# 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?

Result

A =  13.557 °

#### Solution:

$r=20 \ \text{m} \ \\ v=25 \ km/h=25 / 3.6 \ m/s=6.94444 \ m/s \ \\ g=10 \ \text{m/s}^2 \ \\ \ \\ F_o + F_g=F \ \\ F_g=m \cdot \ g \ \\ F_o=m v^2 / r \ \\ \ \\ \tan A=\dfrac{ F_o }{ F_g } \ \\ \tan A=\dfrac{ m v^2 / r }{ m \cdot \ g } \ \\ \tan A=\dfrac{ v^2 }{ r \cdot \ g } \ \\ \ \\ A_{1}=\arctan(\dfrac{ v^2 }{ r \cdot \ g } )=\arctan(\dfrac{ 6.9444^2 }{ 20 \cdot \ 10 } ) \doteq 0.2366 \ \text{rad} \ \\ \ \\ A=A_{1} \rightarrow \ ^\circ =A_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =13.55675 \ \ ^\circ =13.557 ^\circ =13^\circ 33'24"$

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Matematik
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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