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Tension in a rotating ring

  1. Apr 9, 2015 #1
    How do we find out the tension in a rotating ring of density ρ, radius r which is rotating with an angular velocity of ω and area of cross section A? And how do we know that this tension would be constant throughout the ring?
    Last edited: Apr 9, 2015
  2. jcsd
  3. Apr 9, 2015 #2


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    Pick out a very small segment of the ring, calculate its radial acceleration. You have the acceleration of that segment and you have its mass, so you should be able to calculate the net force on it. The only forces acting on that segment come from the tension in the ring.
  4. Apr 9, 2015 #3
    Okay. Help me a bit on this. Lets pick an element of length dx. Then volume of this element is Adx and mass of this element is ρAdx. So the momentum of this particle is ρAdx(rω). How do we go about calculating the tension or even some force from here?
  5. Apr 10, 2015 #4


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    Let the tension be [itex]\vec{T_1} , \vec{T_2}[/itex] at the two edges of dx. Then suppose [itex]d\theta[/itex] is the angle between T1 and dx (same as between T2 and dx due to symmetry). Because there is no tangential acceleration it will be [itex] T_1cos(d\theta)-T_2cos(d\theta)=0[/itex] from which [itex]T_1=T_2=T[/itex] because [itex]cos(d\theta)\to 1[/itex]. Now what holds for the force [itex]T_1sin(d\theta)+T_2sin(d\theta)=2Tsin(d\theta)[/itex]? (in what follows you should take notice that [itex]sin(d\theta)=\frac{dx}{2R}[/itex])
  6. Apr 10, 2015 #5
    That [itex]T_1sin(d\theta)+T_2sin(d\theta)=2Tsin(d\theta)[/itex] is the centripetal force and it should be equal to dmω2r = ρAdxω2r ⇒ 2Tdx/2r = ρAdxω2r ⇒ T = ρAω2r2.

    But why is sindθ = dx/2R? Is it because of (dx/2)/R, i.e. we consider half the length of the element while measuring the sine of the angle because the triangle made involves only half the length dx and not full dx? Can you draw a diagram for this please? Also , does cosdθ → 1 or does it just cancel out? Also, how can we be so sure that the tension would exist, i.e. how can we say that each of the particles would definitely experience a pulling force from either side?
    Last edited: Apr 10, 2015
  7. Apr 10, 2015 #6


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    Sorry i cant draw a diagram. It is because the triangle made when we take the vertical from the center of the circle to dx, cuts dx in the middle. The angle made by the radius r and this vertical l is [itex]d\theta[/itex] because the sides of this angle are perpendicular to the sides of [itex]d\theta[/itex](radius r is perpendicular to [itex]T_1[/itex] and l is perpendicular to dx, and angles with perpendicular sides are equal). [itex]cos(d\theta)[/itex] cancels out because we know as [itex]d\theta \to 0, cos(d\theta) \to 1[/itex]. The tension has to exist , otherwise we would have centripetal acceleration without centripetal force, that is violation of newton's 2nd law.
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