Tension of string and circular motion question

In summary, the conversation discusses the forces acting on an object of mass M attached to a string and suspended from the ceiling. It is established that the force of gravity on the object at rest is [0,0,-Mg] and the tension force is [0,0,Mg] by Newton's third law. However, when the object is spinning in a horizontal circle at a constant speed v, the direction of the tension force changes. The magnitude of the tension force is determined to be (Mv^2)/R by using Newton's second law and the fact that for uniform circular motion, the acceleration is (v^2)/R. The
  • #1
Signifier
76
0
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!
 
Physics news on Phys.org
  • #2
Signifier said:
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?)
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.
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?
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
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.

Best regrads

Cliowa

Edit: There are always people faster than me, bad luck. I hope it helps anyway,
 

Related to Tension of string and circular motion question

1) What is tension in a string?

Tension in a string is the pulling force exerted by the string on whatever is attached to it. In the context of circular motion, it is the force that keeps an object moving in a circular path.

2) How is tension related to circular motion?

In circular motion, the tension in a string is equal to the centripetal force acting on the object. This force is directed towards the center of the circle and is necessary to keep the object in its circular path.

3) What factors affect the tension in a string during circular motion?

The tension in a string during circular motion depends on the mass of the object, the speed of the object, and the radius of the circular path. As these factors change, the tension in the string will also change.

4) How do you calculate the tension in a string during circular motion?

The tension in a string can be calculated using the formula T = (mv^2)/r, where T is the tension, m is the mass of the object, v is the speed of the object, and r is the radius of the circular path.

5) Can the tension in a string be greater than the weight of the object?

Yes, the tension in a string can be greater than the weight of the object. In fact, in order for an object to move in a circular path, the tension must be greater than the weight to provide the necessary centripetal force.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
247
  • Introductory Physics Homework Help
Replies
8
Views
203
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
534
  • Introductory Physics Homework Help
2
Replies
55
Views
821
Replies
2
Views
1K
Back
Top