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Introductory Physics Homework Help
Dynamics of circular motion. Particle on the top of a circle
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[QUOTE="Zipi Damn, post: 4227261, member: 456610"] [h2]Homework Statement [/h2] 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. [ATTACH=full]158234[/ATTACH] [h2]Homework Equations[/h2] v=ωXr [h2]The Attempt at a Solution[/h2] I used two unitary vectors.. [B]εr[/B] : which direction is the same as the line described by the radio in the separation point. [B]εθ[/B]: perpendicular to εr (the direction of the tangential velocity in the separation point) [B]r[/B]=R [B]er[/B] [B]v[/B]=d[B]r[/B]/dt = Rd[B]er[/B]/dt [B]v=ωXr[/B] (counterclockwise, so I changedthe sign to negative) [B]v[/B]=-R([B]ωXer[/B]) [B]v[/B] = -R(-ω[B]eθ[/B]) = ωR[B]eθ[/B] α=dω/dt [B]a[/B] = d[B]v[/B]/dt = R(α[B]eθ[/B] + ω d[B]eθ[/B]/dt) d[B]eθ[/B]/dt = -(ωX[B]eθ[/B]) = -ω[B]er[/B] a = α R [B]eθ[/B] - ω^2 R [B]er[/B] Forces: Weight, Normal force Centripetal is a resultant? F(eθ)==> mgcosθ = αRm F(er)==> N-mgsenθ= -ω^2Rm So: α=dω/dt= (dω/dθ)(dθ/dt)=ωdω/dθ θ ∫αdθ = ∏/2 ω ∫ωdω ω0 From F(er) : α= (gcosθ)/R θ ∫ (gcosθ)/Rdθ = (gsenθ)/R ∏/2 ω ∫ωdω= (ω^2)/2 ω0ω^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) [/QUOTE]
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Dynamics of circular motion. Particle on the top of a circle
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