A particle of mass M is on the top of a vertical circle without initial velocity. It starts to fall clockwise.
Find the angle with respect to the origin, where the particle leaves the circle.
The Attempt at a Solution
I used two unitary vectors..
εr : which direction is the same as the line described by the radio in the separation point.
εθ: perpendicular to εr (the direction of the tangential velocity in the separation point)
v=dr/dt = Rder/dt
v=ωXr (counterclockwise, so I changedthe sign to negative)
v = -R(-ωeθ) = ωReθ
a = dv/dt = R(αeθ + ω deθ/dt)
deθ/dt = -(ωXeθ) = -ωer
a = α R eθ - ω^2 R er
Weight, Normal force
Centripetal is a resultant?
F(eθ)==> mgcosθ = αRm
F(er)==> N-mgsenθ= -ω^2Rm
From F(er) : α= (gcosθ)/R
∫ (gcosθ)/Rdθ = (gsenθ)/R
ω^2 = (2gsenθ)/R
I come back to the force equations, (with zero normal force) and replace ω^2
And I get nonsenses like 2=1. Did I do something wrong?
Solution: 48,19º (by the teacher)