To find the increase in the radius of a rotating ring

[Kaushik]

Homework Statement

A thin ring of radius 'r' is made of material which has density $\rho$ and Young's Modulus '$Y$'. If the ring is rotated about its centre and in its own plane with angular velocity $\omega$, find the small increase in radius.

Relevant Equations

Centripetal Force $= m(\omega)^2r$
$\sigma=Y(\epsilon)$

• The increase in radius is due to the centripetal force acting on the ring. The centripetal force acting on each point of the ring is directed towards its center.
• We can find force using $F_c = M(\omega)^2R$
• We can use this $F_c$ in the equation of Hooke's Law to find the elongation

Could you please help me in progressing with this problem?
Thanks!

 

[mfig]

Can you find the component of the tension that balances the centrifugal force on a differential element?

 

[Kaushik]

Can you find the component of the tension that balances the centrifugal force on a differential element?

$(dm)\omega^2r = T$
But $dm = \rho(dV)$ .
So $T = \rho(dV)\omega^2r$

Is $dV = \pi(dr)^2R(d\theta)$ ?

 

[haruspex]

$(dm)\omega^2r = T$
But $dm = \rho(dV)$ .
So $T = \rho(dV)\omega^2r$

Is $dV = \pi(dr)^2R(d\theta)$ ?

None of that makes algebraic sense. You cannot have more infinitesimals in one term than in another, e.g. one in $(dm)\omega^2r$ and none in $T$.
If the arc element is length rdθ, what forces act on it and in what directions?
Did you draw an FBD?

 

[collinsmark]

This is an interesting problem. It's making me think.

@Kaushik, you are correct that the centripetal force on differential element of mass, $dm$ is

$dF = (dm) \omega^2 r$

(Same thing as $dF = dm \frac{v^2}{r}$)

But the thousand dollar question is, "Is that the same as the tension?"

And the answer is "no," at least not according to my calculations. But fortunately, they are related.

While I won't tell you what that tention is, I'll explain how I found a formula for the tension.

Instead of the ring being a continuous band, model it as 4 pieces of mass, each with mass $M/4$, held together by massless strings. The four masses are rotating around their center with angular velocity $\omega$.

Calculate the centripital force on one of the masses. You know that that force must come from the the two strings (one on either side): Specifically, the component of the tensions that are directed toward the center. You can then use algebra to solve for the tension, $T$.

Now move on to more masses. For example, 5 masses. Do the same thing. The two things that change is the individual mass is now smaller (M/5 instead of M/4), and the angle becomes smaller too. Look for a pattern here.

Now try to find the formula for the tension $T$ for a system of $n$ masses.

Once you have a formula for the tension with $n$ masses, take the limit as $n \rightarrow \infty$.

The result should be rather pleasing, although not too terribly surprising.

[Edit: Hint: although I used degrees to represent the angles in the above figures, you might find it advantagous to use radians.]

 

[mfig]

...Did you draw an FBD?

Yes, a FBD is essential to one approach for solving this. Perhaps the below image will help. It shows a small length $2rd\theta$ of the ring. Now equate the tension forces with the centrifugal force, use the small angle approximation for the sin function, and then solve for T. Once you have T you can solve for the change in length by using Hooke's law.

 

[Kaushik]

None of that makes algebraic sense. You cannot have more infinitesimals in one term than in another, e.g. one in $(dm)\omega^2r$ and none in $T$.
If the arc element is length rdθ, what forces act on it and in what directions?
Did you draw an FBD?

As per the figure provided by @mfig, Tension is tangential to the point where the $dm$ mass is considered. These are the only forces that act on $dm$. The component of Tension towards the center provides the centripetal acceleration.

Is this tension developed only because of rotating of the ring or can it also be there if the ring is not rotating (to hold each particle together)?

 

[collinsmark]

Is this tension developed only because of rotating of the ring or can it also be there if the ring is not rotating (to hold each particle together)?

I don't think it really matters. But let's just assume that the tension is 0 when the ring is not rotating. The ring is going to act like spring of sorts (well, sort of). Treat the "spring" as being in its equilibrium position when the ring is not rotating.

 

[haruspex]

figure provided by @mfig,

Yes, @mfig has been unduly helpful. That should have been an exercise for you.

The component of Tension towards the center provides the centripetal acceleration.

Ok, but what is the algebraic expression for that component?

 

[Kaushik]

Ok, but what is the algebraic expression for that component?

From this figure,

$2T\sin(d\theta) = (dm)\omega^2r$

 

[Kaushik]

Is this tension developed only because of rotating of the ring or can it also be there if the ring is not rotating (to hold each particle together)?

Although @collinsmark as already answered this question, I also wanted to know what would your (@haruspex) answer be to this question.

 

[haruspex]

From this figure,
View attachment 252304

$2T\sin(d\theta) = (dm)\omega^2r$

Ok, but how can you simplify $\sin(d\theta)$ and how can you express dm in terms of dθ?

 

[haruspex]

Although @collinsmark as already answered this question, I also wanted to know what would your (@haruspex) answer be to this question.

In the absence of any other forces threatening to pull it apart, only because of the rotation.

 

[Kaushik]

Ok, but how can you simplify $\sin(d\theta)$ and how can you express dm in terms of dθ?

As $d(\theta)$ is infinitesimally small, $\sin(d\theta) \approx d\theta$

Also, Is $dm = \frac{Md\theta}{2\pi}$

 

[collinsmark]

As $d(\theta)$ is infinitesimally small, $\sin(d\theta) \approx d\theta$

That's right.

Also, Is $dm = \frac{Md\theta}{2\pi}$

Be careful here. Take a look at @mfig's diagram carefully. What is the total angle that the small section of ring extends?

 

[Kaushik]

Be careful here. Take a look at @mfig's diagram carefully. What is the total angle that the small section of ring extends?

Oops! I drew a different diagram on my notebook.

According to the figure provided by @mfig, is $dm = \frac{Md\theta}{\pi}$?

 

[Chestermiller]

The tension in the ring is in the circumferential direction, and is given by $Ti_{\theta}$, where $i_{\theta}$ is the unit vector in the circumferential direction. If we do a force balance on the section of the ring between $\theta$ and $\theta + \Delta \theta$, we obtain $$[Ti_{\theta}]_{\theta + \Delta \theta}-[Ti_{\theta}]_{\theta}=-\rho A\Delta \theta \omega^2 r i_r$$
where $i_r$ is the unit vector in the radial direction. Dividing by $\Delta \theta$ and taking the limit as $\Delta \theta$ approaches zero gives:$$T\frac{di_{\theta}}{d\theta}=-\rho A\omega^2 r i_r$$Geometrically, the derivative of $i_{\theta}$ with respect to $\theta$ is given by: $$\frac{di_{\theta}}{d\theta}=-i_r$$Therefore, the tension in the ring is given by:$$T=\rho A\omega^2 r $$

The next step is to determine the strain in the circumferential direction. How is the strain in the circumferential direction related to the radial displacement $u_r$?

 

[mfig]

According to the figure provided by @mfig, is $dm = \frac{Md\theta}{\pi}$?

Your problem statement did not mention M anywhere, so why introduce it? I see a density, which I assume is a linear density (kg/m) since we are told the ring is thin. So from the FBD I posted,

$dm = 2 \rho rd\theta$

Now plug this into the expression you got previously ($2Td\theta = r\omega^2dm$) to yield an expression for the tension in terms of your original problem parameters:

$T = \rho r^2\omega^2$

 

[Chestermiller]

Your problem statement did not mention M anywhere, so why introduce it? I see a density, which I assume is a linear density (kg/m) since we are told the ring is thin. So from the FBD I posted,

$dm = 2 \rho rd\theta$

Now plug this into the expression you got previously ($2Td\theta = r\omega^2dm$) to yield an expression for the tension in terms of your original problem parameters:

$T = \rho r^2\omega^2$

The right side should have a factor of the cross sectional area. As written, the right hand side is the tensile hoop stress.

 

[mfig]

The right side should have a factor of the cross sectional area. As written, the right hand side is the tensile hoop stress.

Not if the units of $\rho$ are, as I said, kg/m.

 

[Chestermiller]

Not if the units of $\rho$ are, as I said, kg/m.

Ah. But to get the stress and radial displacement, one needs to divide T by the area, which brings us back to the volume density.

 

[mfig]

Ah. But to get the stress and radial displacement, one needs to divide T by the area, which brings us back to the volume density.

I guess it will be up to Kaushik to understand his problem as stated. The first time I saw this was in high school physics class a long time ago, and we used linear density with a simple linear Hooke's law. As he is in high school, that is how I interpret the problem. Either way, he has both interpretations now.