## Bending of a cyclist in circular motion

Bending of a cyclist in circular motion : 1. A cyclist while going round a curve on a horizontal track has to bend himself a little from his vertical position in order to avoided overturning. When he bends himself inward, a component of the reaction of road provides him the necessary centripetal force for circular motion.

2. In order to find the angle through which a cyclist has to bend himself to take a safe turn, let us assume that

m= combined mass of cycle and cyclist

v= uniform speed with which the cyclist takes the turn

r=radius of circular track

And θ = angle through which the cyclist bend from vertical

3. Suppose G represents the centre of gravity of the system (following figure shows). If force of friction between the cycle and the road is not taken into account, then following two forces act on the system at the point G.

a) Weight mg of the system acting vertically downloads through the centre of the gravity of the system.

b) Reaction R exerted by the road on the system acting along GR at the angle θ with the vertical.

4. The reaction R can be resolved into two rectangular components.

a) R cosθ, acting vertically upwards,

b) R sinθ, acting horizontally i.e, along the radius and towards the centre of the circular track horizontal component R sin θ provides the necessary centripetal force. Hence:

R cos θ=mg………………(1)

Rsin θ = mv^{2}/r ………..(2)

Dividing eq.(2) by eq. (1), we have

R sin θ/R cos θ=(mv^{2}/r)/mg or tan θ v^{2}/rg

Or

** θ = tan ^{-1}(v^{2}/rg)**