Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that I have a point mass attached to a massless string and I am rotating it vertically. That means The mass is in uniform circular motion and the path of its motion is vertical circle. How does the tension change with respect to the position of the mass. More specifically is the tension in the string is only due to circular motion ($mv^2/R$) or gravity plays a part in it ($mv^2/R$ + something due to weight)?

share|cite|improve this question
If by "uniform circular motion" you mean moving in a circle at a constant angular velocity, then there must be additional forces besides gravity and tension acting on the point mass... – DJohnM Nov 4 '13 at 8:42
Definitely ,gravity plays a role in the analysis. – Sahil Chadha Nov 4 '13 at 8:55
Then what will be the tension in the string at a certain point when the radius is in some angle theta with vertical downward radius? mv^2/R + mgcos(theta) ? – Md. Taufique Hussain Nov 4 '13 at 9:07
I strongly suggest drawing a free body diagram or two. Choose a couple of places of interest in the system, then resolve the forces present, remembering the thing is by definition travelling in a circle... – Nic Nov 4 '13 at 13:05
up vote 5 down vote accepted

Assume that the point mass, $m$ has two tiny thrusters, mounted so as to exert purely tangential force in the plane of the circular motion, one clockwise, and the other counter-clockwise.

The magnitude of the constant velocity of the mass is $v$, and the radius of the circle is $r$.

Measure the position of the point mass in the standard Cartesian coordinate way: angles are measured from the positive X-axis, counter-clockwise positive.

At the point where the mass is at a position angle $\theta$. the total radial force inward on the mass, $F_R$ is given by the centripetal force equation:$$F_R=\frac{mv^2}{r}$$

There are two forces that supply this radial force: the tension, $T$ in the string, and the inward radial component of the force of gravity:$$F_{G-R}=mg\sin(\theta)$$So:$$\frac{mv^2}{r}=T+mg\sin(\theta)$$and:$$T=\frac{mv^2}{r}-mg\sin(\theta)$$Note that this implies that:$$v >=\sqrt{rg}$$ or the string tension will become negative near the top of the circle, an impossibility.

The conditions of the question also require that at all times the net tangential force, $F_T$, be zero. The tangential component of the force of gravity, $F_{G-T}$ is given by:$$F_{G-T}=-mg\cos(\theta)$$where a positive force implies counter-clockwise force. The thrusters are needed to supply the exact opposite force to the mass at all times.

share|cite|improve this answer
Also keep in mind that you have one equation and two unknown variables: $T$ and $v$. By using conservation of energy or solving a differential equation, you can find $v$. – fibonatic Nov 5 '13 at 14:50
How can you find $v$? It's a given parameter of the motion for a particular case. And conservation of energy would require knowing the energy input from whatever acts to keep the mass moving at that velocity... – DJohnM Nov 5 '13 at 23:35
$v$ will not be constant, because the point mass will move up and down in the gravitational field. – fibonatic Nov 6 '13 at 17:33
The OP explicitly said "uniform circular motion". This means constant radius, constant linear velocity and constant angular velocity. My answer included a way to achieve this condition. – DJohnM Nov 6 '13 at 17:50
@User58220: If there were no tiny thrusters, would the tangential component of the force of gravity F=−mgcos(θ) cause the magnitude of the velocity to change otherwise? Also, in your answer, is the tension being varied in the string? – Eliza Dec 2 '13 at 4:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.