### Video Transcript

question number 12 s. What is the coefficient of friction between a coin and a turntable if the coin only slides off when the turntable is spinning at a rate of 36 revolutions per moment. so to illustrate this, let ‘s create our turning board right here. You know, assume it ‘s something bland like this with the coin at one edge, the mint is at a distance. R, from the center of the turntable in the turntable is rotating at a speed V. now the impel of friction will be acting in this direction, keeping the coin from sliding off and the normal force will be act upwards. Well, I m thousand peer to the coins ways Who will be act downwards immediately ? It ‘s the clash between the turntable and coin that causes centripetal wedge to occur. The mint only slides off when the wedge of clash is n’t big enough to keep that in apparent motion. consequently, we consent the effect of clash equal to the rotational pull. In other words, we can say the coefficient of friction times The normal force is equal to our rotational push in V squared over R. now the normal power is equal to M times G The coins wait so we can substitute magnesium in for FC N and say that New Times M gram is adequate toe meter b-complex vitamin squared over r. The aggregate can be cancelled out from both sides to allow us to write this as mu times G is equal to the square over our No, we want to find mu hera, the coefficient of friction. So let ‘s divide both sides by gram t. Get mu on its own That will give us mu is equal to b squared over g times are now. There are a few values we know here where the mint sits on. The turntable is 11 centimeters from the center of it. Oh are 0.11 meters and the turntable is turning at a speed of 36 r p m. Before the coin slides off of it. How ? Because all our other constants use meters as their units. We ‘re gon na need to convert our rotation ‘s permitted into meters per second, so we take 36 revolutions over one minute, reproduce by one infinitesimal over 60 seconds and then be multiplied by the circumference of the coins path in one rotation two times by times 20.11 meters in one revolution. Canceling out all of that we can we find that our revolutions per minute hera is 0.415 meters per second base. now we confined our clash coefficient that will be equal to the speed we found 0.415 meters per second squared, divided by are 9.8 meters per moment squared. That ‘s G prison term 0.11 meters they are. This gives a measure for our friction coefficient of 0.16