Suppose a ball is going straight along a fixed line without rotating. Now you consider a point which is not along the trajectory of the ball. If I want to consider the angular momentum about that point, as you can understand that the $r$ and $v$ vector aren't in the same direction, so $r\times v $ should have a value. So this object isn't rotating but has an angular momentum. Can I conclude that the ball has angular momentum without rotating ?
[Physics] Can a non-rotating ball have angular momentum
angular momentumnewtonian-mechanicsreference frames
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My initial reasoning was wrong, sorry, I can't justify the use of angular momentum conservation in that non-inertial frame. One can write a balance equation for the angular momentum, but after all it should be the same as solution №2.
But ok, let's choose a single inertial frame. Let it be the one moving with the ultimate ball velocity $v_f$. This will lead the ball having some initial angular momentum $L_0$ and recalculated value for added momentum J. Then the conservation of angular momentum will be
$$(\beta+1) r (J + mvr) + L_0 = \frac{7}{5} m r^2 \omega$$
where
$$ \begin{multline} L_0 = - \int_0^{2r} \rho z v \, \pi \left[r^2 - (r-z)^2\right] dz = \\ - v \frac{m}{\frac{4}{3}\pi r^3} \pi \int_0^{2r} z \left[r^2 - (r-z)^2\right] \, dz = -mvr \end{multline} $$
Hence
$$(\beta+1) r J = \frac{7}{5} m r^2 \omega$$
Instead of direct integration one could use the expression for angular momentum in a moving frame, but I don't remember how to do it.
Yes. For any system of particles, the following statement is true:
If the net torque on a system of particles is zero, and if the interactions between particles of the system point along the lines joining them, then the total angular momentum of the system is conserved.
The proof in the context of classical mechanics is below.
For the ball on the string example, if you are only considering the ball, then there is an external torque on the ball: that of the string. One subtlety is that if you pick the origin of your coordinates to be the center of the circle about which it rotates, then in that case there is no torque and the angular momentum of the ball is, in fact, conserved. However, if you pick a different point as your origin, then it's not the case that the position vector is always along the line of the tension vector, and therefore there will be a nonzero torque. Remember that when you calculate the angular momentum and the torque, you need to use the same origin for both to be consistent.
For the orbiting example, you need to consider the system consisting of both planets, then there is no external torque on this system and the total angular momentum is conserved.
proof.
Let $m_i$ denote the mass of particle $i$ and let $\mathbf x_i$ denote the position of particle $i$, then the total angular momentum of the system is defined as $$ \mathbf L = \sum_i \mathbf x_i\times\mathbf (m_i \dot{\mathbf x}_i) $$ Taking a time-derivative gives $$ \dot{\mathbf L} = \sum_i\Big(m_i\dot{\mathbf x}_i\times\dot{\mathbf x_i} + \mathbf x_i\times(m_i\ddot{\mathbf{x}}_i)\Big) = \sum_i\mathbf x_i\times\mathbf F_i $$ where $\mathbf F_i$ is the net force on each particle. Now split the force on each particle into the net external force $\mathbf F_i^e$ and the net force due to all of the other particles in the system $$ \mathbf F_i = \mathbf F_i^e + \sum_j \mathbf f_{ij} $$ where $\mathbf f_{ij}$ denotes the force on particle $i$ due to particle $j$. Then we have $$ \dot{\mathbf L} = \sum_i\mathbf x_i\times\mathbf F_i^e + \sum_{ij}\mathbf x_i\times\mathbf f_{ij} $$ by Newton's third law, we have $\mathbf f_{ij} = -\mathbf f_{ji}$ which causes the last sum to vanish provided that the interactions between particles point along the lines connecting them, we are left with $$ \dot{\mathbf L} = \sum_i\mathbf x_i\times\mathbf F_i^e $$ where the expression on the right is precisely the net external torque on the system.
Note. The necessary assumption that the interactions between particles need to point along the lines joining them is often reasonable because in classical mechanical systems in the real world, these forces are often the coulomb or gravitational interactions which have this property.
Best Answer
Yes, it does. It may seem a bit more intuitive if you imagine a line connecting your reference point and the centre of the ball: as the ball moves, the line sweeps out an angle across the ball, so there should be some angular momentum. If you think in polar coordinates instead of cartesian, the ball is rotating.