Since the tension force is a central acting force, the torque on an orbiting ball about that center is zero. But if the rope is cut down during motion the torque would still remain zero. This would mean that angular momentum of the orbiting ball should be conserved, but I find everywhere that ball will move in a straight line tangentially. What is wrong in my reasoning?


2 Answers 2


The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.

Angular momentum is given by$^*$ $$\mathbf L=\mathbf r\times\mathbf p$$ Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum $$L=rp\sin\theta$$ where $\theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).

Now, since there are no forces acting on our object, $p$ is constant. Also, $r\sin\theta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.

This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.

$^*$Note that this applies to any type of motion, not just circular motion.

  • $\begingroup$ So, angular momentum can be applied for all motions and not just circular type? $\endgroup$
    – Pirateking
    Jan 5, 2019 at 19:55
  • $\begingroup$ @PiyushGalav Yes. Angular momentum about some point is $\mathbf L=\mathbf r\times\mathbf p$. It can be applied to any motion. $\endgroup$ Jan 5, 2019 at 20:39
  • 3
    $\begingroup$ Straight lines are just circles about the point at infinity... $\endgroup$ Jan 5, 2019 at 20:49

The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = \vec r \times \vec p$. $r \sin \theta$ and $mv$ remain constant after such an event.

The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.

  • $\begingroup$ What do you mean by conical? $\endgroup$
    – Karthik
    Jan 5, 2019 at 16:32
  • 1
    $\begingroup$ @KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity. $\endgroup$
    – Vaelus
    Jan 5, 2019 at 16:41
  • 1
    $\begingroup$ Note: the question has been edited to refer to a less dramatic angular momentum scenario. $\endgroup$
    – rob
    Jan 6, 2019 at 5:52

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