Finding tension as a function of distance from the center of rotation

[vparam]

Homework Statement

Consider a mass m attached to a rope of mass M and length L. The rope has constant mass per length. The mass and rope are spun around the end opposite the mass at angular frequency ω. Find the tension in the rope as a function of distance along the rope measured from the center of rotation. You can neglect gravity.

Relevant Equations

a = v^2/r

I'm not too sure how to account for both the mass and the rope at once.
I think the following are true for the two individually:
For the mass at the end, $T = m ω^2 L$, following from $a = v^2/r$and $v=ωr$.
For the rope, $dT = ω^2 r dM$, where $dM = λ dr$ and λ is the mass per unit length = M/L.
Therefore, $dT = λ ω^2 r dr$, so $T = \frac{1} {2} λ ω^2 r^2$ integrating from 0 to r on the right-hand side.

Knowing this, I'm not sure how to put this information together because it seems like having the mass at the end might change how to solve the problem, but I'm not sure how to go about approaching this.

 

[Orodruin]

For the rope, dT=ω2rdM, where dM=λdr and λ is the mass per unit length = M/L.
Therefore, dT=λω2rdr, so T=12λω2r2 integrating from 0 to r on the right-hand side.

This would mean higher tension the further out you get and zero tension at r=0. Is this reasonable?

 

[vparam]

This would mean higher tension the further out you get and zero tension at r=0. Is this reasonable?

I think so, but I'm not so sure. If the rope undergoes uniform circular motion, then the elements further out have to experience a greater centripetal acceleration based on $a=\omega ^2 r$, right?

 

[Delta2]

You forgot the constant of integration. Other than that I can't see any mistake in your reasoning. You can determine the constant from the boundary condition $$T(L)=m\omega^2L\Rightarrow \frac{1}{2}\lambda\omega^2L^2+C=m\omega^2L$$##

 

[Orodruin]

I think so, but I'm not so sure. If the rope undergoes uniform circular motion, then the elements further out have to experience a greater centripetal acceleration based on $a=\omega ^2 r$, right?

Try to do a free body diagram for a small part of the rope. On which side does the tension need to be larger?

 

[Orodruin]

You forgot the constant of integration. Other than that I can't see any mistake in your reasoning. You can determine the constant from the boundary condition $$T(L)=m\omega^2L\Rightarrow \frac{1}{2}\lambda\omega^2L^2+C=m\omega^2L$$##

This is not sufficient. There are other errors.

 

[Delta2]

This is not sufficient. There are other errors.

Hmmmm ok, now that I think of it a bit more carefully there should be a minus sign **somewhere** but other than that I can't see any other error.

 

[Orodruin]

there should be a minus sign **somewhere**

This is the realization that I was trying to guide the OP towards …

 

[vparam]

Try to do a free body diagram for a small part of the rope. On which side does the tension need to be larger?

There has to be some net center-acting force to allow for centripetal acceleration, so the tension pointing inward should be greater than the tension going outward, right? Then, their difference would be dT. Wouldn't dT need to grow as you move further outward because of the need for the net center seeking force?

 

[Orodruin]

There has to be some net center-acting force to allow for centripetal acceleration, so the tension pointing inward should be greater than the tension going outward, right? Then, their difference would be dT. Wouldn't dT need to grow as you move further outward because of the need for the net center seeking force?

It does indeed. But look at what your definition of dT is. If the tension at r is T, then the tension at r+dr is T+dT. So what is dT if you want a net force pointing inwards?

 

[vparam]

It does indeed. But look at what your definition of dT is. If the tension at r is T, then the tension at r+dr is T+dT. So what is dT if you want a net force pointing inwards?

dT should be negative (or at least the opposite sign of dr since they should point in opposite directions as vectors).

 

[Orodruin]

dT should be negative (or at least the opposite sign of dr since they should point in opposite directions as vectors).

… and therefore …

 

[Delta2]

dT should be negative (or at least the opposite sign of dr since they should point in opposite directions as vectors).

This statement seems correct, but in your analysis you consider dr as scalar.

The mystery of the minus sign unfolds if you think that $$dT=T(r+dr)-T(r)$$ and that T(r+dr) and T(r) have opposite directions hence opposite signs and also that $dT$ as the infinitesimal centripetal force on the infinitesimal segment $dr$ must point inwards, hence must have the sign of T(r).

 

[Delta2]

Actually sorry didn't express myself correctly. The whole point is that the infinitesimal centripetal force is $-dT$ and not just $dT$ but I leave you alone to think the "why" of this. It would help greatly if you **carefully** do as @Orodruin suggested at post #5

 

[vparam]

… and therefore …

$dT=-\lambda\omega^2rdr\Rightarrow T=-\frac{1}{2}\lambda\omega^2r^2+C$

Could I then use $T=m\omega^2L$, the tension that is needed for the block to find C?

i.e.:
$T(L)=-m\omega^2L=-\frac{1}{2}\lambda\omega^2L^2+C =-\frac{1}{2}(\frac{M}{L})\omega^2L^2+C =-\frac{1}{2}M\omega^2L+C$

 

[Delta2]

I believe the boundary condition is $T(L)=m\omega^2L$ in your analysis you "silently" assumed that the positive direction is inwards, that is towards the center.

 

[vparam]

This statement seems correct, but in your analysis you consider dr as scalar.

The mystery of the minus sign unfolds if you think that $$dT=T(r+dr)-T(r)$$ and that T(r+dr) and T(r) have opposite directions hence opposite signs and also that $dT$ as the infinitesimal centripetal force on the infinitesimal segment $dr$ must point inwards, hence must have the sign of T(r).

Oh I see, this explanation makes sense. I think in my confusion I wasn't identifying that the tensions are acting a distance dr apart (which makes sense b/c the tensions acting in either direction should be equal at one point). Thank you for clarifying this!

 

[vparam]

I believe the boundary condition is $T(L)=m\omega^2L$ in your analysis you "silently" assumed that the positive direction is inwards, that is towards the center.

Got it! Ok, so correcting for this: $m\omega^2L=-\frac{1}{2}M\omega^2L+C \Rightarrow C=m\omega^2L+\frac{1}{2}M\omega^2L=\omega^2L(m+\frac{1}{2}M)$.

So the final answer would be: $T(r)=\omega^2L(m+\frac{1}{2}M)-\frac{M}{2L}\omega^2r^2$?

 

[Orodruin]

I believe the boundary condition is $T(L)=m\omega^2L$ in your analysis you "silently" assumed that the positive direction is inwards, that is towards the center.

Tension is well defined and is a scalar quantity, not a vector quantity, it does not have direction as such. It is positive when the string is taut. The force on an object is proportional to the tension multiplied by the unit vector of the cut of the string when you make a free-body diagram. This means that it needs to be positive in order to keep the mass from flying away as the unit vector is in the negative r direction.

Compare with the generalisation to a continuum, which is the rank two stress tensor.

 

[Delta2]

Tension is well defined and is a scalar quantity, not a vector quantity, it does not have direction as such. It is positive when the string is taut. The force on an object is proportional to the tension multiplied by the unit vector of the cut of the string when you make a free-body diagram. This means that it needs to be positive in order to keep the mass from flying away as the unit vector is in the negative r direction.

Compare with the generalisation to a continuum, which is the rank two stress tensor.

Hmm, are you saying that Tension can be only positive or zero?

Got it! Ok, so correcting for this: $m\omega^2L=-\frac{1}{2}M\omega^2L+C \Rightarrow C=m\omega^2L+\frac{1}{2}M\omega^2L=\omega^2L(m+\frac{1}{2}M)$.

So the final answer would be: $T(r)=\omega^2L(m+\frac{1}{2}M)-\frac{M}{2L}\omega^2r^2$?

Yes now I think you are absolutely correct.

 

[Orodruin]

So the final answer would be: $T(r)=\omega^2L(m+\frac{1}{2}M)-\frac{M}{2L}\omega^2r^2$?

I would probably group it as follows
$$
T(r) = m\omega^2 L + \frac{M}{2}\omega^2 L\left(1 - \frac{r^2}{L^2}\right),
$$
but yes.

So what is the tension at $r = 0$? Does this result seem compatible with what you would experience if you were to swing an object attached to a string around over your head?

 

[Orodruin]

Hmm, are you saying that Tension can be only positive or zero?

If you have a string that cannot support compressing forces, yes.

 

[vparam]

I would probably group it as follows
$$
T(r) = m\omega^2 L + \frac{M}{2}\omega^2 L\left(1 - \frac{r^2}{L^2}\right),
$$
but yes.

So what is the tension at $r = 0$? Does this result seem compatible with what you would experience if you were to swing an object attached to a string around over your head?

Evaluating at r=0, $T(0)=m\omega^2L+\frac{M}{2}\omega^2L$. I see what you mean now: yes, it makes sense that the tension is at a maximum at r=0 based on experience.

Thank you, @Orodruin and @Delta2 for all the help!

 

[Orodruin]

yes, it makes sense that the tension is at a maximum at r=0 based on experience.

Tripple olympic hammer throw champion Anita Włodarczyk agrees. Tension is not zero at $r = 0$

Seriously, check out that woman's championship record ... she's been dominating the sport for 10 years and won every single championship during that time ... She also hold the world record, having thrown just above 2.5 meters further than anyone else in history.