Angled cyclist turn

The cyclist passes through a curve with a radius of 20 m at 25 km/h. How much angle does it have to bend from the vertical inward to the turn?

Correct result:

A =  13.5567 °


r=20 m v=25 km/h m/s=25/3.6  m/s=6.94444 m/s g=10 m/s2  Fo+Fg=F Fg=m g Fo=mv2/r  tanA=FoFg tanA=mv2/rm g tanA=v2r g  A1=arctan(v2r g)=arctan(6.9444220 10)0.2366 rad  A=A1  =A1 180π   =0.236609896438 180π   =13.557  =13.5567=133324"

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Showing 1 comment:
A cyclist has to bend slightly towards the center of the circular track in order to make a safe turn without slipping.

Let m be the mass of the cyclist along with the bicycle and v, the velocity. When the cyclist negotiates the curve, he bends inwards from the vertical, by an angle θ. Let R be the reaction of the ground on the cyclist. The reaction R may be resolved into two components:

(i) the component R sin θ, acting towards the center of the curve providing necessary centripetal force for circular motion and
(ii) the component R cos θ, balancing the weight of the cyclist along with the bicycle.

Thus for less bending of the cyclist (i.e for θ to be small), the velocity v should be smaller and radius r should be larger. let h be the elevation of the outer edge of the road above the inner
edge and l be the width of the road then,


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