I don't understand why the Tension $T$ on a conical pendulum and a simple pendulum are different.

In a simple pendulum, one would say that the tension of the rope is $T=mg \cos(\theta)$.

simple pendulum http://n8ppq.net/site_n8ppq/Physics/pendulum_files/image001.gif

However, in a conical pendulum (describing a circular motion), $mg=T \cos(\theta)$.

conical pendulum

The only difference I see in the set up of the two cases is that in the second one there is a velocity component that makes the bob go around in a circle.

I know that in the conical pendulum, the component $T \sin(\theta)$ would give the centripetal acceleration of the circular motion.

I've seen this everywhere. The two cases look pretty much the same to me, so I would be tempted to say one of them (rather the second one) is wrong.


3 Answers 3


A very nice thing you pointed out there which many people tend to skip...
Actually the fundamental rule of taking components of forces is that the coordinate axes should be perpendicular to the instantaneous direction of velocity or you can say the instantaneous direction of motion.
As the tension force is already perpendicular to the direction of motion, we resolve the $mg$ force (weight) into two components.
Hope this helps!

  • $\begingroup$ Isn't it much more useful to find components of forces parallel to and perpendicular to the acceleration of the object? $\endgroup$
    – M. Enns
    Commented Mar 30, 2017 at 2:16

I'm not sure why you believe they should be the same. They have different net forces and therefore different acceleration and motion. You can view the conical pendulum as a 2D superposition of orthogonal pendulum modes that are in phase.


Both pendulums are correct in their respective situation. We must remember that Newton's second law dictates that the vector sum of forces on an object must be equal to the mass of the object times the acceleration of the object.

\begin{equation} \sum_n F_n = ma \end{equation}

In the first pendulum the object is swinging side to side so we know that the acceleration of the object is orthogonal to the arm of the pendulum pointing at an angle $\theta$ below the horizontal towards the center of oscillation of the pendulum. This means that the forces in line with the arm of the pendulum must be equal and opposite since there is no motion in this direction and we see that $T=mg\cos{\theta}$ is true for this pendulum.

For the second (conical) pendulum the object is moving at the same vertical height in a circular path of radius $r$. This tells us that the acceleration of the object points horizontally inward at an angle of $\pi/2-\theta$ with respect to the arm of the pendulum. We also know that for circular motion Newton's second law can be rewritten as

\begin{equation} \sum_n F_n = \frac{mv^2}{r} \end{equation}

Since there is no downward acceleration in the conical pendulum the vertical forces must be in equilibrium such that $T\cos{\theta}=mg$.

Moral of the story? Your choice of coordinate axes is important and net acceleration must be accounted for when making free-body diagrams.

  • $\begingroup$ +1, this is the correct answer. I initially made the rather embarrassing mistake of not realizing that in the first picture the motion is purely vertical and in the second the motion is purely horizontal, and posted a wrong answer (which the OP unfortunately marked as the accepted answer moments before I noticed the error). $\endgroup$ Commented Mar 16, 2014 at 19:18
  • 1
    $\begingroup$ -1, this is not the correct answer. The pendulum is not only acceleration perpendicular to the string, it is also has centripetal acceleration towards the centre. The tension for the simple pendulum would be $F_N=mg\cos\theta+mv^2/r$ $\endgroup$
    – M. Enns
    Commented Mar 30, 2017 at 2:43
  • 1
    $\begingroup$ -1 This answer has (at least) two critical problems: (1) It defines $\theta$ as the angle between the arm and the horizontal, but then uses it as if it were defined more traditionally (as the angle to the vertical). (2) It fails to account for the centripetal acceleration, as discussed in M. Enns’s comment. $\endgroup$ Commented Apr 3, 2022 at 12:54

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