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When a cyclist negotiates a circular pat...

When a cyclist negotiates a circular path of radius .r. with velocity .v., making an angle `theta` with the vertical, show that `tantheta=v^(2)/(rg)`.

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Motion of a cyclist on a circular path :
If a cyclist takes a turn of the road he leans towards the centre of circular path.

If the cyclist leans towards the centre of turn making an angle `theta` with the vertical, his weight mg and force of reaction R = mg form an anti-clockwise couple which balances the clockwise couple formed by the forces `F_(1) and F_(2)`. Thus in the position of equilibrium :
Couple formed by mg and R = Couple formed by `F_(1) and F_(2) (=F)`
`" "mg times theta sinR=F times Rcostheta`
`" "tantheta=F/(mg)`
`" "tantheta=(mv^(2))/(r times mg)" "[F=(mv^(2))/r]`
`" "tantheta=v^(2)/(gr)`.
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