A 140 g coin sits on a horizontally rotating turntable. the turntable makes one revolution each 1 s. the coin is located 14 cm from the axis of rotation of the turntable. the acceleration of gravity is 9.81 m/s2. what is the frictional force acting on the coin? answer in units of n. the coin will slide off the turntable if it is located more than 24 cm from the axis of rotation. what is the coefficient of static friction?

The frictional force experienced on the coin, when it is at a distance of 14 cm is 0.77N and the coefficient of static friction when the coin slides off will be 0.964.

To find the answer, we need to know more about the friction.

How to find the coefficient of static friction?

We have given with the following values,

[tex]m=140*10^{-3} kg\\w=1 rev/s=2\pi rad/s\\r_1=14*10^{-2}m\\g=9.81m/s^2\\[/tex]

When the coin is at a distance of 14cm from the axis of rotation, it will experience a centripetal force inward, and which is in opposite direction of the frictional force.
Thus, the frictional force experience on the coin will be equal to the centripetal force.

[tex]f=F_c=ma_c=m*r_1w^2\\f=140*10^{-3}*14*10^{-2}*4*\pi ^2=0.77N[/tex]

The coin will slide off the turntable, if it is located at a distance of 24cm.
We have to find the coefficient of static friction when it is at a distance of 24cm from the axis of rotation.
We have the expression for static friction as,

[tex]f_s=kN,\\\\Where, \\N=mg\\\\k=\frac{f_s}{mg}[/tex]

We have to find the value of static friction at a distance 24 from the center of the table. This will be equal to the Fc at 24cm.

[tex]F_c=f_s=mrw^2=140*10^{-3}*24*10^{-2}*4\pi ^2\\f_s=1.325N[/tex]

Thus, the coefficient of static friction will be,

[tex]k=\frac{1.325}{140*10^{-3}*9.81}=0.964[/tex]

Thus, we can conclude that, the frictional force experienced on the coin, when it is at a distance of 14 cm is 0.77N and the coefficient of static friction when the coin slides off will be 0.964.

Learn more about the friction here:

https://brainly.com/question/20129398

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