = 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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# Homework Help: Mass Slides Inside a Hoop

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