1
$\begingroup$

So imagine a ball is attached to a string and is circling around a fixed axis. If we calculate angular momentum from the center, it will be constant. However, if we moved to an arbitrary point, it would be constantly changing, which means there is external torque. But how is now tension of string producing an external torque, and at the center internal? Are angular momentum and torque just measures for how much things "want" to rotate around you, for example when $\vec{L}=\vec{r}\times\vec{p}=\vec{0}$ it is because during that $dt$ it has no "rotational" component? And when $|\vec{L}|=rp$ it is because it's as though at that moment it was rotating?

$\endgroup$
0

2 Answers 2

0
$\begingroup$

In the first case, the tension is not providing an 'internal' torque, but rather no torque at all, as the force passes through the origin.

If you instead consider another arbitrary point as the origin, then the tensile force does not pass through the origin, and $ \vec F \times \vec r $ is no longer zero. If you take the ball as your system, then the string is providing this external torque, while if you take the ball and string as your question, then the point of attachment of the string to the axis is providing this external torque to the system.

I didn't really understand what you are trying to ask in the second part about angular momentum, but what you've written seems to be correct. A particle traveling in a straight line will also have angular momentum about a point which is not on that line.

$\endgroup$
0
$\begingroup$

A better way to build intuition about angular momentum is not to think directly in terms of rotation but in terms of surface area.

Indeed, it is quite simple to see that ${\bf L} \Delta t = {\bf r}\times m {\bf v} \Delta t$ is directly proportional to the area spanned by the position vector ${\bf r}$ in a short time $\Delta t$. Here, short time means a $\Delta t$ small enough to allow to use the segment ${\bf v}(t) \tau$, with $0 \le \tau \le \Delta t$, as a good approximation of the trajectory between $t$ and $t+\Delta t$.

Therefore, it becomes immediate to see that the choice of the origin changes the angular momentum. In the case of a rotation around a point with uniform speed, a change of origin transforms a motion at constant angular momentum (when the origin is the center of the circle) into a motion where ${\bf L}$ is varying. The change of torque is entirely due to the change of the position vector as a consequence of the change of origin.

The answer to your last question is also simple, by using this interpretation of the angular momentum. The "rotation" is just a byproduct of the fact that if $ \left| {\bf L} \right| \neq 0$ consecutive small time intervals $\Delta t$ correspond to a sequence of triangles with vertex at the origin, two sides made by ${\bf r}(t)$ and ${\bf r}(t+\Delta t)$ and the third represented by $ {\bf v} \Delta t$. The area of such a triangle depends on the lengths of two sides and one angle. By choosing the angle at the origin, we can easily recover the connection between "rotation" (of the position vector around the origin) and angular momentum. Such a description shows clearly that angular momentum shouldn't be connected necessarily with a bending trajectory. Even a uniform motion on a straight line can be described as a constant angular momentum motion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.