Rotational Motion in Newtonian Mechanics

[Student149]

Homework Statement

Given an plate that has 2 pins attached to it. Each pin has a single balanced circular disc firmly attached to it. Each pin can rotate on their axis in both clockwise and anti clockwise direction. (See image below)

Now, assume the discs are rotating with the same angular velocity in opposite direction (they are same in every possible way). Since, the discs are rotating with the same angular velocity and are same in size, weight etc. there will be no change in center of mass, thus no motion in the plate and the plate itself will be at rest.

Now, suppose we choose 1 point on each disc's circumference such that the line joining them is parallel to x-axis. Moreover both points are traveling in +y direction at the instance we chose them (with obviously the same angular velocity, so the line joining them is always parallel to x axis).

We now apply an external tangential force (from some unspecified external source) at only both these chosen points only when the points are traveling in +y direction with the rotation of discs. The direction of the force keeps adjusting so the force stays tangential to these points or in other words is pointing exactly opposite to the velocity vectors at these points.

Would this external force only slow down the rotation of the discs while the platform remains at rest or also generate motion in the plate in -y direction?

In the image below blue arrows represent the rough direction of the chosen points (in yellow) along the circumference (as the discs rotate in opposite directions). The opposing force (opposite to the velocity vectors) at those points, at a particular time instance (shown in red arrows).

Homework Equations

The Attempt at a Solution

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My understanding is since the force is always tangential to the circumference points or in other words is exactly opposite to the velocity vectors at those points, there should be no motion in the plate as such, but only the angular velocity of the discs would be reduced.

 

[Orodruin]

Hint: Conservation of total linear momentum.

 

[Student149]

Hint: Conservation of total linear momentum.

I understand that will apply. But, my understanding is this:

As the force is exactly opposite to the velocity vector of the point it is going to be applied at and this continues as the point rotates with the disc the two are going to cancel out. And that would slow down the rotation.

Moreover, as the velocity vector and the force vector are exactly opposite to each other (both would make a 90 degree angle with the line joining the point with the pivot) that ensures the force component at 90 degrees or towards the pivot would be 0). Thus, pivot would not experience any force and thus the platform would remain at rest as discs slow down.

I am not sure how both of these are incorrect if at all?

 

[Orodruin]

I understand that will apply. But, my understanding is this:

As the force is exactly opposite to the velocity vector of the point it is going to be applied at and this continues as the point rotates with the disc the two are going to cancel out. And that would slow down the rotation.

Moreover, as the velocity vector and the force vector are exactly opposite to each other (both would make a 90 degree angle with the line joining the point with the pivot) that ensures the force component at 90 degrees or towards the pivot would be 0). Thus, pivot would not experience any force and thus the platform would remain at rest as discs slow down.

I am not sure how both of these are incorrect if at all?

The conclusions you draw are simply incorrect. It is unclear to me why you think they would be true. This is a very simple matter of applying Newton's second law to the entire system.

 

[Student149]

The conclusions you draw are simply incorrect. It is unclear to me why you think they would be true. This is a very simple matter of applying Newton's second law to the entire system.

Lets assume two points on the disc and mark them with a yellow marker.

Now, as the points rotate on the discs the velocity vector at those two points moves too. In the problem statement we have specified that the force is applied at only those points on the discs so as the points rotate, the point at which the force is being applied is moved along too. Moreover the direction of force is opposite to the direction of velocity.

So, shouldn't these opposite vectors cancel each other and slow down the discs?

 

[Orodruin]

Moreover the direction of force is opposite to the direction of velocity.

This has absolutely no bearing on Newton's second law. The second law relates force and acceleration. Can you describe what vectors you think would cancel?

Let us go to a simpler scenario. Just consider a single disk moving freely. It is rotating with its centre at reat and you apply a force in the opposite direction to the tangential velocity at the periphery. What is the acceleration of the disk's centre?

 

[Orodruin]

Also, the rotation will slow down because you are applying a torque. However, that does not rule out an overall linear acceleration. The linear acceleration is zero only if the force sum is zero.

 

[Student149]

Also, the rotation will slow down because you are applying a torque. However, that does not rule out an overall linear acceleration. The linear acceleration is zero only if the force sum is zero.

This was the point I was trying to make. Essentially, what I meant was if the force was applied to the rotating disk's circumference tangentially it would be spent in slowing down the disc. So, as compared to the second scenario when the same force was applied to the platform directly.

So, my understanding was the platform would have different final velocities in both cases (The second case would have give more velocity to the platform in -y direction as in first a part of force was spent in stopping/slowing the rotating discs)?

Apologies, I have no formal argument for that thus this query in the first place.

 

[Orodruin]

Essentially, what I meant was if the force was applied to the rotating disk's circumference tangentially it would be spent in slowing down the disc.

There is no such thing as "spending" a force. You can consider the work done by the force and how it relates to the energy of the disk, but that not only depends on the force, but also on the displacement.

So, my understanding was the platform would have different final velocities in both cases (The second case would have give more velocity to the platform in -y direction as in first a part of force was spent in stopping/slowing the rotating discs)?

What does Newton's second law tell you?

 

[Student149]

There is no such thing as "spending" a force. You can consider the work done by the force and how it relates to the energy of the disk, but that not only depends on the force, but also on the displacement.What does Newton's second law tell you?

The second Law simply states F=m*a. In other words force is proportional to acceleration.

So, as I understand in the two cases above, the final velocity of the system in -y direction as well as the angular velocity of the two discs would be same if the force was applied (as described) on disc's circumference or on the platform directly?

 

[Student149]

There is no such thing as "spending" a force. You can consider the work done by the force and how it relates to the energy of the disk, but that not only depends on the force, but also on the displacement.What does Newton's second law tell you?

I would be thankful if you could clear this up (confirm my understanding in the last comment) ?

 

[Orodruin]

The second Law simply states F=m*a. In other words force is proportional to acceleration.

So, as I understand in the two cases above, the final velocity of the system in -y direction as well as the angular velocity of the two discs would be same if the force was applied (as described) on disc's circumference or on the platform directly?

As long as the force sum is the same, it does not matter (for the CoM acceleration) where on the system it is applied.