Question

A car is moving on a circular level road of curvature 300m . If the coefficient of friction is 0.3 and acceleration due to gravity is 10m//s^(2) , the maximum speed of the car be

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

A car is moving on a circular level road of curvature 300m . If the coefficient of friction is 0.3 and acceleration due to gravity is 10m/s² , the maximum speed of the car be

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Radius = 300 m
• Friction (μ) = 0.3
• g = 10 m/s²

$\large\underline{\underline{\sf To\:Find:}}$

• Maximum speed (v)

We know,

❏$\large{\boxed{\bf \blue{\omega_{max}=\sqrt{\dfrac{\mu g}{r}}} }}$

$\implies{\sf \sqrt{\dfrac{0.3×10}{300}} }$

$\implies{\sf \sqrt{\dfrac{1}{100}}}$

$\implies{\sf \omega_{max}=\dfrac{1}{10}rad/s}$

Now ,

We also know

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀❏$\large{\boxed{\bf \blue{v=\omega × r} }}$

$\implies{\sf v = \dfrac{1}{10}×300}$

$\implies{\bf \red{Maximum\:Speed( v )= 30\:m/s}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Maximum speed of the car be ${\bf \red{30m/s}}$.

Answer 2

$\huge \fcolorbox{red}{pink}{Solution :)}$

Given ,

• Radius of the circular road = 300 m
• Coefficient of friction b/w road and tyres = 0.3
• Acceleration due to gravity = 10 m/s²

We know that , the maximum speed that the car can have to negotiate the turn without skidding is given by

$\mathtt{ \large \fbox{v = \sqrt{\mu \times R \times g \: } }}$

Substitute the known values , we get

$</p><p> \sf \hookrightarrow v = \sqrt{0.3 \times 300 \times 10} \\ \\ \sf \hookrightarrow v = \sqrt{3 \times 300} \\ \\ \sf \hookrightarrow v = \sqrt{900} \\ \\ \sf \hookrightarrow</p><p>v = 30 \: \: m/s$

Hence , the required value of maximum speed is 30 m/s

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