Question

A motar cyclist at a speed of 5m/s is describing circle of radius 25m his inclination with vertical is

Answer 1

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GiveN :-

• Speed of the motorcyclist(v) = 5 m/s

• Radius of the circle(r) = 25 m

• Acceleration due to gravity(g) = 10 m/s²

TO FinD :-

The angle of inclination of the motorcyclist with the vertical.

ConsideratioN :-

Let angle of inclination with the vertical be θ.

AcknowledgemenT :-

$\displaystyle{\sf{tan\theta=\frac{v^2}{r \times g} }}$

CalculatioN :-

$\displaystyle{\sf{tan\theta=\frac{(5)^2}{25 \times 10} }}\\\\\\\displaystyle{\sf{\longrightarrow tan\theta=\frac{25}{25 \times 10} }}\\\\\\\displaystyle{\sf{\longrightarrow tan\theta=\frac{1}{10} }}\\\\\\\displaystyle{\sf{\longrightarrow \theta=tan^{-1}(\frac{1}{10} )}}\\\\\\\displaystyle{\sf{\longrightarrow \theta=tan^{-1}(0.1)}}\\\\\\\boxed{\underline{\boxed{\displaystyle{\sf{\longrightarrow \theta=5.71^\circ }}}}}$

∵ So, the angle of inclination of the motorcyclist with the vertical is 5.71°.

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With regards from :-

CunningKing :)

Answer 2

His inclination with vertical is 0.102.

Explanation :

Given :

Speed of motorcycle , v = 5 m/s  

Radius of circle, r = 25 m

Acceleration due to gravity , g = 9.8 m/s²

Consider the :

Angle of inclination as θ.

Coefficient of friction as μ.

So,

$\implies \sf \theta=tan^{-1} \bigg(\dfrac{r}{gv^2}\bigg)\\ \\ \\ \implies \sf tan^{-1}\bigg(\dfrac{5^2}{25\times 9.8}\bigg)\\ \\ \\ \implies \sf tan ^{-1} (0.1020)=5^051'$

We know :

$\implies \sf F = \dfrac{mv^2}{r}=\mu\;mg\\ \\ \\\implies \sf \mu=\dfrac{v^2}{2g}$

$\implies \sf \dfrac{5^2}{2 \times 9.8}\\ \\ \\\implies \bf 0.102$