A racing car travels on a track (without banking) ABCDEFA (Fig). ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 m s–1. Find the minimum time for completing one round.

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alok kumar singh | Contributor-Level 10

3 months ago

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – centripetal force, F= mv²/r= f= $μ N = μ m g$

V= $\sqrt[]{μ r g}$

For path ABC path length =3/4 (2 $π 2 R$ )= $3 π R = 3 π (100) = 300 π$

V₁= $\sqrt[]{μ 2 R g} = \sqrt[]{0.1 \times 100 \times 10} = 1.414 m / s$

So t= 300 $\frac{π}{1.414} = 66.6 s$

For path DEF path length = $\frac{1}{4} (2 π R) = 50 π$

V₂= $\sqrt[]{μ R g} = \sqrt[]{0.1 \times 100 \times 10} = \frac{10 m}{s}$

So t₂= 50 $\frac{π}{10}$ = 15.7s

For path CD and FA

Path length =R+R= 200m

T= 200/50= 4s

Total time = 66.6+15.7+4= 86.3s

This is a Long Answer type Questions as classified in NCERT ExemplarExplanation – centripetal force, F= mv2/r= f= <math> <mi>μ</mi> <mi>N</mi> <mo>=</mo> <mi>μ</mi> <mi>m</mi> <mi>g</mi> </math> V= <math> <mroot> <mrow> <mrow> <mi>μ</mi> <mi>r</mi> <mi>g</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> For path ABC path length =3/4 (2 <math> <mi>π</mi> <mn>2</mn> <mi>R</mi> </math> )= <math> <mn>3</mn> <mi>π</mi> <mi>R</mi> <mo>=</mo> <mn>3</mn> <mi>π</mi> <mfenced separators="|"> <mrow> <mrow> <mn>100</mn> </mrow> </mrow> </mfenced> <mo>=</mo> <mn>300</mn> <mi>π</mi> </math> V1= <math> <mroot> <mrow> <mrow> <mi>μ</mi> <mn>2</mn> <mi>R</mi> <mi>g</mi> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <mn>0.1</mn> <mo>×</mo> <mn>100</mn> <mo>×</mo> <mn>10</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>1.414</mn> <mi>m</mi> <mo>/</mo> <mi>s</mi> </math> So t= 300 <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>1.414</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>66.6</mn> <mi>s</mi> </math> For path DEF  path length = <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mn>2</mn> <mi>π</mi> <mi>R</mi> </mrow> </mrow> </mfenced> <mo>=</mo> <mn>50</mn> <mi>π</mi> </math> V2= <math> <mroot> <mrow> <mrow> <mi>μ</mi> <mi>R</mi> <mi>g</mi> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <mn>0.1</mn> <mo>×</mo> <mn>100</mn> <mo>×</mo> <mn>10</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>10</mn> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> </mfrac> </math> So t2= 50 <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> </mfrac> </math> = 15.7sFor path CD and FAPath length =R+R= 200mT= 200/50= 4sTotal time = 66.6+15.7+4= 86.3s

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