# Can the angular momentum be calculated from a point not lying on the rotational axis

I have seen that angular momentum of a rigid body about any point lying on its axis of rotation is moment of inertial times angular velocity, I'm clear about it, my worry is if we will calculate the angular momentum about any other point lying outside the axis of rotation. Can we calculate it? if so will its value remain the same? When I tried to calculate from an outside point, I found that the perpendicular distance between the particles and the point changes, also the the direction of velocities of the particles are different, so I'm unable to use the basic meaning of angular momentum of $\vec r \times \vec v$.

will its value remain the same

No, because you are actually calculating something completely different when you change the rotation point.

• A rotating rigid object does not have an angular momentum.
• It has an angular momentum around a specific point, and another angular momentum around another point, and a third angular momentum around a third point...

Think of torque. The torque around the screw when turning a wrench is different than the torque around the middle of that wrench and different from the torque around the end of the wrench where you are holding (here it will be zero).

All the parameters of the rotational world - such as angular momentum $\vec L$, torque $\vec \tau$, moment of inertia $I$ and also angular acceleration $\vec \alpha$, angular velocity $\vec \omega$ etc. - depend on the rotation point. An object does not "have" angular velocity $\vec \omega$ in the same way that it has (linear) velocity $\vec v$, it does not "have" moment of inertia $I$ in the same way that it has mass $m$ - and it does not "have" angular momentum $\vec L$ in the same way as it has (linear) momentum $\vec p$.

Also note, when you calculate angular momentum $\vec L$ with formulae such as

$$\vec L=I\vec \omega$$ $$\vec L=\vec p \times \vec r$$

then you must be sure that all the variables included are calculated with respect to the same rotation point. You cannot input an $I$ and an $\vec \omega$ into the first-mentioned formula if they are not themselves calculated about the same point.

• to your reply "It has an angular momentum around a specific point, and another angular momentum around another point, and a third angular momentum around a third point...",does this mean that i just cant calculate the angular momentum in any way through a different point,cant we calculate with out taking angular velocity into account. – sachinrath123 Jan 30 '18 at 14:00
• but when we are applying conservation of angular momentum we say it can be conserved about any point of choice,how is that possible. – sachinrath123 Jan 30 '18 at 14:06
• @sachinrath123 "does this mean that i just cant calculate the angular momentum in any way through a different point" Yes, you can. You just have to redo the calculation with that new distance. "cant we calculate with out taking angular velocity into account." Sure we can. I wrote two different formulas in the answer here. Both are fine to use. If you don't know the angular velocity, then pick the other formula. – Steeven Jan 30 '18 at 14:19
• @sachinrath123 "but when we are applying conservation of angular momentum we say it can be conserved about any point of choice,how is that possible" That angular momentum is conserved means that $$\vec L_{before}=\vec L_{after}$$The value of $\vec L$ might be different around different points, yes. But whatever that value is, it is constant. The same before and after any impact. It is conserved across different points in time, but not across different points on the geometry of the object. – Steeven Jan 30 '18 at 14:22
• what exactly u meant by "A rotating rigid object does not have an angular momentum." – sachinrath123 Jan 30 '18 at 14:29