Laws Of Motion Question 142

Question: A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a uniform angular speed 2 rad $ {{s}^{-1}} $ . The radius of the disc is 50 cm. Find the minimum coefficient of friction between disc and coin so that the coin does not slip $ (g=10m{{s}^{-2}}) $ .

Options:

A) 0.1

B) 0.2

C) 0.3

D) 0.4

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ f _{1}=\mu mg, $ friction will provide the necessary centripetal force $ f=m{{\omega }^{2}}r $

$ f\le f _{1}\Rightarrow m{{\omega }^{2}}r\le \mu mg $

$ \Rightarrow \mu \ge \frac{{{\omega }^{2}}r}{g}=\frac{2^{2}\times 50/100}{10} $

$ \Rightarrow \mu \ge 0.2 $



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