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Find the angle through which a cyclist has to bend himself to take safe turn.

Sarthaks Test For Class 7-12

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When a cyclist negotiated a curve, he bends slightly from his vertical position (Figure). The various forces acting on the system (the cycle and man) are:

(i) Weight (mg) of the system acts vertically downward through the centre of gravity of the system.

(ii) Normal reaction (R) offered by the road to the system and acts at an angle 'θ' with the vertical. Resolving 'R' into two components:

(a) R cosθ which is equal and opposite to the weight (mg),

i.e., R cosθ = mg    ...(i)

(b) R sinθ which is directed towards the centre of the circular path and provides the necessary centripetal force ({mv2}/{r}) to the system to remain in the circular path i.e.,

R sinθ = {mv2}/{r}    ...(ii)

Dividing (ii) by (i), we get

{R sinθ}/{R cosθ} = {mv2}/{r} x {1}/{mg} ⇒ tanθ = {v2}/{rg}

or, θ = tan-1({v2}/{rg})

Clearly, θ depends on 'v' and 'r' at a place. For safe turn, θ should be small. θ will be small if v will be small and r is large. Thus, turning should be at slow speed and along track of larger radius.

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