# 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 =  13.557 °

#### Solution:

$r=20 \ \text{m} \ \\ v=25 \ km/h \rightarrow m/s=25 / 3.6 \ m/s=6.94444 \ m/s \ \\ g=10 \ \text{m/s}^2 \ \\ \ \\ F_o + F_g=F \ \\ F_g=m \cdot \ g \ \\ F_o=m v^2 / r \ \\ \ \\ \tan A=\dfrac{ F_o }{ F_g } \ \\ \tan A=\dfrac{ m v^2 / r }{ m \cdot \ g } \ \\ \tan A=\dfrac{ v^2 }{ r \cdot \ g } \ \\ \ \\ A_{1}=\arctan(\dfrac{ v^2 }{ r \cdot \ g } )=\arctan(\dfrac{ 6.9444^2 }{ 20 \cdot \ 10 } ) \doteq 0.2366 \ \text{rad} \ \\ \ \\ A=A_{1} \rightarrow \ ^\circ =A_{1} \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =0.236609896438 \cdot \ \dfrac{ 180 }{ \pi } \ \ ^\circ =13.557 \ \ ^\circ =13.557 ^\circ =13^\circ 33'24"$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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Showing 1 comment: 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, Tips to related online calculators
Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.
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