Problem

The hammer throw is a track-and-field event in wh…

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Problem 20 Medium Difficulty

Multiple-Concept Example 7 deals with the concepts that are important in this problem. A penny is placed at the outer edge of a disk (radius 0.150 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.80 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.

Answer

0.187

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Physics 101 Mechanics

Physics

Chapter 5

Dynamics of Uniform Circular Motion

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Video Transcript

here we have to find the minimum coefficient of friction necessary. Give it penny indicated by the red dot rotating with me discredited scorn and not just lay off. So let's start by finding the angular velocity in this problem, we know that he is equal to two pi over Omega. So that means that omega must be too piratey and given team. This problem, so we have to do is put the quantity we get that omega is to pie over 1.80 seconds. And since the only above his unit ingredients, we get that omega equal to 10 pie ninth. Look, that looks like a G ninth red per second. Okay, now, from there we can analyze our forces. So since our professional force is the only force pointing inward toward the center of the circle, we know that the fresh Nall force must be responsible for a centripetal force as well. Now, the first of the fictional Force Correctional is given by F N times. The coefficient of friction, which is what we're looking for in this problem. And Evan must be equal to M G. Since, as you can see in the free body diagram it's opposing the gravitational force, so now we can right M J Times The coefficient of friction is M B squared over R and R centripetal force. We can also rewrite in terms of omega and just get and Omega Square are if you wanted to, you could also have sold for V Street from the equation or period. It really doesn't make a difference with where you approve it, because in the end it's all just known quantities. So we get MG cooperation. Friction equals and will make the square are and our enemies cancel out. Now we can right our co efficient in terms of we'll make a squared R and G and then all that is left to do is plug in. So we just calculated our omega, which is when we square it is ah 100 pi squared over anyone readings per second multiplied by our radius with 0.15 meters and that is all over our No, our co are gravitational acceleration which is 9.8 meters per second squared and who me duel that out? Plummeting toward calculator, we find that our coefficient of friction is equal to zero point 1865 that is, the minimum coefficient of friction required a pretty penny to broke eight with the circle, and that is our final answer.

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