Circular MotionHard
Question
If a cyclist moving with a speed of 4.9 m/s on levelled road can take a sharp circular turn of radius 4m, them the coefficient of friction between cycle there and road will be
Options
A.0.81
B.0.41
C.0.71
D.0.61
Solution
For movement on circular path, cantripetal force = force of friction
= μR = μmg
V2 = μrg ⇒ μ =
= 0.61
= μR = μmgV2 = μrg ⇒ μ =
= 0.61Create a free account to view solution
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