Bead threaded on a smooth rod - why does spinning it at constant $\omega$ cause the bead to slide radially outward?

Question

We over a thought experiment of sorts in class today, and I'm trying to look for some help to better wrap my head around it. For this experiment, I ask you to image a frictionless rod rotating horizontal (oriented on the x-axis and spinning "in and out of the page") with a bead threaded around it, its distance from the origin some amount $r_0$ of $r$. It spins at a constant velocity $\omega$. I can be convinced that at $t=0$, the only force on the bead is the rod, parallel to $\hat \theta$. Since it's not really an extension of the rod but threaded around it, it doesn't feel a central force and is just pushed by the rod in a circle.

Now, according to my lecturer, at later $t$ the bead starts to slide off the rod. This is anecdotally plausible, but proving it using mathematics seems to cause me some confusion. If the force vector is always perpendicular to the rod, how could it possibly slide off?

Here's the explanation the lecturer gave:

Use polar coordinates, so the force on an object can be expressed by the following:

$$\vec F = \hat r \vec F_r + \hat \theta F_{\theta}$$

The lecturer then notes $\hat r \vec F_r = 0$. I assume this is because the rod is smooth, so there is no source for a force to occur radially.

$$\vec F = \hat \theta F_{\theta}$$

And, since $\vec F_r = 0$,

$$\hat r a_r = 0$$ $$a_r = \ddot r-r \dot \theta ^2$$

$$ r \dot \theta ^2 = \ddot r$$

Thus, according to the fact that $a_r =0$, if there is no friction on this object it will move outward. Now, other than making deductions using the equations, I have no clue as to why this is true. Sure, according to the equations this makes sense, but intuitively, I have no clue how the bead can mov outward if the force on it is always perpendicular to the rod.

Answer 1

Place yourself in the frame of the rotating rod. In this frame of reference there is a centrifugal force on the bead, causing it to accelerate radially outwards. This is the force which "pushes" you outwards in a car which is cornering. The origin of this fictitious force is the bead's tendency to oppose any change in its momentum, ie its inertia. In the case of the car, the door prevents you from moving outwards. In the ground frame it provides the centripetal force to keep you moving in a circle. Because there is no friction with the rod, there is nothing to provide a centripetal force to oppose the fictitious centrifugal force.

The bead also experiences a sideways Coriolis force which pushes it against the rod. This fictitious force is opposed by the normal reaction of the rod. If the bead did not move, it would not experience any sideways Coriolis force, nor any tangential reaction force from the rod.

Both of these fictitious forces are invoked to explain the motion of an object which is not constrained by any forces, from the point of view of an observer rotating with the rod. From the point of view of the ground (see below) the object moves in a straight line. From the point of view of an observer on the rod the object moves outward and sideways (opposite to the direction of rotation). If the object (the bead) is constrained to move on the rotating rod, then it only moves outwards. The Coriolis force is balanced by the normal reaction from the rod.

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In the inertial ground frame of reference, the bead moves outwards because of inertia, its tendency to move in a straight line.

Suppose the bead at P is moving along with a rod OP which is rotating anticlockwise in the ground frame, as in the diagram below.

From the point of view of the ground a free bead released at P tends to move in a straight line along a tangent. The rod rotates from OP to OP1. Meanwhile a free bead has moved from P to P1' where PP1' equals arc length PP1.

Relative to the moving rod the free bead has moved outwards (to a larger distance from O) and sideways (to the right of P1 looking outwards from O). In the ground reference frame the straight-line motion of the free bead is explained by Newton's 1st Law : it is not necessary to invoke any forces. In the rotating reference frame this change of motion is explained by fictitious forces, according to Newton's 2nd Law : the outward acceleration is attributed to the centrifugal force, and the sideways acceleration to the Coriolis force.

If the bead is confined to the rod then the sideways motion is prevented by the normal reaction force provided by the rod, which is equal and opposite to the Coriolis force. If the rod is frictionless there is nothing to pull the bead back inwards to keep it a fixed distance from O. The bead moves outwards not because of a force but because of the absence of any centripetal force which would keep it moving in a circle.

Answer 2

I have no clue how the bead can move outward if the force on it is always perpendicular to the rod.

True, but the rod (and thus the bead) are turning.

Think of it this way, the rod starts moving the bead up from the $x$-axis. Both objects have a velocity in the $y$ direction. Since the rod is perpendicular to this axis, the bead cannot move along it.

Immediately after, the rod has moved upward slightly. The rod (and thus the bead) rotate. As the rod rotates, the pivot constrains it to reduce the speed in the $y$ direction. But the rotated rod cannot apply a similar force on the bead to reduce its $y$ velocity. So the bead continues to move in that direction (now along the rod), and is simultaneously accelerates in the $x$ direction by the rotating rod.

Answer 3

It is because force does not generate velocity, it generates acceleration. For instance, to keep the Moon on its orbit, Earth pulls it. The Moon moves tangentially but Earth is pulling radially. The motion of the bead is somewhat the converse: the beam pushes tangentially but the bead moves radially. I encourage you to draw a diagram of $\vec F(t)$, $\vec a(t)$, $\vec v(t)$ and $\vec v(t+\delta t)$ for the bead and for the moon and check the difference.

Answer 4

The claim that you make "The bead will not have a radial acceleration since there is no force in that direction" is wrong. The radial direction is not a fixed direction that, so you can't really claim that force in radial direction is rate of change of radial momentum. To see this just imagine the rod oriented in aparticular direction, say $x$ direction at $t=0$. Now at a later time $t$ the rod comes back to the orientation along $x$ direction. In this time duration $t=0$ to $t=t$ the bead has experienced a net force in $x$ direction and so will experience a change in momentum in that direction, hence justifying the outward motion. Now to see intuitively the value of acceleration: Consider an instant of time $t=t_0$ when the rod is in a certain orientation and the bead is at distance $r$ from the center, moving with a velocity $v_r$ in the current direction of the rod and a velocity $\omega r$ in the direction perpendicular. Since a this instant the force on the bead is perpendicular to the current orientation of the rod, so it will show no change in momentum in the direction of the currect orientation of the rod. Hence \begin{equation} v_r+dv_r - \omega (r+dr)d\theta = v_r \end{equation} \begin{equation} dv_r = \omega r d\theta \end{equation} \begin{equation} \frac{dv_r}{dt} = \omega r \frac{d\theta}{dt} = \omega^2 r \end{equation} \begin{equation} \end{equation}