Question:

If a cyclist moving with a speed of $4.9\, m / s$ on a level road can take a sharp circular turn of radius $4 \,m$, then coefficient of friction between the cycle tyres and road is

Updated On: Jul 28, 2022
  • 0.51
  • 0.41
  • 0.71
  • 0.61
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The Correct Option is D

Solution and Explanation

We know that while a cyclist moving with a speed $v$ takes a sharp turn on a circular track of radius $r$, the coefficient of friction is given by $\mu=\tan \,\theta=\frac{v^{2}}{r g}$ Here $v=4.9\, m / s$ $r=4 \,m$ and $g=9.8\, m / s ^{2}$ $\therefore \,\,\,\,\,\mu=\frac{4.9 \times 4.9}{4 \times 9.8}=0.61$
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The laws of motion, which are the keystone of classical mechanics, are three statements that defined the relationships between the forces acting on a body and its motion. They were first disclosed by English physicist and mathematician Isaac Newton.

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