# HELP stuck on this problem

1. Mar 15, 2005

### Sesj

HELP!!!! stuck on this problem

A circle of rope of total mass 3.14kg and a radius of 2 meters is spinning at an angular velocity of 1 rad/sec about an axis through the center of the circle. What is the tension in the rope?

any help or solution would be great!!!

2. Mar 16, 2005

### Berislav

I'm not a physicist (and as such not a competant advisor), but it seems to me that it would be equal to the tangenitial component of the force,

$$F_t=m\alpha r$$.

Someone should possibly confirm or deny.

Last edited: Mar 16, 2005
3. Mar 16, 2005

### ramollari

Berislav, there is no angular acceleration $$\alpha$$ because the motion is uniformly circular.

Also the question is not as easy as it seems. Because the rope has mass, the tension varies (probably non-linearly) along the length of the rope. So it depends on which point of the rope is asked the value of tension T. It can be expressed as a formula, instead.

4. Mar 16, 2005

### Berislav

Ah, yes. I misread the question. I thought that it said angular acceleration instead of angular velocity. Sorry for posting the wrong answer.

EDIT:

This is probably wrong as well, but since I'm planning to study physics at University I would like to try to solve as much exercies as possible:

What about finding the tension by comparing the differance of the x-th component of two centripical forces acting in two infitesmaly close point and then integrating over the enitire rope.

Last edited: Mar 16, 2005
5. Mar 16, 2005

### ramollari

Yes, you are close to the solution. The tension at the very end is zero. If you sum up (or rather integrate) the differences in tension up to the point we are interested in, you come up with the resultant tension.
A better approach IMO is to divide the rope into tiny pieces of mass dm and find the tension on each piece as

$$dT = \omega ^2rdm$$,

Knowing that the tension is transmitted along all the rope, we add the tensions of all the pieces with integration from $$m_0$$ the mass at the extreme (= 0), to $$m_f$$, which is the mass from the point we are interested in to the extreme. The variable r can be expressed in terms of the mass from a specific point to the extreme of the rope, since the latter is uniformly distributed.

$$T = \int dT = \omega ^2 \int_{m_0 = 0}^{m_f} r(m)dm$$,

PS: I may also have misunderstood the question because maybe the rope is in circular shape rotating about an axis passing through its centre.

Expressing r in terms of m is left to the guy who started the thread.

Last edited: Mar 16, 2005