1
$\begingroup$

If a coin is placed on a slanted surface and the surface is rotated about some center located away from the coin, the coin will overcome friction and slide up the surface. However, the force acting on it is centripetal force directed toward the center of the rotation. If you decompose this into normal and parallel components, the parallel component points down the surface, not up. So what force is responsible for it moving upward?

In the following, every real force has no component directed up the plane.

enter image description here

Improve this question
$\endgroup$
2
  • $\begingroup$ You might try thinking why the same thing would happen on a flat surface. $\endgroup$
    – BowlOfRed
    Commented Nov 2, 2017 at 3:58
  • $\begingroup$ @BowlOfRed On a flat surface, I think I can get the idea better: The centripetal force induces a normal force, and friction pointing upward can be strong enough to resist gravity. The only force that points upward is a balancing force and so the coin could not actually move upward. But on a slanted surface, the coin could actually move up the surface. $\endgroup$
    – Addem
    Commented Nov 2, 2017 at 5:02

2 Answers 2

Reset to default
1
$\begingroup$

Your hypothesis is correct. If you adopt a frame of reference which is rotating with the tilted surface, then there is a fictitious centrifugal force which is $F_a=mr\omega^2$ in your diagram. ($\omega$ is the angular velocity of rotation and $r$ is the radius of rotation.) This acts radially outwards and has a component up the plane. If this component exceeds the component of the weight of the coin down the plane (plus any friction force) then the coin moves up the plane.

In an inertial frame, centripetal force is not a force itself, it is the component of real forces which cause circular motion.

As you are aware, the only real forces acting on the coin are gravity $W$ and the normal reaction $N$ anf friction $F$ from the tilted surface. The horizontal componenents of $N$ and $F$ provide the centripetal force $m r\omega^2$. If the vertical component of $N$ is greater than $W$ and the vertical component of $F$ then the coin will move vertically upwards. As it does so, it leaves contact with the surface and inertia causes it to move tangentially in a straight line. This motion has a radial component which carries the coin horizontally back to the inclined plane. So it is the combination of the normal force $N$ and inertia which carries the coin upwards and outwards along the plane.

Improve this answer
$\endgroup$
0
$\begingroup$

Centripetal force would keep the coin from moving up. If, and only if, the coin is not moving up the surface, there is (apparently) a sufficient centripetal force to keep it in place. But there is no guarantee that such a force would exist. In the scenario you are describing, the friction between the coin and the surface is not sufficient to keep the coin in a circular trajectory - so it slides up.

In a rotating frame of reference, there is a "fictitious force" called the centrifugal force - this is a force that seems to act on an object, and it is needed to explain why you need to apply a centripetal force on the object to keep it from moving (in the rotating frame).

There is no guarantee, in general, that there will be a sufficient centripetal force. But the centrifugal force is always acting in a rotating frame of reference. And that's the force that, in your example, is pushing the coin up the slope.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So in the scenario, all true forces acting on the coin are weight, the normal force, and friction. The weight generates a normal force that allows friction to keep the coin in place granted the slope is gentle enough. The friction acts sorta in the opposite of the direction of circular motion to resist slippage between the coin and the surface, but also a little down the surface. So of all these forces, weight, the normal force, and friction, they all point down the surface. So what I don't get is how there's this fictitious force--what is it? If I'm solving a physics problem do I use it? $\endgroup$
    – Addem
    Commented Nov 2, 2017 at 4:54
  • $\begingroup$ To put it another way, the only "true" force that points up the plane is friction, which is a balancing force and cannot actually cause the object to move upward. My instinct tells me that my problem with this scenario is the same as it would be if, say, you put the coin on the slanted surface and then accelerated it (through empty space) in a straight line fast enough to overcome friction. Intuitively at a certain acceleration this would happen. But formally, I don't see where the force comes from, which would cause it to slip up the plane. $\endgroup$
    – Addem
    Commented Nov 2, 2017 at 5:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.