Respuesta :
Answer: The coefficient of static friction between the coin and the turntable is 0.13.
Explanation:
As we know that,
Centripetal force = static frictional force
[tex]\frac{mv^{2}}{r} = F_{s}[/tex]
or, [tex]\frac{mv^{2}}{r} = \mu_{s} \times m \times g[/tex]
v = [tex]\sqrt{\mu_{s} \times r \times g}[/tex]
or, [tex]\mu_{s} = \frac{v^{2}}{rg}[/tex] ......... (1)
Here, it is given that
r = 17 cm, [tex]\omega[/tex] = 26 rpm,
and v = [tex]r \omega[/tex] ..........(2)
Putting equation (2) in equation (1) we get the following.
[tex]\mu_{s} = \frac{r^{2}\omega^{2}}{rg}[/tex]
= [tex]\frac{17 \times 10^{-2} \times (26 \times [\frac{2 \times \pi}{60}]^{2})}{9.8}[/tex]
= 0.128
= 0.13 (approx)
Thus, we can conclude that the coefficient of static friction between the coin and the turntable is 0.13.
