- #1

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I'm new here, and I'm learning physics by myself. Please, if the answer to my question is too obvious, be kind hee hee, I'm a 13 and trying to learn physics by myself.

So, I know that we have

## \textbf{L}= \textbf{r} \times \textbf{p} ##

Then, we can write

## \frac{d}{dt} \textbf{L} = \dot{\textbf{L}} = \frac{d}{dt} \left( \textbf{r} \times \textbf{p} \right) = \frac{d}{dt} \left( \textbf{r} \times m \textbf{v} \right) ##

So, this is

## \dot{\textbf{L}} = \frac{d}{dt} \textbf{r} \times m \textbf{v} + \textbf{r} \times \frac{d}{dt} \left( m \textbf{v} \right) ##

This yields

## \dot{\textbf{L}} = \textbf{v} \times m \textbf{v} + \textbf{r} \times \textbf{F} ##

The first term on the RHS vanishes, since

## \textbf {v} \times \textbf{v} = (v)(v) \sin \theta = (v)(v) \sin 0 = (v)(v) 0 = 0##

So, we have, then

## \dot{\textbf{L}} = \textbf{r} \times \textbf{F} = \textbf{N} ##

Where ## \textbf{N} ## is the torque (a pseudovector by the way)

Then, it follows that if the total torque is zero, then angular momentum is conserved.

So far so good.

My questions is, then, if a particle is moving about a point, the linear momentum of the particle is changing indeed, so is its position vector. Which then made me wonder certain things, then I realized that the only way for the angular momentum to be conserved is that the force must be pointing in the direction of the position vector, that way ## \textbf{r} \times \textbf{F} = 0 ##

Now, here's my question, can I state that "

*If the angular momentum is conserved, it implies that the sum of all torques is zero, but, also, conservation of angular momentum do implies the presence of a force directed along the same line of the radius vector*## \textbf{F} ## ", in other words, "

*In order to have angular momentum conservation THERE MUST EXIST A FORCE acting on the particle*"?

Am I right?

Thanks in advance for your help!