SOLVED: To coin Ig A coin of mass m is on rigid disk at distance d from the center of the disk. There is friction between the coin and the disk. The coefficient of static friction is /s. At time t = 0

Video Transcript

so in this interrogate we are given a situation mrs puff over hera. Looking at the problem from the top scene, we have a big magnetic disk on which a mint placed in addition to d from the center of the large magnetic disk. And this coin has a aggregate of m uh there is friction between the harrow and the coin which is us. And the acceleration due to graveness is vertically down. so since we ‘re looking at this from top see, the graveness vector basically goes into the airplane of the paper on which we are writing and there is, we ‘re told that at meter t equal to zero, peer to zero, the body starts to rotate with an angular acceleration of alpha. so given this, we are asked a couple of questions. So the first part of the question asks us to write, okay what the frictional violence will be ? What is that ? Okay let ‘s destroy release body diagram of the body sol that we can talk a little by rights. so this lashkar-e-taiba ‘s say so did I say central. somewhere over hera. So there will be a centrifugal force out. There will be a frictional push, let ‘s call it F. F centrifugal wedge might be from the center of the coin and there will be A. M. Times G. And there will besides be a normal force out. So these are the four forces that are acting on the consistency as it is rotating. The first partially says, what is the frictional military unit while the mint is at stay, end force as a function of fourth dimension while the mint is at rest. so while the coin is at lie, we can say that the frictional force out will completely counter the centrifugal force because that ‘s all it is doing so while it is remainder, it precisely wholly balances centrifugal impel. thus all we need to do is write a formula for the center which for a deal consistency in this font would be ah millimeter. Omega square. Sorry, omega squared r. There are many guys the instantaneous ah angular speed. so instantaneous angular speed respect to time would be equal to alpha times T. Or write the proper formalizes W. not plus E. And since W. Is zero because the consistency starting from respite, indeed Minka becomes self identity. We can put that in. So we get M alpha T. Square. And the radius for this soundbox is basically because that does the distance from the center. So that becomes trench. So the firm love for the friction as a officiate of meter is equal to M. D. Alpha square T. Square. This is your answer for apartment and part B asks us at what omega uh will the coin begin to slide ? No. thus let ‘s call it south begins to slide. And at this point the centrifugal force will be equal to the utmost clash coerce and maximum friction force. From the free body diagram we can say is adequate to uh mu times the utmost uh normal force and the utmost normal force is peer to mass times gravity or the burden of the forced to do weight of the torso. There ‘s no more force that we can no higher force than this coin can produce. sol this becomes our maximum friction wedge. so once the cardinal pull matches this uh coerce value right after this period the coins has to slide. Because after this the centripetal force will overcome the maximum frictional impel. So we can just write down the frictional violence centripetal force recipe. Again it ‘s um omega feather are or in this case it is D. The distance from the center. And since it has not started slither, the distance is still the same and that is equal to mu. S. Us here. I ‘m good-for-nothing us times M. Times G. So the mask is canceled out. And we see that the omega becomes square root of New Times S. G. By D. And this is the answer for. That ‘s your answer.

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