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A particle is rotating in a conical pendulum with help of a string of length $\ell$. The speed of the particle is constant and angle theta is also constant with time. Why isn't the net torque on the particle about the point of suspension 0? As the torque given by mg is 0 at the axis of rotation?

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  • $\begingroup$ @FGSUZ It's a conical pendulum, which rotates around a vertical axis in a circular path. $\endgroup$ – Bill N Oct 6 '18 at 22:01
  • $\begingroup$ Ok I hadn't read that, sorry. $\endgroup$ – FGSUZ Oct 7 '18 at 11:48
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The torque due to the weight of the object is not $0$ about the point where the string is attached to the point of suspension. This can be seen with the diagram below. $\mathbf r$ is the vector pointing from the point where we are calculating the torque about to the point where the weight force is applied.

torque

Torque is given by $\mathbf\tau=\mathbf r \times \mathbf F$, which is $0$ when $\mathbf r$ and $\mathbf F$ are either parallel or anti-parallel, assuming none of the vectors are $0$. You can tell by the diagram that the two vectors do not fall into either of these categories. Therefore, the torque is non-zero.

Contrast this with the tension force $\mathbf T$. In this case $\mathbf T$ and $\mathbf r$ are anti-parallel, so the torque produced by the tension about the point of suspension is $0$.

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When dealing with forces, torques, momentum, and angular momentum, one must remember that they are not scalars. Force and momentum are vectors, and torque and angular momentum are axial vectors (pseudovectors ??). That means you have to pay attention to the directions as well as the magnitude. Even if the magnitude remains constant, the directions may be changing.

In your case, you need to consider the instantaneous direction of the angular momentum about the suspension point. Is it constant? If not, then there must be a torque. $$\vec{L}=\vec{r}\times\vec{p}$$. I think you will see that this direction changes, and you should also see that the torque vector is perpendicular to it, so that only the direction of $\vec{L}$ (the angular momentum) is changing.

Edit replay to comment question: The torque about the support point is $$\vec{\tau}=\vec{r}\times m\vec{g}$$ where $\vec{r}$ points from the support to the particle and $\vec{g}$ is down, so the angle between those vectors is the angle the string makes with the vertical. So the magnitude is $mgr\sin\theta$, either parallel or anti-parallel to the velocity.

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  • $\begingroup$ I get it that there is a torque, but where is the torque coming from? $\endgroup$ – John Smith Oct 6 '18 at 15:23
  • $\begingroup$ @JohnSmith My answer shows exactly where the torque is coming from :) $\endgroup$ – BioPhysicist Oct 6 '18 at 15:39
  • $\begingroup$ @JohnSmith See my edit. The gravitational force does not have a line of action through the support point, so it produces a torque about that point. It's direction is perpendicular to the axis of rotation, so there's no component along that axis, but there is still a torque changing the direction of the angular momentum vector. $\endgroup$ – Bill N Oct 6 '18 at 18:40

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