Question

a metal ring of mass M and radius R is placed on the smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v .find the tension in the ring

Answer 1

Your answer appears to have a tiny flaw Sarang Sharma.

At the beginning we can picture given ring as a finit point system held in place by tension forces and afterwards explore what happens in infinitesimal case. Let us say α is the angle between two adjacent points of that finit point ring as seen from center of the ring (finit point ring is actually an n-sided polygon). Considering that we can choose one discrete point of the ring and say T1 and T2 are the only forces acting upon this chosen point by two adjacent points.

Correct reasoning would then be:
T1+T2=Fcp
T1sinα2+T2sinα2=mv2R
If the ring is homogeneous then |T1|=|T2|=T:
2Tsinα2=mv2R

where m is the mass of that one point of the ring and in case of an n-point ring m=Mn.
However, if increase the number of discrete ring points n→∞ we get α→dα and m→dm:
2Tsindα2=dmv2R

Assuming uniform distribution of mass in the ring it can be said dm=M2Rπdl and also in infinitesimal case sindα=dα:
Tdα=M2Rπv2Rdl

by the definition of radian it obvious thatdα=dlR
Tdα=Mv22Rπdα

which finally yields the ring tension equation:
T=Mv22Rπ

Hope it helps
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Answer 2

Answer:

your answer is mv^2/2πr

Explanation:

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