I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simply applying conservation of angular momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.


1 Answer 1


The point is that the object does clearly not move on a circle if you pull the string and decrease its distance from the center hole. Its path is rather some sort of spiral. Although this will barely be noticeable in a real experiment, the local center of curvature of this spiral will be changing and be different from the center of the hole, and similarly, the motion of the object will not be strictly orthogonal to the string.

Actually, you will have to exert a force while you pull the string, and since the motion is frictionless, this work has to go somewhere, namely into the kinetic energy of the rotating object.


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