Conceptual difficulty in a rotational dynamics

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.