Object flying off a rotating disc

In summary, as the speed of rotation on a disc increases, the net inwards force on the disc and the object on the disc increases as well. This force, known as the centrifugal force in the reference frame of the body, is a pseudo force due to the non-inertial nature of the body's frame. Initially, the friction between the object and the disc is able to provide the necessary force to keep it in a circular path. However, as the speed increases, the friction force becomes insufficient and the object flies off the disc, following a straight line tangential to the circular path. This is due to the fact that the force required for the object to accelerate at the same rate as the disc is greater than the maximum friction
  • #1
quark001
44
0
Why does an object on a rotating disc fly off the disc as the speed of rotation is increased?

To accelerate, the object must experience a net force in the direction of acceleration -- in this case, away from the disc, perpendicular to the object's velocity vector. But what is this force?

When the speed of the disc is small, and the object remains stationary relative to the disc, it still needs an inwards force in order for it to be accelerating relative to me, right? Although relative to the disc, it doesn't feel any forces?

Anyway, as the speed of rotation increases, so does the net inwards force on the disc and the net inwards force on the object on the disc (Fnet = v^2*m/r). Why shouldn't the object just remain in position, then?
 
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  • #2
quark001 said:
Why does an object on a rotating disc fly off the disc as the speed of rotation is increased?

To accelerate, the object must experience a net force in the direction of acceleration -- in this case, away from the disc, perpendicular to the object's velocity vector. But what is this force?

This force (in the reference frame of the body) is called the centrifugal force. It is a pseudo force because the body's frame is non-inertial.

quark001 said:
When the speed of the disc is small, and the object remains stationary relative to the disc, it still needs an inwards force in order for it to be accelerating relative to me, right? Although relative to the disc, it doesn't feel any forces?

It is at rest relative to the disc as friction balances the centrifugal force. In your frame of reference, the friction is responsible for keeping it in a circular path (it is the source of centripetal acceleration).
quark001 said:
Anyway, as the speed of rotation increases, so does the net inwards force on the disc and the net inwards force on the object on the disc (Fnet = v^2*m/r). Why shouldn't the object just remain in position, then?

You see, as I told earlier, the mv2/r force (centripetal force) is provided by friction. But friction can only increase to a certain limit (μmg). If v increases further, friction is no longer able to provide the necessary force and hence it flies away (up to a point where mv2/r is again equal to the maximum friction force possible)
 
  • #3
Sunil Simha said:
This force (in the reference frame of the body) is called the centrifugal force. It is a pseudo force because the body's frame is non-inertial.

I probably need to do some reading about the centrifugal force since this is the first time I'm dealing with it, and I've never heard of "pseudo forces" and don't know what "non-inertial" means. I understand why a centrifugal force is necessary but I don't understand where it comes from (if there is anything to understand about it). But I understand the rest of the explanation. So, thanks. :smile:
 
  • #4
quark001 said:
Why does an object on a rotating disc fly off the disc as the speed of rotation is increased?

To accelerate, the object must experience a net force in the direction of acceleration -- in this case, away from the disc, perpendicular to the object's velocity vector. But what is this force?

When the speed of the disc is small, and the object remains stationary relative to the disc, it still needs an inwards force in order for it to be accelerating relative to me, right? Although relative to the disc, it doesn't feel any forces?

Anyway, as the speed of rotation increases, so does the net inwards force on the disc and the net inwards force on the object on the disc (Fnet = v^2*m/r). Why shouldn't the object just remain in position, then?

Consider a book left on the roof of a car (in error) and you drive away.
If you accelerate slowly, the book remains in position on the roof.
If you then increased your acceleration, the book may fall off the back of the car.
If someone saw you do this, they would see the book fall off the roof, but they would also observe that the book was - at all times - traveling forward.
When accelerating slowly, the friction between the car roof and the book was sufficient to cause the book to accelerate forward at the same rate as the car.
When you increased your acceleration, the friction force was insufficient to have the book accelerate at the same rate as the car - so the book falls behind: just as you would if you lined up to start a 100m race against Usain Bolt. You would not be accelerating in the opposite direction - merely accelerating forward at a lower rate than him.

The situation described above is motion in a single dimension.

Your mass on the disc is 2-dimensional motion - but the analysis is the same.

When a body travels in a circle, it is accelerating towards the centre of that circle.

To see what is happening you need to draw a picture, using a compass.

Mark out a disc of radius 10cm.

Now draw the circle, with a common centre to your disc, of radius 5cm.
That represents the path taken by your mass as the disc rotates slowly, and the friction force is sufficient to cause the mass to follow that circular path.

Now draw another circle, of radius 15cm. Place the point of the compass on the circumference of the disc, while the pencil is on the your first circle on the other side of the centre.

That circle might show the path your mass would take if the disc rotated quickly. The friction force is sufficient only to cause a circular path of radius 15cm. The mass moves further and further from the centre (but still follows a circular path) until it reaches the rim. At that point, the friction would cease, and the mass would continue in a straight line, tangential to your 15cm circle.

Note that the mass does not accelerate away from the disk [your sentence I showed red above] it merely accelerates towards the centre at a lower rate that the material of the disc is accelerating inwards.

When a car "spins out" on a race track, it merely means its circular path happens to have a larger radius that the race track - so the car eventually arrives at the out side of the track (where it often hits a barrier or slides off across some very slippery grass)
 
  • #5
THANKS. This really helped to clarify many things for me.

My question has now changed to why there is a frictional force on the object on the disc. To make it simpler, I'm reverting back to the book and car example. Say the car accelerates to the left and the book remains in position (accelerating at the same rate as the car relative to me).

The book must be experiencing a net force to the left that allows it to accelerate at the same rate as the car. This is a frictional force. Frictional forces arise because they work against a force in the opposite direction, so there must a force acting on the book towards the right. This force to the right has to be much smaller than the frictional force in order for the book to experience a net force to the left.

So, what is this force to the right that gives rise to the friction?
 
  • #6
quark001 said:
THANKS. This really helped to clarify many things for me.

My question has now changed to why there is a frictional force on the object on the disc. To make it simpler, I'm reverting back to the book and car example. Say the car accelerates to the left and the book remains in position (accelerating at the same rate as the car relative to me).

The book must be experiencing a net force to the left that allows it to accelerate at the same rate as the car. This is a frictional force. Frictional forces arise because they work against a force in the opposite direction, so there must a force acting on the book towards the right. This force to the right has to be much smaller than the frictional force in order for the book to experience a net force to the left.

So, what is this force to the right that gives rise to the friction?


Not quite. Friction prevents, or attempts to prevent, relative motion (slipping).

When the car accelerates to the left, the book will tend to remain where it was - which can only be achieved by sliding across a moving roof.

Friction might prevent that slippage - meaning the book will accelerate with the car.

The maximum possible friction force is μN, and if that is insufficient to cause an acceleration as large as the car's, then the book will fail to keep up (it will slip off the back of the car).

Note: When the car accelerates to the left, you as a passenger would tend to be left behind - except that the seat you are sitting on pushes you forward, so you keep up.
If the car accelerates at a great rate, the seat pushes you really hard. You will likely think "wow this engine must be really powerful - it is pushing we way back into the seat" when in fact the engine is not even touching you, and indeed NOTHING is pushing you BACK into the seat, it is all the seat pushing you forward to keep up with the car.
 
  • #7
quark001 said:
THANKS. This really helped to clarify many things for me.

My question has now changed to why there is a frictional force on the object on the disc. To make it simpler, I'm reverting back to the book and car example. Say the car accelerates to the left and the book remains in position (accelerating at the same rate as the car relative to me).

The book must be experiencing a net force to the left that allows it to accelerate at the same rate as the car. This is a frictional force. Frictional forces arise because they work against a force in the opposite direction, so there must a force acting on the book towards the right. This force to the right has to be much smaller than the frictional force in order for the book to experience a net force to the left.

So, what is this force to the right that gives rise to the friction?

To give another example to counter the phrase in red above.

Suppose you are running to a lecture, and a pen falls from your pocket. It will be traveling along with you at the few m/s you were running at.

When that pen reaches the ground, friction acts in a direction against the MOTION of the pen and eventually brings it to a halt.

BUT, there was no forward force acting on that pen. There may have been when the pocket was in your pocket, but once it came free from your body, there was no longer a lateral force - until it reached the ground and friction began.
 

Related to Object flying off a rotating disc

1. What causes an object to fly off a rotating disc?

The centrifugal force, which is a result of the object's inertia, causes it to move in a straight line tangent to the circle of rotation. This force becomes greater as the speed of rotation increases, eventually causing the object to overcome the centripetal force that keeps it on the disc and fly off.

2. How does the mass of the object affect its tendency to fly off the disc?

The mass of the object has a direct effect on its inertia, which is a crucial factor in determining the amount of centrifugal force it experiences on a rotating disc. A heavier object will have a greater tendency to remain on the disc compared to a lighter object at the same speed of rotation.

3. Is there a specific speed at which an object will always fly off a rotating disc?

Yes, there is a critical speed at which the centrifugal force experienced by the object becomes greater than the centripetal force, causing it to fly off the disc. This critical speed is determined by the mass and size of the object, as well as the rate of rotation.

4. Can the direction of rotation affect the trajectory of an object flying off a disc?

Yes, the direction of rotation can affect the trajectory of an object flying off a disc. When the disc is rotating clockwise, the object will fly off in a clockwise direction as well. Similarly, if the disc is rotating counterclockwise, the object will fly off in a counterclockwise direction.

5. How can the shape of the rotating disc impact the flight of an object?

The shape of the rotating disc can have a significant impact on the flight of an object. A disc with a larger radius will result in a higher critical speed, making it more difficult for an object to fly off. Additionally, a disc with a concave shape will create a centripetal force that can help keep the object on the disc, while a convex shape will have the opposite effect.

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