Tension in a Massive Rotating Rope with an Object

[AzimD]

You have $dT=\omega^2r.dm_1$, which you turned into an integral as $T(r)-T(0)=\int_0^r\omega^2r.dm_1$. That is not really a valid way to write it. The variable of integration is the thing after the 'd', so in your integral it is $m_1$, but the bounds should refer to that variable, whereas your bounds refer to r.
You then wrote it as $\omega^2r\int.dm_1$, but $dm_1=\rho.dr$, where $\rho$ is the linear density of the rope. You cannot move outside the integral terms which vary through the integral.
You need to solve $T(r)-T(0)=\int_0^r\omega^2r\rho .dr$

For this portion of the problem, I managed to get T(r)-T(0) =(m_1* omega^2 * r^2)/2l

 

[haruspex]

For this portion of the problem, I managed to get T(r)-T(0) =(m_1* omega^2 * r^2)/2l

Looks right. Looks even better with proper use of parentheses, dropping the redundant "*"s, adding the bracing pairs of "#" and "\" in front of "omega":
$T(r)-T(0) =(m_1\omega^2 r^2)/(2l)$

 

[AzimD]

Looks right. Looks even better with proper use of parentheses, dropping the redundant "*"s, adding the bracing pairs of "#" and "\" in front of "omega":
$T(r)-T(0) =(m_1\omega^2 r^2)/(2l)$

Thanks for checking my answer! Yep, I recently found the LaTeX Guide! I'll be attempting to use it from here on out.

 

[AzimD]

Looks right. Looks even better with proper use of parentheses, dropping the redundant "*"s, adding the bracing pairs of "#" and "\" in front of "omega":
$T(r)-T(0) =(m_1\omega^2 r^2)/(2l)$

So going from here then: $T(r)=m_1\omega^2(\frac{r^2}{2l} -l)$

Then considering the Tension caused by the ball $T_b=-m_2\omega^2l$

We can combine the two tensions and get the actual T(r) to be $T(r)=m_1\omega^2(\frac{r^2}{2l} -l) - m_2\omega^2l$

Am I conceptualizing this right? Or does there seem to be an error in my thought process? Am I making it to simple?

 

[haruspex]

So going from here then: $T(r)=m_1\omega^2(\frac{r^2}{2l} -l)$

Looking back through the thread, unless I misunderstand how the variables are defined, the equation of mine you quoted in post #31 is wrong. It has a sign error.
See if you can correct it.
(I believe T(0) should be the tension at the pole if we ignore the point mass.)

Also, what do you have for T(0)?

 

[AzimD]

Looking back through the thread, unless I misunderstand how the variables are defined, the equation of mine you quoted in post #31 is wrong. It has a sign error.
See if you can correct it.
(I believe T(0) should be the tension at the pole if we ignore the point mass.)

Also, what do you have for T(0)?

For T(0) I have $T(0)=-m_1\omega^2l$
Should this be a positive value?

 

[jbriggs444]

For T(0) I have $T(0)=-m_1\omega^2l$
Should this be a positive value?

As I understand it, we are considering "tension" $T(r)$ in the sense of "the force from the anchored direction on the remainder of the cable from point $r$ on out"

As I understand it, we are adopting a sign convention in which outward is positive and inward is negative.

If so, then $T(0)$ would properly be negative. The rope is being pulled inward.

However, $-m_1 \omega^2 l$ would be correct for a mass rotating in a circular trajectory at radius $l$. What is the relevant rotational radius for mass $m_1$ here?

 

[haruspex]

As I understand it, we are considering "tension" $T(r)$ in the sense of "the force from the anchored direction on the remainder of the cable from point $r$ on out"

As I understand it, we are adopting a sign convention in which outward is positive and inward is negative.

If so, then $T(0)$ would properly be negative. The rope is being pulled inward.

That works, but then I would not call it tension (nor label it T). To me, tension is a state, not a force. It is more like a pair of equal and opposite forces at a point, or a distribution of such pairs along a body. A suitable sign convention for that would be positive for a tension and negative for a compression - or the converse.

 

[jbriggs444]

That works, but then I would not call it tension (nor label it T).

Yes. I'd drafted a paragraph pontificating on how tension is more properly a scalar or a portion of a tensor rather than a vector-valued force, but then discarded it as more confusing than helpful.

 

[AzimD]

As I understand it, we are considering "tension" $T(r)$ in the sense of "the force from the anchored direction on the remainder of the cable from point $r$ on out"

As I understand it, we are adopting a sign convention in which outward is positive and inward is negative.

If so, then $T(0)$ would properly be negative. The rope is being pulled inward.

However, $-m_1 \omega^2 l$ would be correct for a mass rotating in a circular trajectory at radius $l$. What is the relevant rotational radius for mass $m_1$ here?

$T(0)$ for $m_1$ would be at length $l$ no? Of course, the force caused by $m_2$ isn't added in because I come back to add that in later.