# What force push ball outward in rotating tube in inertial frame? [duplicate]

Ball is free to move in hollow tube, if you rotate tube ball will move outward. What force push ball outward in inertial frame and how ball trajectory looks?

(I know that centrifugal force don't exist in inertial frame, but without this force I can't explain what force move ball outward...)

• Check out the diagram in @Farcher's answer here: link Commented Nov 6, 2023 at 1:13
• Note that at position 1 in the referenced diagram, the distance from the origin to the mass is greater than at position 0. Commented Nov 6, 2023 at 1:26
• I’m sure this has been asked before. Commented Nov 6, 2023 at 4:40
• Commented Nov 6, 2023 at 7:15
• @Not_Einstein Farcher answer is the best one! Commented Nov 6, 2023 at 15:16

There is no force required to make the ball move outwards in the tube.

Suppose the tube did not exist, and the ball was being swung round in a circle at the end of a rope. If the rope was cut, the ball would move along a straight line tangential to the circle (because of Newton's First Law). Its distance from the centre of the circle would be increasing - but no force is required to achieve that.

With the tube in place, the ball does not move in a straight line (in the inertial reference frame) - but this is only because of the tangential force exerted on it by the sidewall of the tube. Once again, no radially outward force is required to change its distance from the centre of the circle. Rather the opposite - if the ball's distance from the centre of the circle was constant then a radially inward force (centripetal force) would be needed.

The apparent problem arises from the pre-Newtonian prejudice that the motion of a body should be along the direction of the total force exerted on the body in all cases.

That is not the case as it happens for instance for planets around the sun. What matters in giving rise to the actual motion are the applied force(s) and the initial conditions.

Here, the force due to the constraint and the initial data give rise to the expulsion motion in the inertial reference frame.

Here are the details.

Let us assume that the tube is frictionless, so that the reaction $$\vec{\phi}$$ is normal to the tube -- no force along the tube exists -- and that the angular velocity is constant $$\Omega>0$$.

I will take advantage of a polar coordinate system $$r, \theta$$ in the plane containing the rotating tube centered on the rotational axis.

Newton equations read, decomposing (*) the motion along the mobile basis $$\vec{e}_r, \vec{e}_\theta$$: $$m\left(\frac{d^2r}{dt^2}- r(t) \Omega^2\right)=0$$ $$m\left(2\Omega \frac{dr}{dt} \right) = \phi$$ where I used $$\frac{d\theta}{dt}= \Omega$$ constant and $$\vec{\phi}= \phi \vec{e}_\theta$$.

Notice that no inertial forces take place here as we are dealing with an inertial reference frame.

The first equation determines the motion $$r=r(t)$$ according to the initial data, and the second one determines $$\phi(t)$$ accordingly.

The first equation has general solution $$r(t) = A \sinh (\Omega t) + B \cosh (\Omega t)$$

Let us assume that the ball is at rest in the tube at $$t=0$$ at distance $$r_0>0$$ from the axis.

As a consequence, $$r_0 = B$$ and $$0=\frac{dr}{dt}(0)= A\Omega$$ so that $$A=0$$. The solution is $$r(t) = r_0 \cosh(\Omega t)\:.$$ The force due to the tube on the ball is $$\phi \vec{e}_\theta$$ with $$\phi(t) = 2mr_0\Omega^2 \sinh(\Omega t) \:.$$

In Cartesian coordinates centered on the rotational axis with $$z$$ parallel to it, the (curved) trajectory has therefore the form $$\vec{x}(t) = r_0 \cosh(\Omega t) \cos(\Omega t) \vec{e}_x + r_0 \cosh(\Omega t) \sin(\Omega t) \vec{e}_y\:.$$

(*) As is well known in that representation, the acceleration has the form $$\ddot{\vec{x}} = \left(\ddot{r}-r\dot{\theta}^2 \right) \vec{e}_r + \left( r\ddot{\theta}+ 2 \dot{r}\dot{\theta}\right)\vec{e}_\theta\:,$$ and thus 2nd Newton's law reads $$\vec{F} = m\left(\ddot{r}-r\dot{\theta}^2 \right) \vec{e}_r + m\left( r\ddot{\theta}+ 2 \dot{r}\dot{\theta}\right)\vec{e}_\theta\:.$$

• Do you mean $r \dot{\theta}^2$ instead of $r^2 \dot{\theta}$ in your last questions? Commented Nov 6, 2023 at 5:47
• Yes, thank you for pointing out the typos. Corrected. Commented Nov 6, 2023 at 5:58
• @ValterMoretti I don understand this "math hieroglyphs" , but I understand your first sentence, I find it very important, because this is root of confusion, so I accept this answer. Commented Nov 6, 2023 at 19:30

It's the inertia of the ball, i.e. the lack of a centripetal force keeping it along a curved (accelerated) trajectory.

We can project the velocity at one point along the trajectory (e.g. parallel to the x-axis)

$$\vec{v} = \left(\begin{matrix}v \\ 0 \\ 0\end{matrix}\right)$$

into cylindrical coordinates through the transforms

$$x = r \cdot sin(\theta)$$

$$y = r \cdot cos(\theta)$$

Which yields

$$\dot x = \dot r \cdot sin(\theta) + r \cdot cos(\theta)\dot \theta$$

$$\dot y = \dot r \cdot cos(\theta) + r \cdot sin(\theta) \dot \theta$$

Plugging in our value for $$\vec{v}$$ at $$\theta = 0$$, we obtain

$$v = r \cdot cos(0) \dot \theta = r \dot \theta$$

$$0 = \dot r.$$

So clearly initially, we are moving only tangentially (without any radial motion). However, as the angle now increases, we next consider a small angle $$\delta \theta$$. Then we may take $$sin(\delta \theta) \simeq \delta \theta$$ and $$cos(\theta) \simeq 1$$. This yields:

$$\dot x = r \dot \theta + \dot r \delta \theta$$

As well as

$$\dot y = \dot r + r \delta \theta \dot \theta.$$

This finally yields

$$\dot \theta \simeq v / r$$

And

$$\dot r = v / \delta \theta.$$

We now have radial velocity! Note that this is an approximate treatment trying to keep the arguments as simple as possible, but it illustrates why the ball moves outwards. In reality, there is no centrifugal force, only conservation of momentum, which means that the ball resists any acceleration (which includes changing direction).

I hope this helps!

• even in frictionless case, ball path will be curved, that is weird case how curved path can exist even without centripetal force! Commented Nov 5, 2023 at 11:19
• The situation is still not entirely force-less. The tube still imparts a force in the angular direction, which will move the ball along a curve. My answer wasn't a full treatment of the problem, but rather an illustration of the underlying physics. Commented Nov 5, 2023 at 11:31

I basically agree existing answers but would like to offer what I hope is a useful insight.

Imagine looking down on the apparatus from above. As the tube moves, the sides of the tube make the ball move. Suppose for example the side of the tube is moving in $$y$$ direction at some instant. Then the ball must move in that direction. As the tube next swings around, the ball simply carries on moving in the same direction it was already moving in! At least that is what it does apart from forces from the sides of the tube. So you see there's no need to push the ball outwards. Outwards is the direction it is heading in already! (I mean outwards in the sense of increasing distance from the point about which the tube rotates). The tube simply pushes the ball in a sideways direction, away from the purely straight-line motion it would follow in the absence of force.

due to the rotation of the tube, you obtain two forces on the mass

the centrifugal force

$$F_r=m\omega^2\,r\,\hat{\mathbf{e}}_r$$

and the coriolis force

$$F_\theta=-2\,m\omega\,\dot{r}\,\hat{\mathbf{e}}_\theta$$

from here the EOM

$$m\,\ddot r=F_r$$

the solution $$~r(t)~$$ with the initial conditions $$~r(0)=L~,\dot r(0)=0~$$ is

$$r(t)=\frac L2\,(e^{\omega\,t} +e^{-\omega\,t})$$

the path of the mass in inertial system is:

$$x=\cos(\omega\,t)\,r(t)\quad, y=\sin(\omega\,t)\,r(t)$$

• centrifugal force dont exit in inertial frame Commented Nov 5, 2023 at 11:48
• This is true, but the mass is in the rotating system
– Eli
Commented Nov 5, 2023 at 12:30
• But my question is about inertial frame Commented Nov 5, 2023 at 12:33