# Transformation of linear into angular momentum

Imagine you have this system:

Composed by a disk that can rotate around a fixed axes $$O$$,and a particle with mass $$m$$ and velocity $$v_0$$ crushes with the disk at a distance $$D$$ from the center of the disk.

At first glance I thought that the movement would decompose into a tangential velocity and into a normal velocity (which I assume would be absorbed by the system). Then you could easily see:

$$\sin(\theta) = \frac{d}{r} \\ v_t = \omega r = v_0 \frac{d}{r} \\ \omega = v_0 \frac{d}{r^2}$$

My question is: Is this line of reasoning right? And how would you express this as a conservation of momentums, taking into account that the particle crashes at a distance $$d$$? Something like this: $$m v_0 = M \omega r + m v_0'$$

In absence of friction at the axis of rotation, the hinge at the axis of rotation only introduces a force constraint in the system as an external force to the system composed by the mass and the disk.

Thus, you can readily use the conservation of angular momentum of the system, w.r.t. the pole $$O$$. Comparing the state before the collision, with the condition after the collision we get

$$m v_0 d = I_{disk} \omega + m r v_{\theta}$$,

with $$r$$ the distance from the point $$O$$ of the point where the mass stops into the disk after the collision, and $$v_{\theta} = r \omega$$ the velocity (only tangential, due to the rotational motion) of the mass rotating with the disk. If the mass sticks to the outer surface of the disk, we get

$$m v_0 d = I_{disk} \omega + m R^2 \omega = [I_{disk} + m R^2 ] \omega$$

Particular cases. Let's have a look at two particular cases:

1. the mass hits the disk with a trajectory passing through the point $$O$$, and thus $$d=0$$. It's easy to verify that $$\omega = 0$$ and the disk doesn't rotate.
2. mass hits the disk at $$d = R$$ of a disk with no inertia, $$I_{disk}=0$$. The angular velocity reads $$\omega = \frac{v_0}{R}$$ and the tangential velocity reads $$v_{\theta} = v_0$$.
• Why $m v_0 d$? Is this an angular momentum? Of what exactly? Of the particle? Oct 15, 2022 at 18:46
• of the particle. You can consider the system comprised by the particle and the disk. If the disk is initially at rest, its angular momentum is zero, and the total angular momentum of the system is only the angular momentum of the particle Oct 15, 2022 at 18:48
• But the particle isn't moving in a circular trajectory? Oct 15, 2022 at 18:56
• who cares? laws of physics talk about momentum and angular momentum balance, not only about rotation Oct 15, 2022 at 19:46
• A particle moving in a straight line in one inertial reference frame is moving in a straight line in any other inertial reference frame. When it is viewed at a point not on the line it is moving in a way so that its viewing angle relative to the point is changing continuously from $-180^{\circ}$ to $+180^{\circ}$, that is a "rotation" and the corresponding angular momentum is conserved. Oct 15, 2022 at 22:09