1
$\begingroup$

enter image description here

I have this ideal system. There is a bar with negligible mass and a ring(a point) with mass $m$. Bar is rotating with angular speed $\omega$.

If I see that system from Non-inertial reference system I get a Centrifugal force and then all is clear the ring go away from rotation axis.

enter image description here

Now I'm going to see this system from an inertial reference system.

enter image description here

I don't understand how can I deduct thanks this view that mass go away long the bar.

In a linear case I understand concept of Fictitious force very well so I can switch non-inertial system with inertial system easly. In the rotational case I'm not able to do this; my brain see only non-inertianal cases with fictitious force. In my view, as the real forces are positioned (inertial system), what happens instead and what my experience tells me should not happen. Obviously I know I'm wrong!

A simple intuitive explanation would also be fine, I thank everyone in advance.

$\endgroup$
3
$\begingroup$

Your mistake in the inertial frame is viewing the length along the rod as a coordinate, $q$, to which we can simply apply $F_q = m \ddot q$. This is not valid, since the direction $\hat q$ is not constant!

$\endgroup$
  • $\begingroup$ Ok, but in the direction q my ring is moving and I don't understand why $\endgroup$ – ABC Sep 10 at 15:13
  • $\begingroup$ So first off, you should not draw $m\omega^2 r$ as a force in the inertial case -- that is incorrect. What you have is $R_n$ and $mg$. You can determine that the net force in the vertical direction is nonzero. (Even as the stick spins, the vertical direction is constant in time.) Based on the constraints of the ring and stick, you can see that it must move along the stick in order to have a nonzero vertical acceleration. $\endgroup$ – Danny Sep 11 at 16:25
  • $\begingroup$ Sure $m\omega^2r$ as force is wrong. However, even taking a simpler situation: the same situation as above, but with the $\alpha = 90 °$ angle, I would have the same situation but with the horizontal pole. Then $ mg = R $. Suppose there is no friction. How is it possible that my object (has no structure is comparable to a point), fly away from the bar? $\endgroup$ – ABC Sep 11 at 19:50
  • $\begingroup$ Ah, you are right in pointing out the flaw in my reasoning. The correct answer is then that the answer is not obvious apart from the math, when working in an inertial coordinate system. You need to calculate $\ddot r$, which involves differentiating $\hat r$ and $\hat \theta$ with respect to time. $\endgroup$ – Danny Sep 12 at 1:56
  • $\begingroup$ We are at the starting point. I don't know what to do to solve my problem. $\endgroup$ – ABC Sep 12 at 7:08
0
$\begingroup$

My intuition on the problem is the following:

Consider a rotating horizontal bar. If there is no friction along the bar, and since the ring has no vertical movement, the only horizontal force applied by the bar to the ring has to be perpendicular to the bar in the direction of rotation. This force has to exist, since the ring is not moving with a constant velocity vector.

Since in this case, there is no inward force along the axis of the bar, the distance to the centre of rotation is increasing.

If the bar is inclined, we have to consider the balance between effect of the force the bar is applying to the ring (that increases the distance to the origin) and the gravity projected along the axis of the bar (that pulls the ring to the origin). If the angular velocity and the distance to the centre of rotation are large enough to overcome the effect of the gravity, then the ring will slide outwards, otherwise, it will slide inwards.


In more detail, lets assume that the bar starts at the origin $(0,0,0)$. In Cartesian coordinates, the position $\vec{r}$ of the ring at time $t$ when it is at a distance $d(t)$ from the origin, is given by $(x, y, z)$, with:

$$ \begin{align} x(t) & = d(t) \sin \alpha \ \cos(\omega t + \phi)\\ y(t) & = d(t) \sin \alpha \ \sin(\omega t + \phi)\\ z(t) & = d(t) \cos \alpha \end{align} $$

The velocity $\dot{\vec{r}} = \vec{v}$ is:

$$ \begin{align} \dot{x}(t) & = \dot{d}(t) \sin \alpha \ \cos(\omega t + \phi) - \omega \ d(t) \sin \alpha \ \sin(\omega t + \phi)\\ \dot{y}(t) & = \dot{d}(t) \sin \alpha \ \sin(\omega t + \phi) + \omega \ d(t) \sin \alpha \ \cos(\omega t + \phi)\\ \dot{z}(t) & = \dot{d}(t) \cos \alpha \end{align} $$

And the acceleration $\ddot{\vec{r}} = \dot{\vec{v}} = \vec{a}$ is:

$$ \begin{align} \ddot{x}(t) & = \ddot{d}(t) \sin \alpha \ \cos(\omega t + \phi) - 2 \ \omega \ \dot{d}(t) \sin \alpha \ \sin(\omega t + \phi) - \omega^2 \ d(t) \sin \alpha \ \cos(\omega t + \phi)\\ \ddot{y}(t) & = \ddot{d}(t) \sin \alpha \ \sin(\omega t + \phi) + 2 \ \omega \ \dot{d}(t) \sin \alpha \ \cos(\omega t + \phi) - \omega^2 \ d(t) \sin \alpha \ \sin(\omega t + \phi)\\ \ddot{z}(t) & = \ddot{d}(t) \cos \alpha \end{align} $$

In order to simplify the analysis, let's assume that $t$ is such that $\cos(\omega t + \phi) = 1$ and, thus, $\sin(\omega t + \phi) = 0$. Then the acceleration would be:

$$ \begin{align} \ddot{x}(t) & = \ddot{d}(t) \sin \alpha - \omega^2 \ d(t) \sin \alpha\\ \ddot{y}(t) & = 2 \ \omega \ \dot{d}(t) \sin \alpha\\ \ddot{z}(t) & = \ddot{d}(t) \cos \alpha \end{align} $$

Simplifying once more, let's start with the case where the bar is horizontal. In this case, $\sin \alpha = 1$ and $\cos \alpha = 0$ and the acceleration becomes:

$$ \begin{align} \ddot{x}(t) & = \ddot{d}(t) - \omega^2 \ d(t) \\ \ddot{y}(t) & = 2 \ \omega \ \dot{d}(t) \\ \ddot{z}(t) & = 0 \end{align} $$

If we think about the system in this case, the gravity will be ignored, because it will be cancelled out by the vertical reaction of the bar. Regarding the horizontal force the bar is applying to the ring, we can imagine that is along the $y$ axis and is zero along the $x$ axis. It is zero along the $x$ axis because I'm assuming there is no friction in this direction. So we can obtain $d(t)$ by solving the acceleration equation along the $x$ axis: $$ 0 = \ddot{d}(t) - \omega^2 \ d(t) $$ Therefore we get: $$ d(t) = k_1 e^{\omega t} + k_2 e^{-\omega t} $$ for some constants $k_1$ and $k_2$.

If we go back to the inclined bar, we may consider an inclined coordinate system that, at time $t$ has the $x$ axis aligned with the bar. In this case, we have to consider the component of the gravity along this $x$ axis and the acceleration along this axis is: $$ - g \cos \alpha = \ddot{d}(t) - \omega^2 \ d(t) $$ Which results in: $$ d(t) = k_3 e^{\omega t} + k_4 e^{-\omega t} + \frac{g \cos \alpha}{\omega ^ 2} $$ for some constants $k_3$ and $k_4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.