A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of mu(s) = 0.80 and mu(k) = 0.50. The turntable very slowly speeds up to 60 rpm. Does the coin slide off?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A binder clip of mass 2 g is placed six centimeters from the axis of rotation of a spin coaster. The coefficient of static and kinetic friction between the binder clip and the spin coaster surface is us equals 0.95 and mu K equals 0.64 respectively. The spin coaster starts from rest and accelerates until it reaches a speed of 20 rotations per minute, identify the maximum angular velocity and determine whether the binder clip will stay on the spin coaster or slide off. Awesome. So it looks like our end goal is we're trying to determine the maximum angular velocity. And we're also asked to figure out whether or not the binder clip will stay on the spin coaster or it will slide off as it's spinning. Awesome. So that's ultimately what we're trying to solve for is we're trying to figure out the maximum angular velocity and we're also trying to figure out if the binder clip will stay on the spin coaster or come off. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer pair might be. A is 114.4 rotations per minute won't slide off. B is 168.2 rotations per minute slide off. C is 88.4 rotations per minute slide off. And finally D is 118.9 aren't rotations per minute won't slide off. Awesome. So first off, let us try picture to help us better visualize this problem. So I've gone ahead and put together a fancy diagram for us to refer to, to help us visualize what's going on in this problem. So it says paper clip, but it's, it's looks like a binder clip. So it's paper clip, binder clip, it doesn't really matter, but we have our binder clip represented in red and it's clipped to the spin coaster and from the binder clip to the center of our spin coaster, that's our radius which is six centimeters and it's spinning at some angular velocity which is represented by Omega which is going counter clockwise it looks like. And we have our spin coaster which is represented by this gray circle. Awesome. So firstly, now that we have created our diagram picture to help us better visualize what's going on. So by using our picture, we can also create a free body diagram. So let's do that. So we have our normal force pointing upwards which we could denote as capital F subscript capital N. So that's our normal force. And then going down, we have our gravitational force which is denoted as F subscript G and then going to the left fighting, fighting. This motion is our static friction. And then we have going to the right is our radius R and going up towards the normal force. This would be Z awesome being as a three dimensional plane. So once again, FS is our static friction. FN is our normal force. FG is the force of gravity and R is the radius. Awesome. So moving right along. So let us also note that the maximum centripetal force experienced by the binder clip will be equal to the force of the static friction. For example, FC, the centrifugal force is equal to the static friction. So we can write the equation for the centrifugal force as FC is equal to the mass multiplied by the acceleration of the centripetal force where AC is the acceleration of the centripetal force. And let's note that AC is equal to V squared divided by R. Therefore, we could write that FC is equal to M multiplied by V squared, all divided by R. And we can also write that FS is equal to M multiplied by V squared all divided by R. Where once again the repeat FS is the force of the static friction. And we can write that FS is equal to mu s multiplied by F. Thus, we can go ahead and write that us which is the coefficient of static friction multiplied by the normal force is equal to mass multiplied by V squared all divided by R. And let's call this equation one, we need to note however, that we do not know the value of FN. So now we need to recall and apply Newton's second law of motion to the Z axis which states that the sum of FZ is equal to M multiplied by A. So let us note also that since the binder clip has no vertical motion that will mean and let's make a note here in blue that the acceleration will equal zero. Thus, we can write that the sum of FZ is equal to zero. And we can write that FN minus FG equals zero. So we can take it even further and write that FN equals FG. So we can write that FN equals mass multiplied by gravity. So now we can substitute this into equation one. So let's do that. So MS or sorry, us, us multiplied by M multiplied by G is equal to M multiplied by V squared divided by R which we can simplify and write that us multiplied by G is equal to V squared divided by R. Awesome. So note that the tangential velocity is written as and let's write this in blue V equals are multiplied by omega. So we can write that MS multiplied by G is equal to R multiplied by Omega max squared. Where when we rearrange the isolating sulfur omega max, we'll get that Omega max is equal to the square root of us multiplied by G. So mute once again. So this is MS divided by R. OK. So the Koi fish, so the coefficient, a static friction multiplied by gravity divided by R and this is all the square root of mm us multiplied by G all divided by R. So now we can plug in all of our known variables and solve for our final answer. So let's do that shall we? So we'll determine that Omega max to the max angular, the angular subventions were ultimately trying to figure out the maximum angular velocity. So the maximum angular velocity is equal to the square root of 0.95 multiplied by 9.8 which is gravity meters per second squared or the value for gravity, I should say all divided by 0.06 m. And note that we have to convert all of our centimeter values to meters. Awesome, which is equal to 12.46. Yeah, so four. So it's 12.456 which will round to 12.45 when we round to two decimal places, radians per second. Awesome. But however, we need to convert radiance per second to rotations per minute. R PM. So to do that, we need to note that. And let's make another note here in blue that in one radiant per second there is 9.55 RP MS. So with that in mind, let us determine that our maximum angular velocity is equal to 12.45 radiance per second, multiplied by 9.55 are PMS in one radon second. So our radiance per second cancels out leaving us with just RP MS, which is what we want our final U units to be in. So when we plug that into a calculator, we will get 118.9 RP MS when we round to one decimal place. And this is our final answer. Hooray, we did it. The max speed of the binder clip is 118.9 rotations per minute and it turns out that the binder clip will not slide off, so not slide off. Awesome. So let's go look at our multiple choice answers really quick and determine which answer corresponds with the answers that we found together. And it turns out the correct answer has to be the letter D 118.9 RP MS rotations per minute and it won't slide off. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.