- #1
Staerke
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Homework Statement
A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 4.25 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 44.0°.
Homework Equations
E = Pe + Ke
Pe = m * g * h
Ke = 1/2 * m * v^2
F=mv^2/r
The Attempt at a Solution
Alright so energy at the top = Pe + Ke
So m*g*h + 1/2*m*v^2 = 11.2226 + 4.69625 = 15.9189
The energy in the bottom is going to be equal to the energy at the top
The height at the bottom is R-Rcos(44) = .295511
15.9189 = .52*9.81*.29511 + .5*.52*v^2
V=7.44503
Centripital force = mv^2/r, so .52*7.44503^2/1.1 = 26.2026 N
But this is wrong. Now I thought it might be reduced due to gravity.
so F=26.2026-m*g*cos(44) = 22.5331
Neither answer works.
Help?