3
$\begingroup$

If a mass $m$ is put on a stationary table, and then we start rotating the table counterclockwise, the mass will fall from the table, as can be seen in this video https://youtu.be/IOcrHOc23N4?t=187.

If I understand correctly, this happens due to friction $f$ between the mass and table. When the table starts rotating, the mass wants to stay inertial, and thus the table applies a force on it, which causes it to fall from the table. I've added a diagram:

Diagram

  1. Is my analysis correct? is the direction of the force correct?
  2. This force pushes the ball out, thus it acts centrifugally (away from center). However, of course, centrifugal force are fictitious! So is this not a centrifugal force? (I'm in an inertial frame, outside the table).

I've read previous questions and couldn't understand them.

$\endgroup$
3
  • $\begingroup$ shouldn't friction be pointing in the same direction as the rotation of the plate? Since the plate wants to accelerate the mass. The reason why the mass won't remain on the plate is because friction does not provide enough centripetal force for the mass to remain in a circular path. $\endgroup$ Commented Jan 2, 2022 at 10:02
  • 1
    $\begingroup$ No, table does not apply force to an object due to which it falls down, because centrifugal force is pseudo force, i.e. imaginary force which has no real physical causes (or no real physical field which would resulted in such action, like gravity, electrostatics, inter-molecular interaction, etc). All pseudo forces are generated only because object "wants" to stay in an inertial reference frame, but non-inertial frame "opposes" to this wish, hence object responds with a pseudo force. With respect to ground object actually IS at rest, as it was before movement, assuming zero friction. $\endgroup$ Commented Jan 2, 2022 at 10:05
  • 2
    $\begingroup$ @AgniusVasiliauskas I'm in an inertial frame, outside the table. Once the table start spinning, I see the mass falls from the table. Thus the mass accelerates, therefore there is a force acting on it. My question is: what force is this, and what is its direction? $\endgroup$
    – blz
    Commented Jan 2, 2022 at 10:15

2 Answers 2

4
$\begingroup$
  1. Is my analysis correct?

Almost. A correction: Friction initiates the motion. Without friction, the object would never move and would never start sliding. Friction causes the object to speed up.

But from here on, there is no further need for friction. An object with a speed will keep that speed until stopped. This is what is meant by inertia and described via Newton's 1st law. So, simply by having a speed, the object will eventually fall off the table (if smooth).

There is a force present trying to alter the object's route towards the table edge, though, and that is friction along the centripetal direction (towards the centre). This friction comes into existence gradually as the object's straight-line motion due is no longer tangential but now slides over the spinning surface once again. So, we should keep two friction directions separated in our heads: tangential friction that causes speeding up, and centripetal friction that causes turning.

All in all, we are dealing with two types of effects simultaneously here: the tendency of objects to move in straight lines (the kinematics), and the forces that create motion (the dynamics). Each topic can easily be analysed separately - together, they easily become more confusing to overview what causes what, and we have to describe and explain both the kinematics and the dynamics at the same time as here.

is the direction of the force correct?

No. The rotation will cause the surface under the mass to move rightwards. The friction will pull rightwards as well. Not leftwards. Remember that friction tries to prevent sliding - it does so by pulling the object along with the moving surface towards the right.

Via Newton's 3rd law, all forces come in pairs. As the object is pulled rightwards by friction, simultaneously the spinning surface is pulled leftwards by that same friction force, counteracting the spinning a bit. (But I don't think that is what you meant.)

This force pushes the ball out, thus it acts centrifugally (away from center).

No, it acts tangentially. At first sight it might look like the object is moving straight, perpendicularly away from the centre. But it actually isn't. It starts out tangentially and ideally continues in a straight line from here which is away from the circle - realistically, the mentioned inwards friction will appear and caues the object's path to deviate and follow the circle a little bit while at each moment moving a bit further away alongs its instantaneous straight-line direction.

However, of course, centrifugal force are fictitious! So is this not a centrifugal force?

Indeed, no centrifugal force exists. Only a centripetal force does. As described, there is no force pushing the object straight outwards. Rather, the object will - when in motion, so due to its inertia - want to keep its straight-line motion which in this case started out tangentially due to a tangential force. And should the object ever fully leave the spinning surface, then it will (ideally) continue moving farther and farther away at the same, constant speed. Until something stops it. There is no force pushing it outwards; there are only forces pulling inwards and tangentially.

$\endgroup$
4
$\begingroup$

Friction always acts in a direction that opposes any relative motion between the object and the table. If initially the object is at rest (in an inertial reference frame) and the table is rotating anti-clockwise then friction will also initially act anti-clockwise on the object i.e. from left to right in your diagram. (There is an equal and opposite force acting on the table, but if we assume that some mechanism keeps the table rotating at constant angular speed then we can ignore this reaction force).

So friction starts the object moving from left to right. If friction were turned off after a small interval of time then the object would continue moving in a straight line (in an inertial reference frame) and would eventually fall off the table. Note that once the object starts to move no further force is required to make it fall off the table. Instead, the object falls off the table due to the absence of a centripetal force to keep it moving in a circle.

If friction is not turned off then it will continue to act on the object so as to reduce relative motion between the object and the table. Even if the object reaches the same (local) speed as the table, friction will still act inwards towards the hub of the table because the table’s local velocity vector is changing over time. If the coefficient of friction is high enough and the object is close enough to the hub of the table then friction will provide enough centripetal force to keep the object moving in a circle at the same speed as the table. If not, then the object will slide off the table.

So it is not friction that makes the object slide off the table, it is the object’s inertia combined with insufficient friction.

$\endgroup$
2
  • $\begingroup$ Thanks. If I understand - friction initially accelerates the mass (from left to right). then, friction decelerates the mass in a centipetal force. If the friction is sufficient, then the mass will remain stationary w.r.t. to the center of rotation - i.e. it will rotate in the same speed and direction of the table. When the mass reaches that stationary state - is it true to say that no force is acting on it, since it's stationary? $\endgroup$
    – blz
    Commented Jan 2, 2022 at 10:48
  • 1
    $\begingroup$ @blz No. If the object is stationary with respect to the table then it is moving in a circle in an inertial reference frame. A centripetal force is needed to keep the object moving in a circle. Friction provides this centripetal force. Friction still acts on the object because the direction of the table’s velocity vector (at the point where the object is) is changing even though its magnitude is constant. $\endgroup$
    – gandalf61
    Commented Jan 2, 2022 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.