# Is the angular momentum in uniform circular motion zero?

In uniform circular motion the direction of linear velocity changes (although magnitude is constant), so linear velocity is not constant and we know that the formula of angular momentum contains linear velocity i-e L=mvr. Why the angular momentum is zero although the direction of velocity changes?

• Why do you think it is zero? – RunMachine_Kohli May 9 '19 at 9:51
• If your book says the angular momentum is zero (as you state in comments tothe answers), then you should mention that in the question. That makes for a much better question because it tells us the source of your confusion. – Bill N May 9 '19 at 15:14

The angular momentum is not zero.
It is equal to $$\vec r \times (m\vec v)$$ and so has a constant magnitude $$rmv$$ and a constant direction given by the right hand rule.

• thanks! it's mistake in my book – yaseen wazir May 9 '19 at 11:58
• so you mean that the change in direction of linear velocity does not affect the direction of angular momentum? – yaseen wazir May 9 '19 at 12:02
• @yaseenwazir Correct, the direction of the angular momentum vector is perpendicular to the plane containing the trajectory of the mass undergoing uniform circular motion. – Farcher May 9 '19 at 12:05
• hmmm...thank you! – yaseen wazir May 9 '19 at 12:50

Angular momentum of the particle executing uniform circular motion is not zero. But the angular momentum of the particle executing circular motion remains conserved about the axis of rotation as torque is in the radial direction. So you can say change in angular momentum is zero.

• thanks! so my book is wrong! – yaseen wazir May 9 '19 at 11:57

If you take into account that the modulus of the vector product $$\vec a \times \vec b$$ between two vectors $$\vec a$$ and $$\vec b$$ is twice the area of the triangle where $$\vec a$$ and $$\vec b$$ coincide with two sides, you can immediately visualize why, in the case of uniform circular motion this modulus not only is not zero but it is constant.

I also just want to add that angular momentum must always be taken with respect to a certain point; the $$\vec{r}$$ is defined to point from an origin of convenience. There will always exist a coordinate system wherein the angular momentum is zero. Angular momentum in this case is zero with respect to the co-moving frame. With respect to the center, as other answers have mentioned, it has magnitude $$|\vec{r} \times \vec{p}|$$.