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Suppose we have a disc which is placed in horizontal plane and have radius R. Disc is rotating and rotation axis is passing through center of disc and in vertical plane . Now we place a coin a on disc and disc instantly feels a force and come outward. Now if we are observing it from outside the disc then there is no force which is acting on coin but why it is coming out.

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  • $\begingroup$ Is the disc smooth ? $\endgroup$ – Aditya Garg Jun 2 at 9:06
  • $\begingroup$ Yes disc is smooth $\endgroup$ – Sanjay Jun 2 at 9:08
  • $\begingroup$ It is easy it would stay still you can see that from it's fbd. $\endgroup$ – Aditya Garg Jun 2 at 13:14
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I would take the liberty to point out that there will be no 'force' the way you expect on the coin when you drop it on disk , as you've mentioned , the disk is smooth so if you are going to put it on the disk without a prior velocity in the plane of disk , the coin is just going to stay there still and the disk will keep on rotating beneath it. Smoothness of disk settles the fact that no frictional interaction should be there between disk and coin and the only other forces being normal forces act in a perpendicular direction to plane have no effect on the coin's outward journey.

After saying all this I am just trying to imply that for the coin to move out like the way you describe , there must be friction on it. Now , when you place the coin on the disk ,friction speeds it up and the coin starts moving in circles , when the tangential velocity is high enough the centrifugal force on the coin surpasses the force of limiting friction between the coin and disk and as a result ,the coin slides outwards.

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  • $\begingroup$ What will happen if disc is not smooth . in daily life we see it throw outward but why it will happen $\endgroup$ – Sanjay Jun 2 at 9:49
  • $\begingroup$ the above explanation is exactly for a non-smooth disk $\endgroup$ – ADITYA PRAKASH Jun 2 at 9:52
  • $\begingroup$ the second paragraph $\endgroup$ – ADITYA PRAKASH Jun 2 at 9:52

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