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 $
BETA
