Physics, asked by Atlas99, 1 month ago


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A circular track of radius 220 m is banked at an angle of 60°. If the coefficient of friction between the wheels of a car and the road is 0.5, then what is the (i) optimum speed of the car to avoid wear and tear on its tires, and (ii) maximum permissible speed to avoid slipping?

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Answers

Answered by Anonymous
31

Question :-

  • A circular track of radius 220 m is banked at an angle of 60°. If the coefficient of friction between the wheels of a car and the road is 0.5, then what is the (i) optimum speed of the car to avoid wear and tear on its tires, and (ii) maximum permissible speed to avoid slipping ?

Given :-

  • A circular track of radius 220 m is banked at an angle of 60°.
  • The coefficient of friction between the wheels of a car and the road is 0.5.

To Find :-

  • What is the optimum speed of the car to avoid wear and tear on its tire.
  • What is the maximum permissible speed to avoid slipping.

Solution :-

First, we have to find the optimum speed of the car to avoid wear and tear on its tire :

As we know that,

\bigstar\: \: \rm\boxed{\bold{\pink{tan\: \theta =\: \dfrac{v^2}{rg}}}}\: \: \bigstar

Given :

  • Radius (r) = 220 m
  • \rm\theta = 60°
  • Acceleration due to gravity (g) = 9.8 m/

According to the question by using the formula we get,

\longrightarrow \bf tan\: \theta =\: \dfrac{v^2}{rg}

By doing cross multiplication we get,

\longrightarrow \sf\bold{\purple{v^2 =\: rg\: tan\: \theta}}

\longrightarrow \sf v^2 =\: (220)(9.8) \times tan\: 60^{\circ}

As we know that :

\leadsto \bf tan\: 60^{\circ} =\: \sqrt{3}

\longrightarrow \sf v^2 =\: 220 \times 9.8 \times \sqrt{3}

\longrightarrow \sf v^2 =\: 2156 \times 1.732

\longrightarrow \sf v^2 =\: 3734.192

\longrightarrow \sf v =\: \sqrt{3734.192}

\longrightarrow \sf\bold{\red{v =\: 61.11\: m/s}}

{\footnotesize{\bold{\underline{\therefore\: The\: optimum\: speed\: of\: the\: car\: to\: avoid\: wear\: and\: tear\: on\: its\: tires\: is\: 61.11\: m/s\: .}}}}

Now, we have to find the maximum permissible speed to avoid slipping :

As we know that :

\bigstar\: \: \rm\boxed{\bold{\pink{v_{(maximum)} =\: \sqrt{\dfrac{rg(tan\: \theta + \mu)}{1 - \mu\: tan\: \theta}}}}}\: \: \bigstar

Given :

  • Radius (r) = 220 m
  • Acceleration due to gravity (g) = 9.8 m/
  • \rm \theta = 60°
  • \rm \mu = 0.5

According to the question by using the formula we get,

\longrightarrow \sf\bold{\purple{v_{(maximum)} =\: \sqrt{\dfrac{rg(tan\: \theta + \mu)}{1 - \mu\: tan\: \theta}}}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{(220)(9.8)\{tan\: 60^{\circ} + 0.5\}}{1 - (0.5 \times tan\: 60^{\circ})}}

As we know that :

\leadsto \bf tan\: 60^{\circ} =\: \sqrt{3}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{220 \times 9.8(\sqrt{3} + 0.5)}{1 - (0.5 \times \sqrt{3})}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{2156(1.732 + 0.5)}{1 - (0.5 \times 1.732)}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{2156(2.232)}{1 - (0.866)}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{2156 \times 2.232}{1 - 0.866}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{\dfrac{4812.192}{0.134}}

\longrightarrow \sf v_{(maximum)} =\: \sqrt{35911.88}

\longrightarrow \sf\bold{\red{v_{(maximum)} =\: 189.50\: m/s}}

{\footnotesize{\bold{\underline{\therefore\: The\: maximum\: permissible\: speed\: to\: avoid\: slipping\: is\: 189.50\: m/s\: .}}}}

Answered by Prince063867
1

Explanation:

A circular track of radius 220 m is banked at an angle of 60°. If the coefficient of friction between the wheels of a car and the road is 0.5, then what is the (i) optimum speed of the car to avoid wear and tear on its tires, and (ii) maximum permissible speed to avoid slipping ?

Given :-

A circular track of radius 220 m is banked at an angle of 60°.

The coefficient of friction between the wheels of a car and the

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