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Karnataka Board PUCPUC Science Class 11

A racing car travels on a track (without banking) ABCDEFA (Figure). 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.

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Question

A racing car travels on a track (without banking) ABCDEFA (Figure). 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 co-efficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 ms–1. Find the minimum time for completing one round.

Long Answer
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Solution

Balancing frictional force for centripetal force = `(mv^2)/r` = f = μN = μmg

Where N is the normal reaction

∴  `v = sqrt(μrg)`  ....(Where, r is radius of the circular track)

For path ABC, Path length = `3/4 (2π  2R) = 3π R = 3π xx 100`

= 300 πm

`v_1 = sqrt(μ2Rg) = sqrt(0.1 xx 2 xx 100 xx 10)`

= 14.14 m/s

∴ `t_1 = (300π)/(14.14)` = 66.6 s

For path DEF, Path length = `1/4 (2πR) = (π xx 100)/2` = 50π

`v_2 = sqrt(μRg) = sqrt(0.1 xx 100 xx 10)` = 10 m/s

`t_2 = (50π)/10` = 5π s = 15.7 s

For path, CD and FA

Path length = R + R = 2R = 200 m

`t_3 = 200/50` = 4.0 s

∴ Total time for completing one round t = t1 + t2 + t3 = 66.6 + 15.7 + 4.0 = 86.3 s 

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Chapter 5: Laws of Motion - Exercises [Page 36]

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NCERT Exemplar Physics [English] Class 11
Chapter 5 Laws of Motion
Exercises | Q 5.37 | Page 36

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