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November 2nd, 2010, 10:31 PM   #1
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Physics Question.

A coin is placed 13.0cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 34rpm is reached and the coin slides off. What is the coefficient of static friction between the coin and the turntable?
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November 2nd, 2010, 11:30 PM   #2
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Re: Physics Question.

We have that the magnitude of centripetal acceleration for circular motion of radius r and speed v is given by:

Also, we have that the magnitude of the velocity (speed) for circular motion is

where t is the time it takes to make one revolution. We are given a frequency f, or revolutions per time, thus:

We want the frequency in revolutions per second, since that is the unit of time that is used for the constant of gravitational acceleration .

And since frequency is revolutions per time, we may express the rotational speed as:

, and therefore, our centripetal acceleration is:

Now if we equate the maximum for the force of static friction with the centripetal force (using Newton's 2nd law), we have:

Divide through by mass m, and solve for the coefficient of static friction :

Now, using r = 0.13 m., f(rps) = f(rpm)/60 = 34/60 = 17/30, g = 9.81 m/sē, we have

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November 3rd, 2015, 10:47 PM   #3
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The circular velocity part was where I got caught. Very strait forward now, thank you.
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November 12th, 2015, 06:35 AM   #4
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34 rpm means 68pi radians/60 secs that makes angular speed as 1.13pi radians per second.
since coin is placed at a distance of 13 cm so linear velocity at this point will become
v= (13/100)1.13pi m/sec
v= 0.46 m/sec
now centrefugal force on it would be m*v^2/r= m*0.46*0.46*100/13=m1.62 N
frictional force would be Cmg here, c is coefficient of friction
as both forces must be equal for equilibrium
so Cmg=1.62m
this means C=1.62/9.8
here g=9.8 acceleration due to gravity
sp C=0.16
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