It is noticed that when a disk or something of that sort is rotated really fast and a small bead is put on it that bead flies almost radially outwards. Here's my problem:

  1. If there were no friction at all then from an inertial frame of reference the ball would be absolutely stationary. Ok that's fine.

  2. But in the practical case ( that is as we observe generally like I have when I out a small bead on rotating plate) when there is friction (static and dynamic) then the bead seems to be slipping outwards.

Where is this force coming from?

The only inertial (real) force acting is friction; static intially for a short time when the bead just starts to slip away and then dynamic when it should tangentially fly off. (this friction probably curves it a little bit but thats not the issue) But we see that the trajectory is almost radial. There is no radial outwards real force whatsoever so why is this being observed?

My attempt: The bead does actually slip outwards tangentially in a linear trajectory (ignoring small curving due to dynamic friction) but the closer it is to the center the more it seems to be coming from the centre and thus radially outwards. And this is why when we keep the bead to the rim there it does actually looks like flying off tangentially.

So is this attempt of mine correct where the radial outwards motion is just not there but it seems like that to an observer due to the actual tangential motion?

Just to again clearly state my question:

If an observer from an inertial frame observers a bead kept on a rotating frame then does the bead start moving (amsince velocity was alot to overcome any static frictoon)outwards tangentially or radially?

  • 2
    $\begingroup$ When I test it, the bead moves making a spiral trajectory. $\endgroup$ Nov 20 '20 at 1:54
  • $\begingroup$ How spiral is it though? That little spiral is probably due to dynamic friction while in effect it seems to be moving outwards ...is that radial or tangential? $\endgroup$
    – Lost
    Nov 20 '20 at 12:33
  • $\begingroup$ @Lost en.wikipedia.org/wiki/Coriolis_force this might help $\endgroup$
    – Anonymous
    Dec 15 '20 at 10:18
  • $\begingroup$ I know about the coriolis force and its effect. Please read the question again. I am asking about the trajectory from an i ertial frame where no fictitious forces exist. $\endgroup$
    – Lost
    Dec 15 '20 at 17:23
  • 1
    $\begingroup$ @Lost think like this : the disc atoms have enough friction between them to provide neccessary centripetal force while the bead doesn't. So it is the disc that is moving radially and not the bead. From the ground frame it may look that bead is pushed outward and hence it moves in a spiral path. $\endgroup$ Dec 15 '20 at 20:31

Suppose that after being rotating with the disk, the static friction force suddenly vanishes. The bead just follows a tangential path, keeping its momentum of the last time of contact.

The real situation is similar, except for the existence of a kinetic surface friction. But here, instead of the usual kinetic friction force against velocity, it helps to increase it. Once the movement starts, the bead moves to a region of bigger radius. The speed of the disk is bigger here than the bead speed, ($v = \omega r$). So the consequence is to accelerate the bead.

The final effect is not only in the bead speed, but also in its direction. If the disk is turning clockwise for example, the bead will spiral clockwise.

Normally the process is very fast, and the spiral seems a radial movement.

  • $\begingroup$ So as I had stated while asking the question: The fact that it seems to slip away radially is just an observational defect while in reality, the path is tangential as expected (of course with effects of dynamic friction which spiral it)? Is this what you mean while saying "Normally the process is very fast, and the spiral seems a radial movement"? $\endgroup$
    – Lost
    Nov 21 '20 at 7:06
  • $\begingroup$ Yes. Our eyes keep track of the rotating bead before it loses contact. When it hapens, our perception are at a point in the rotating frame. And for that frame the movement is radial. $\endgroup$ Nov 21 '20 at 11:22
  • $\begingroup$ When u say is radial...you mean the perception of it really right? While in reality, that is, if we observe it from a slow motion frame by frame observation it would look tangential ryt? Its no more than an illusion. $\endgroup$
    – Lost
    Nov 21 '20 at 19:52
  • $\begingroup$ I said the if we were in the rotating frame, fixed in a chair near the center for example, any released object would move radially. $\endgroup$ Nov 21 '20 at 19:55
  • $\begingroup$ Yea. That is due to centrifugal that is obvious. As I mentioned in the question I am talking about the observer who is in the inertial frame. $\endgroup$
    – Lost
    Nov 22 '20 at 12:46

Take a rotating disk with speed $\Omega$ and place a rolling sphere of radius $R$ on it. Assume no-slip conditions and track its path using the polar coordinates $r$ for radial position and $\varphi$ for azimuth.

Take at one instant a coordinate system with its x-axis pointing towards the sphere such that

$$ \begin{aligned} \vec{pos} & = \pmatrix{r \\ 0} \\ \vec{vel} & = \pmatrix{\dot{r} \\ r \dot{\varphi} } \\ \vec{acc} & = \pmatrix{\ddot{r}-r \dot{\varphi}^2 \\ r \ddot{\varphi} + 2 \dot{r} \dot{\varphi}} \end{aligned} $$

The sphere has 5 degrees of freedom, which in velocities are expressed as the two velocities above and the following 3 rotational components

$$ \begin{aligned} \vec{omg} & = \pmatrix{ \omega_r \\ \omega_\varphi \\ \omega_z } \\ \vec{alp} & = \tfrac{\rm d}{{\rm d}t} \vec{omg} = \pmatrix{ \alpha_r \\ \alpha_\varphi \\ \alpha_z } \end{aligned} $$

The no-slip condition means that the rotational motion is coupled with the translational motion.

$$ \begin{aligned} \vec{omg} & = \pmatrix{ \tfrac{r (\Omega-\dot{\varphi})}{R} \\ \tfrac{\dot{r}}{R} \\ \omega_z } \\ \vec{alp} & = \pmatrix{ \tfrac{(\Omega - \dot{\varphi}) \dot{r}-r \ddot{\varphi}}{R} \\ \tfrac{\ddot{r}}{R} \\ \alpha_z } \end{aligned} $$

Now that the kinematics are set, we write the equations of motion in terms of two frictional forces at the contact $$ \vec{frc} = \pmatrix{F_r \\ F_\varphi } $$

The five (two translational and three rotational) equations of motion are

$$ \begin{aligned} F_r & = m ( \ddot{r}-r \dot{\varphi}^2) \\ F_\varphi & = m (r \ddot{\varphi} + 2 \dot{r} \dot{\varphi}) \\ R\,F_\varphi & = I_c \left( \tfrac{ \Omega \dot{r}-r \ddot{\varphi} -\dot{r} \dot{\varphi}}{R} \right) \\ -R\,F_r &= I_c \left( \tfrac{\ddot{r}}{R} \right) \\ 0 & = I_c \alpha_z \end{aligned}$$

And since for a sphere $I_c = \tfrac{2}{5} m R^2$, the path the sphere takes is described by the following coupled differential equations

$$ \boxed{ \begin{aligned} \ddot{r} &= \tfrac{5}{7} r \dot{\varphi}^2 \\ \ddot{\varphi} & = \tfrac{2 \dot{r} ( \Omega - 6 \dot{\varphi})}{7\;r} \\ \end{aligned} }$$

And frictional forces

$$ \begin{aligned} F_r & = - \tfrac{2}{7} m r \dot{\varphi}^2 \\ F_\varphi & = \tfrac{2}{7} m \dot{r} ( \Omega + \dot{\varphi}) \end{aligned}$$

The resulting behavior is as follows, given initial conditions of a slight inwards radial speed.


radial acceleration oscillates between zero and some max value, making the radial velocity outwards ratchet in steps. Tangentially it seems to be orbiting in one direction at first, but then it would reverse directions.

  • $\begingroup$ In which frame have you calculated this? and if it's in an inertial frame then how can "(instead of only moving radially outwards)" be true since there is no radial force of any kind whatsoever(friction has no radial component) so it cannot in the first place have any radial motion. $\endgroup$
    – Lost
    Dec 18 '20 at 10:12
  • $\begingroup$ It is an inertial frame that in the particular instant is co-aligned with the location of the sphere. I developed the equation for any azimuth angle $\varphi$ but reported the case when $\varphi=0$, because it simplifies the answers. Anyone is welcome to do a similar model and validate or invalidate my answer. $\endgroup$ Dec 18 '20 at 13:11
  • $\begingroup$ But friction does have a radial component for no-slip as you can see from the 4th of the 5 equations of motion. The radial friction causes tangential torque which drives the rolling in and out but also causes the centripetal acceleration keeping the sphere in orbit. $\endgroup$ Dec 18 '20 at 13:14
  • $\begingroup$ In an inertial frame , the radial component of friction can be inwards which provides centripetal, when the body is performing circular motion without getting tangentially driven off. How can there be a fricitonal force radially outward since friciton always motion? $\endgroup$
    – Lost
    Dec 19 '20 at 16:11
  • $\begingroup$ @Lost - I looked at the problem again (see updates above) and it seems I had made some mistakes before. The radial force now is $$F_r = -\tfrac{2}{7} m r \dot{\varphi}^2$$ and strictly related to orbital motion. $\endgroup$ Dec 19 '20 at 20:04

It will be a combination of both. Friction plays two roles here. It provides both the radial and tangential component of force. There will be centrifugal force to push it out radially and there will also be a tangential force again provided by friction.

The path will be spiral.

If we keep the bead on the rim and rotate the disk really fast then we can say that the angular velocity and thus the centrifugal force is huge. It is not like the tangential component does not exist. The component of the radial force just dominates largely over it. So the bead just appears to fly more in the radial direction.

  • $\begingroup$ I am not able fully understand your question as far as i can i would say it will fly off radially since friction is not able to provide it necessary centripetal acceleration so due to inertia the object has more tendency to move on straight paths rather than in circular $\endgroup$ Nov 20 '20 at 4:31
  • $\begingroup$ @PrateekMourya Centripetal force is what causes the straight path motion I have mentioned that since the centripetal(or centrifugal- depending on your frame of reference) component is much more than the tangential component. $\endgroup$ Nov 20 '20 at 4:35
  • $\begingroup$ @Phy_Amateur No. The centrifugal force does not exist for an inertial observer. Its a fictitious force. $\endgroup$
    – Lost
    Nov 20 '20 at 12:35

Actually, when the bead is put onto the rotating disc, your thinking is absolutely correct. The bead is acted upon by the frictional force which tends to move the bead in the same circular path as the point of contact on the disc is going along. This makes the situation seem like this-
Point 1: the bead is acted upon by a force (frictional) that subsequently gives it a tangential velocity. By the way, I am considering the time interval when slipping has not yet started
Here is where the game begins. As soon as the bead acquires a tangential velocity, it experiences an inward centripetal force (in any inertial frame, to sustain a circular motion, a body 'experiences' a radially inward pull due to its tendency to change the instantaneous velocity vector) due to its acquired motion in the "instantaneously" circular path (I wrote instantaneously because I don't know about the motion of disc or when will the slipping in tangential direction begin). The friction will begin once again, comes to the action and tries to balance this centripetal force by acting in radially outward (centrifugal) direction .
Point 2: Friction is acting in 2 perpendicular directions in this instant, one tangential and one radially outwards.
The friction being the only "real" force, the bead experiences an outward acceleration in the ground frame along with an acceleration in the tangential direction.
Point 3: This would make the path of the bead an "imperfect" spiral (imperfect because friction in both directions can change its value)
Point 4: But in real aspects, when the rpm of the disc is high enough in its magnitude and the radius of the disc is small enough in length that the bead is ejected out long before it can spend a significant time for our weary eyes to notice it's tangential motion. Thus all we notice is a "nearly" radially outward motion of the bead.

  • $\begingroup$ Point 2 : Why is friciton acting radially outwards? $\endgroup$
    – Lost
    Dec 20 '20 at 6:12

Let's start using a small cube rather than a bead. The bead will roll, and it will get an inertial spin, and that complicates things. Let's try to reduce the complication.

Ignoring spin on the cube, friction does two things. One is it provides a tangential force. The other is it opposes the cube's existing velocity in any nontangential direction.

WLOG, say the cube is at position $[0,y]$ on the turntable, and has velocity $[v_{cx},v_{cy}]$. That spot on the turntable is traveling at velocity $[v_{tx},0]$.

The effect of friction is proportional to the cube's mass, so we can fold those two together. The acceleration of the turntable on the cube is $f[v_{tx}-v_{cx},-v_{cy}]$

Where $f$ is a term for friction (or fudge factor). The turntable accelerates tangentially proportional to the difference between the existing cube velocity versus the turntable velocity, and it decelerates the cube proportional to its existing radial velocity.

$f$ is linearly proportional to $y$.

When it's a bead that rolls? I don't want to think about that.


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