# Tension of string and circular motion question

1. Feb 16, 2006

### Signifier

Let's say I hang an object of mass M from a string attached to the ceiling. The force of gravity on this object at rest would be [0,0,-Mg]. The tension force I assume would be [0,0,Mg] by Newton's third law. (Am I right on this?)

Next, assume I start the object spinning in a horizontal circle at a constant speed v. The force of gravity still points down, but the direction of the force of tension is now rotated. Let's say theta is the angle the end of the string (where the object is attached) makes with the horizontal axis, that the string is S meters long and that the circle the object is spinning has a constant radius R.

Now is where I start guessing, essentially. The magnitude of the tension force would be (Mv^2)/R (by using Newton's second law and the fact that for uniform circ. motion the acceleration a = (v^2)/R). The vector of the tension force would be (Mv^2)/R multiplied by [cos(theta),sin(theta),0]. Is this right?

Any help would be greatly appreciated. Thank you!

2. Feb 16, 2006

### arildno

No. Whatever makes you think this has something about Newton's 3.law?
The tension is certainly (0,0, Mg), but that result follws from Newton's 2.law and that the object experiences no acceleration.
No.

1. You need to specify the angle the string makes with the VERTICAL.
Assume that one is fixed. (i.e, the string traces out a cone in 3-D).
see if you can work further on this.

3. Feb 16, 2006

### cliowa

If you have a problem like this, it is actually quite easy to get a feeling for what is happening: you know by experience that the tension in the string will be
- lower?
- higher?
when rotating the object of mass M around the z-axis. Now, your next step is quite alright in general terms. You're assuming constant speed at a fixed radius R. Try drawing the forces acting on the object and the tension in the string. Note that the tension force MUST be in the direction of the string. Then include the force acting on the object due to your circular motion.

The situation should be more clear after doing this, and you shouldn't be in need of a guess.