A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v0. The angle θ with respect to the vertical where it leaves contact with the track is

N+mgcosθ= mv2R(N=0)cosθ= v2rg Conserving energy mg(2r)+12mvo2= mgr(1+cosθ)+12m(rgcosθ) 2gr+12vo2= gr(1+cosθ)+grcosθ2 32rgcos θ= gr+12vo2 cosθ= 23+13vo2rg

A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v0. The angle θ with respect to the vertical where it leaves contact with the track is

N+mgcosθ= mv2R(N=0)cosθ= v2rg Conserving energy mg(2r)+12mvo2= mgr(1+cosθ)+12m(rgcosθ) 2gr+12vo2= gr(1+cosθ)+grcosθ2 32rgcos θ= gr+12vo2 cosθ= 23+13vo2rg

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A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v₀. The angle $θ$ with respect to the vertical where it leaves contact with the track is

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$\sin^{- 1} (\frac{v_{0}^{2}}{rg} + \frac{1}{3})$

$\sin^{- 1} (\frac{v_{0}^{2}}{3 rg} + \frac{2}{3})$

$\cos^{- 1} (\frac{v_{0}^{2}}{rg} + \frac{1}{3})$

$\cos^{- 1} (\frac{v_{0}^{2}}{3 rg} + \frac{2}{3})$

answer is A.

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Detailed Solution

N+mgcos $θ$ = $\frac{m v^{2}}{R}$ (N=0)

$\cos θ = \frac{v^{2}}{r g} C o n s e r v i n g e n e r g y m g (2 r) + \frac{1}{2} m v_{o}^{2} = m g r (1 + \cos θ) + \frac{1}{2} m (r g \cos θ) 2 g r + \frac{1}{2} v_{o}^{2} = g r (1 + \cos θ) + g r \frac{\cos θ}{2} \frac{3}{2} r g \cos θ = g r + \frac{1}{2} v_{o}^{2} \cos θ = \frac{2}{3} + \frac{1}{3} [\frac{v_{o}^{2}}{r g}]$