Tension in a rope wrapped around a shaft, pulling in a mass

[rashida564]

Homework Statement

One end of a uniform rope of mass m1 and length l is attached to a shaft that is
rotating at constant angular velocity of magnitude ω. The other end is attached to a
point-like object of mass m2. Find T(r), the tension in the rope as a function of r, the
distance from the shaft

Relevant Equations

Fc=MV^2/R

Hi everyone, I am thinking of solving it this way. Since Fc=mw^2r. and Fc=T. Then -dT=dmw^2r. where dm=m1(dx/l). and putting the lower at x=l the tension will be T=m2w^2l.
Am I doing things right.

 

[haruspex]

The set up is not clear. Is there a diagram?
What is r?
Is this in a horizontal or vertical plane?
Is the rope wrapped around the shaft (radius?) or... what?

 

[rashida564]

So r is the distance to the rotation rod.

 

[gneill]

So your shaft has radius zero (since the rope has length l and the image shows the distance between the shaft center is also l. Is gravity acting on the rope and point mass, or is this "experiment" taking place in deep space, away from any significant gravitational fields?

 

[rashida564]

The question doesn't mention anything about gravity. But I think their must be no gravity, otherwise This rope won't be able to move in circular motion. "It will having some Angular motion in the plane of the page"

 

[haruspex]

dT=dmw^2r. where dm=m1(dx/l)

Two of those variables are really the same. Fix that and you will be well on your way to the answer.

By the way, your diagram shows that the thread title is misleading. The rope is not wrapped around the shaft and the mass is not being wound in.

 

[jbriggs444]

Hi everyone, I am thinking of solving it this way. Since Fc=mw^2r. and Fc=T. Then -dT=dmw^2r. where dm=m1(dx/l). and putting the lower at x=l the tension will be T=m2w^2l.
Am I doing things right.

Why integrate at all?

It is simple to calculate the mass and center of mass of the portion of rope + mass beyond a given point.

 

[haruspex]

Why integrate at all?

It is simple to calculate the mass and center of mass of the portion of rope + mass beyond a given point.

Yes, but that only works because the centripetal force varies linearly with r (for a given angular speed). For a student that knows how to integrate but is not confident of the linearity argument, taking the long way round seems reasonable.

 

[rashida564]

Why integrate at all?

It is simple to calculate the mass and center of mass of the portion of rope + mass beyond a given point.

Can you explain to me why does logic works

 

[jbriggs444]

Can you explain to me why does logic works

@haruspex has alluded to it -- linearity.

I find it easiest to think of this problem in terms of centrifugal force. Each tiny piece of the rope and the mass on the end is subject to a centrifugal force. That force is, of course, proportional to the mass of the piece. It is also proportional to the distance from the pole.
$$F=m\omega^2 r$$
The key is that it is directly proportional to r -- it is "linear in r".

If we imagine piece of rope with significant length, part of the rope will be closer to the pole. Part of it will be farther away. The part that is close to the pole will have reduced centrifugal force relative to the rope's midpoint. The part that is far from the pole will have increased force relative to the rope's midpoint. It ends up being a wash -- the total force on the rope is the same as if all the mass were concentrated at the midpoint. More accurately, it is the same as if all the mass were concentrated at the position of the center-of-mass.

That is my intuition talking. Let us see if we can formalize it.

If force is proportional to displacement, the total force on a piece of rope is proportional to the sum (or integral if you prefer) of the mass of each piece times the displacement of that piece.Leaving out the constant of proportionality throughout the following, that's:$$F = \sum_i m_i x_i$$The formula for the position of the center of mass is
$$X_{\text{CM}} = \frac{\sum_i m_i x_i}{M}$$
The formula for the force that would be applied if all the mass were positioned at the CM is
$$F = M X_{\text{CM}}$$
If we substitute in the formula for $X_{\text_{CM}}$ it is clear that this will give the same result as adding up the force on each of the incremental mass elements.

Or, to put it differently, since force is proportional to mass times displacement, it does not matter which we integrate. One is just a fixed multiple of the other.

 

[rashida564]

Thanks, now I get it but wouldn't this but the total tension on the rope instead of being the tension at distance r from the rotating rod

 

[jbriggs444]

For the tension at a certain point, you are interested in the force needed to support the mass at the end of the rope plus the portion of the rope further out than that point. Your task is to come up with a formula for that force.

 

[rashida564]

Oh I think I get it, also since the rope is uniform density then, the centre of the mass of the rope will be (r+l)/2. And the portion of the mass will be m1(l-r)/l . I can find the centripetal force of the rope and add it to the mass then that will be the tension, am I right.

 

[jbriggs444]

Oh I think I get it, also since the rope is uniform density then, the centre of the mass of the rope will be (r+l)/2. And the portion of the mass will be m1(l-r)/l . I can find the centripetal force of the rope and add it to the mass then that will be the tension, am I right.

Yes, that looks entirely correct.

 

[haruspex]

since the rope is uniform densit

Only by assuming the strain is small. Note how interesting the problem gets if we have to allow for the non-uniform stretch that would arise.

 

[rashida564]

Only by assuming the strain is small. Note how interesting the problem gets if we have to allow for the non-uniform stretch that would arise.

Way more interesting, so we have to find the centre of the mass of the portion using integration, so we can model it as single particle with the given mass, then we find the centripetal force from the centre of mass to the rotation, will it be solved this way?

 

[haruspex]

Way more interesting, so we have to find the centre of the mass of the portion using integration, so we can model it as single particle with the given mass, then we find the centripetal force from the centre of mass to the rotation, will it be solved this way?

If stretched, the non-uniformity will make it necessary to integrate. The centre of mass will not help.