Answer:
0.43
Explanation:
At the time of slipping , centripetal force becomes equal to the limiting static friction whose value is μ R or μ mg
Centripetal force which makes it rotate is given by
m v² / r where v is velocity of rotation and r is radius of circular orbit
So at limiting equilibrium
m v² /r = μ mg
μ =[tex]\frac{v^2}{rg}[/tex]
μ = [tex]\frac{120\times120}{34\times980}[/tex]
= .43
The coefficient of static friction between the coin and the turntable is 0.43
If a coin is placed 34 cm from the center of a horizontal turntable, initially at rest, at the time of slippage, the centripetal force is always equal to the friction force. Mathematically;
[tex]F_c=F_f\\\frac{mv^2}{r} = \mu mg\\\frac{v^2}{r} = \mu g\\\mu = \frac{v^2}{gr}[/tex]
Given the following parameters
v = 120 cm/s
g = 980 cm/s²
r = 34 cm
Substitute the given parameters into the formula to have:
[tex]\mu = \frac{120^2}{34 \times 980}\\\mu = \frac{14400}{33,320}\\\mu = 0.43[/tex]
Hence the coefficient of static friction between the coin and the turntable is 0.43
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