Question #91706
Question 19
A coin placed 0.4m from the centre of a rotating, horizontal turntable slips when its speed is 0.6 m/s. What is the coefficient of static friction between the coin and turntable?
1
Expert's answer
2019-07-22T11:33:28-0400

At the moment when the coin starts to slide, centripetal force equals to the friction force:


Fcentr=FfricF_{centr}=F_{fric }

That means:


mv2r=μmg\frac{mv^2}{r}=\mu mg

μ=v2rg=0.620.49.8=0.092\mu=\frac{v^2}{rg}=\frac{0.6^2}{0.4*9.8}=0.092

So, the coefficient of friction is 0.092


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