What Happens to a Ring with tangential force?

Question

so if I know the inertia, momentum of the ring, and the force I'm applying to it how do I calculate how much its going to spin and how much its going to translate horizontally? (plz explain at a high school level)

Answer 1

The first step would be to look at this from a center of mass perspective. Don't worry about rotation for a moment. Forces and accelerations. As you have drawn it, there is only one force, the one applied at the bottom of the ring to the right. Thus with $\Sigma F=ma$, and just one force, we can state that $a=\frac F m$. The center of mass of the ring must accelerate rightward with an acceleration of $\frac F m$. No other acceleration will satisfy the equations of motion.

Now we can consider torques to look at rotation. In this case, they are independent from forces. Sometimes when you have a force of friction, there's a bit of interplay between them and you have to compute both simultaneously. However, in your case we can handle them completely separately.

The magnitude of the torque applied is $\tau = F\cdot r$, where $r$ is the distance from the center of mass to the point of application of the force. You say you know the moment of inertia, we'll call it $I$. Just like we have $F=ma$ in translational space, we have $\tau=I\alpha$ in rotational space. So with a division, we get $\alpha=\frac \tau I$, which is your angular acceleration.

Now this is assuming that as the ring starts spinning, you are adjusting how you apply the force so that it continues to be at the bottom of the ring. If you are applying the force with a hand, that force will typically change direction to follow the ring as it rotates until you let go with one hand, grab at the bottom again with the other hand, and so forth. If this is a ring on a treadmill, where the force is being applied by the treadmill moving, then you won't have to do any extra tricks like you would if you applied the force by hand. The treadmill would only be capable of applying the force at the bottom, and it would constantly be touching a different part of the ring as it rotates to ensure the force is still only applied at the bottom.