Angular momentum in different points I have a question about angular momentum:
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?
 A: Consider central-force motion, such as a planet moving around a (very massive) star.  The angular momentum of such a planet is constant if we take the origin as the center of the star.  It is not constant if we take the origin to be any other point.
A: 
Is it possible to have a system where angular momentum is conserved relative to 1 point,but not conserved relative to another?

Total angular momentum will be conserved but the angular momentum of any part of the system will have a value dependent on where you take your base point. 
A: Angular momentum relative to an origin ${\mathcal O_1}$
$$ \mathbf{L_{\mathcal O_1}} = \mathbf{r_{\mathcal O_1} \times p_{\mathcal O_1}}$$
where $\mathbf r_{\mathcal O_1}$ is the position vector to the particle relative to some origin ${\mathcal O_1}$. 
Now suppose that angular momentum is conserved in ${\mathcal O_1}$. Then 
$$ \frac{d \mathbf L_1}{dt} = \mathbf{\dot{r_1} \times p_1} + \mathbf{r_1 \times \dot{p_1}} = \frac{1}{m} \mathbf{p_1 \times p_1} + \mathbf{r_1 \times \dot{p_1}} =0 $$ 
but since the direction of momentum is frame-independent, the first term vanishes (that is, $\mathbf{p_1} = \mathbf{p}$). It then follows that 
$$ \mathbf{r_1 \times F_1} =0 . $$
Now, let's look at some other origin $\mathcal{O}_2$, given that $L$ is conserved in $\mathcal O_1$.  Well the first term much vanish again, that's fine but what about the second term? Does 
$$\mathbf{r_2 \times F_2} \stackrel{?}{=}0. $$
Well, no not necessarily. Namely, just choose an origin in which the force is perpendicular to your position vector. 
