3
$\begingroup$

Here is a question from my book.

enter image description here

I need to know which force is acting on the ball, making it move outwards?

It cannot be centrifugal force, as centrifugal force acts if the particle is moving in a circle. Here it is clearly not. Moreover centrifugal force acts when it is seen from the point of view of the ball itself (pseudo forces act only in an accelerated frame). What if I look at it from the ground? Which force will then be responsible for making it move outwards?

$\endgroup$
  • 2
    $\begingroup$ It is centrifugal force. Imagine yourself seated on the rotating table and holding a ball by a taut string (forget the groove). The string will be radially stretched because of centrifugal force, which is also the force acting on the ball. $\endgroup$ – Deep Sep 22 '16 at 11:26
  • $\begingroup$ Well the particle is moving in a circle so you have an (instantiaous) centrifugal force acting on the particle. However as you don't have any balancing inward force, the radial distance of the particle will increase with time. $\endgroup$ – Mikael Fremling Sep 22 '16 at 11:26
  • $\begingroup$ @Zero, centrifugal force works if it's going in a circle. Clearly it is not $\endgroup$ – Aaryan Dewan Sep 22 '16 at 12:48
  • $\begingroup$ @AaryanDewan, from the point of view of the ground, the ball will move in an outward spiral as it rotates with the disk and moves away from the center. This is effectively moving in a circle whose radius is increasing over time, meaning centripetal/centrifugal forces apply. $\endgroup$ – Nuclear Wang Sep 23 '16 at 19:33
2
$\begingroup$

Here is an attempt to explain what is going on in the (inertial) frame of reference of the world:

enter image description here

The red vector is the force from the side of the groove on the ball: as a result the ball starts to move. Initially, it will get the same lateral speed as the groove - if it's at a distance $r$, and the disk rotates at $\omega$, the velocity will be $v=r~\omega$.

A moment later, the groove will be at a different angle - but the ball tries to keep going in a straight line. It will have moved to a new radial direction, where the groove is going faster than the ball. As a result, it will once again feel a force of the wall, and it will accelerate in a new direction; I tried to indicate the new velocity as the vector sum of the old velocity plus the acceleration.

Obviously you can repeat the diagram for subsequent positions of the disk.

In the rotating frame of reference of the disk, you can describe the same thing in a different way. In a rotating frame of reference, there appear to be two fictitious forces: the centrifugal force that makes the object "want to move away from the center", and the Coriolis force that is only apparent if the object has a velocity in the rotating frame of reference.

When the ball is stationary in the groove (in the rotating frame of reference), the only force it experiences is the centrifugal force (this is right after the initial impulse that will have given the ball the same velocity as the part of the groove where it was placed). As soon as it starts moving outwards (under the influence of the centrifugal force) it will also start to swerve (under the influence of the Coriolis force). The groove will exert a force equal and opposite to the Coriolis force to keep the ball moving in a straight line in the groove.

In the rotating frame of reference, the radial acceleration of the ball can be calculated directly from the centrifugal force. The total velocity can be arrived at by calculating both the radial and tangential components of the velocity (tangential velocity is $r\omega$).

I will leave the details up to you.

$\endgroup$
  • $\begingroup$ Hi. See the ball is not moving in a circle. How can you apply centrifugal force on it, as there's no centripetal for on it too! ( even if you look at it from the centre of the circle ) ? $\endgroup$ – Aaryan Dewan Sep 25 '16 at 20:12
  • 2
    $\begingroup$ @AaryanDewan The centrifugal force is a "fictitious force" that appears in a rotating frame of reference and that only depends on the mass of the object, the rate of rotation of the frame of reference and the distance from the axis of rotation. When you constrain an object to go in a circle you need to apply an equal and opposite centripetal force to balance the centrifugal force - making "no net force" in the rotating frame of reference, so the object appears stationary. But the centrifugal force is there, regardless. $\endgroup$ – Floris Sep 25 '16 at 21:36
  • $\begingroup$ Thanks! But @Floris , can you tell me WHY do we always apply the fictitious force away from the centre, no matter from where are we looking the object from? $\endgroup$ – Aaryan Dewan Sep 26 '16 at 2:05
  • 1
    $\begingroup$ The fictitious force appears ONLY in the rotating frame of reference. Not sure what you mean by "no matter from where we are looking at the object". $\endgroup$ – Floris Sep 26 '16 at 11:37
1
$\begingroup$

Technically it's the groove that's exerting a mechanical force on the ball pushing it in the direction the groove is moving in. Inertia is why the ball travels outward because more of it's momentum is in the outward direction as opposed to the inward direction. The groove constantly applying a force and altering the ball's momentum ensures that the majority of it's momentum will always be in the outward direction.

$\endgroup$
  • $\begingroup$ There is no friction. The question says that the table is smooth. $\endgroup$ – sammy gerbil Sep 22 '16 at 16:43
  • $\begingroup$ Edited my answer $\endgroup$ – Yogi DMT Sep 22 '16 at 17:34
0
$\begingroup$

You could think of this in terms of Fictious Forces.

What if i look at it from the ground? Which force will be then responsible to make it move towards the left?

enter image description here

Source: Coriolis Force

In the inertial frame of reference (upper part of the picture), the black ball moves in a straight line. However, the observer (red dot) who is standing in the rotating/non-inertial frame of reference (lower part of the picture) sees the object as following a curved path due to the Coriolis and centrifugal forces present in this frame.

In your example, the groove constrains the ball from moving sideways.

As this a homework type question, I can give you an outline and references for you to read yourself. We have Newton's laws for when we are examing an inertial (non accelerating) frame of reference. When we move to a rotating frame of reference, we acquire extra forces, called either inertial forces or pseudo forces.

These forces are called the Coriolis Force and Centrifugal Force

You can read up more on how these forces work at Rotating Frames of Reference and Forces Involved in Circular Motion

$\endgroup$
  • $\begingroup$ I don't think coriolis force acts here; since it is along the $\rm{OY}$ axis as is evident from the pic. $\endgroup$ – user36790 Sep 22 '16 at 11:28
  • 1
    $\begingroup$ This answer (and most other comments to the question) do not address the question when viewed from the non-rotating frame. $\endgroup$ – garyp Sep 22 '16 at 11:55
  • 1
    $\begingroup$ I think the fictitious Coriolis Force does act here. However, it causes no tangential motion relative to the disk because the particle is confined to the groove, so the Coriolis Force is opposed by a real force of reaction from the side of the groove. $\endgroup$ – sammy gerbil Sep 22 '16 at 11:59
  • $\begingroup$ @sammygerbil: Yes, there is indeed Coriolis force when viewed with respect to the rotating table. But it is irrelevant here since the particle is constrained to move only along $\rm{OX}\,.$ $\endgroup$ – user36790 Sep 22 '16 at 12:29
  • $\begingroup$ @CountTo10: Yes, I do know there is Coriolis force, but it is totally irrelevant here. It is nullified by the normal forces from the sides of the groove. $\endgroup$ – user36790 Sep 22 '16 at 12:30
0
$\begingroup$

In order for the ball to stay at a constant distance from the center of rotation at a constant angular velocity, the net force on the ball would have to point inwards, from the ball towards the center of rotation (i.e. centripetal force).

However, the only "real" force acting on the ball is the normal force from the side of the groove, which points tangentially. So, the net effect on the ball is that the direction of its velocity vector is constantly being turned to point outward radially.

$\endgroup$
  • $\begingroup$ Can you post a picture of your answer that " normal force from the side of the groove, which points tangentially." $\endgroup$ – Aaryan Dewan Sep 22 '16 at 22:48
  • $\begingroup$ Floris's answer contains the diagram. The picture on the left shows the normal force (red) and the initial direction of the velocity of the ball (green). Note that the groove points radially, and the normal force is always perpendicular to the radius (i.e. tangential). Now look at the solid green arrow on the right image. If you break that into radial and tangential components, you will see that there is a small radial component - hence the ball starts moving outward radially. $\endgroup$ – mbeckish Sep 26 '16 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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