# Problems With Circular Motion

1. Jul 7, 2006

### Delzac

hi i hope that u are able to help me with these concept problems.

Q1. My teacher said that centrifugal force is aa imaginary force and that is does exist, is this true? i quote(from my teacher) " last time people thought that centrifugal force exist, but in actual fact it is a imaginary force which does not exist"

Q2. we know that the earth is orbiting around the sun, circular motion requires that there be a centripetal acceleration towards the sun. If there is a acceleration towards the sun, shouldn't the earth fly towards sun and collide together??

Q3. Say we have a eraser, and we put is on a spinning disc. when the rotation of the disk is slow, the eraser doesn't fly off because of friction. Now the problem arises that my teacher told me that frictional force is actually equal to centripetal force( magnitude and direction), but at any instant, the rubber's velocity is tangential to the radius, so shouldn't there be also a frictional force opposing this velocity in the apposite direction of the velocity? If this is the case, there there will be 2 frictional forces acting on the eraser instead of one.( The one which is Frictional force = centripetal force)

Q4. say we are pendulum, and we make is osscilate in a circular motion, should there be a outwards force? my teacher said that there isn't.( His explaination isn't very comprehensive)

2. Jul 7, 2006

### Hootenanny

Staff Emeritus
Yes, the centrifugal force is a pseudo forces and arises when we consider circular motion from a non-inertial reference frame (such as the frame of the object undergoing circular motion).
Although, the earth is always accelerating towards the sun, the velocity is always perpendicular to the acceleration hence the circular path. (Technically it is an elliptical orbit, but you get the idea). Just because an object is accelerating towards another object, doesn't mean that it must get any closer.
You may want to reconsider this. Is the rubber accelerating tangentially? Or is it travelling at a constant speed?
No, there will never be an outward force (except the weight at the bottom of the oscillation). Ignoring drag, the only to forces acting on a pendulum is its weight and the tension in the string. Draw a free body diagram. Where would this other force originate?

Last edited: Jul 7, 2006
3. Jul 7, 2006

### Delzac

For Qns 3. I meant that at any instant, the velocity of the eraser is tangent to the circle right?( refer to attachment althought it is quite blur)

Since the eraser is in circular motion, there is a centripetal acceleration right?( the "a" in the picture)

and because the eraser doesn't fly off the spinning disk, there is a frictinal force which is equal in direction and magnitude to the centripetal acceleration correct?( or so my notes says) my teacher told that only this frictional force is affecting the eraser.

Now the problem arises that, " any instant, the velocity of the eraser is tangent to the circle ". so naturally shouldn't there be also a frictional force opposing this velocity at that instant?? this in the end makes up 2 frictional force acting on the eraser, instead of the " my teacher told that ONLY this frictional force is affecting the eraser." mention above.

+ hope i explained clearly+

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4. Jul 7, 2006

### Delzac

For Qns 4, let us replace the pendulum bob with a human, i am sure if i am in circular motion i will feel a outwards force........how can we explain this?

5. Jul 7, 2006

### nazzard

You have to make clear about which reference frame you are talking. You are feeling a force pointing outwards because your reference frame is moving with you.

Last edited: Jul 7, 2006
6. Jul 7, 2006

### Delzac

And i still can't get Qns 2. If acceleration is towards the sun, a simple equation $$F=ma$$ will show that a force is exerted on the earth by the sun(towards the direction of the sun), so can't it collide with it?( i just need a better explaination, if there is one :) )

7. Jul 7, 2006

### nazzard

The force you are describing will lead to a change of direction of your tangential velocity.

The applet on this website might be helpful to visualize to process: http://highered.mcgraw-hill.com/sites/0078458137/student_view0/chapter6/circular_motion_applet.html

There's a vertical scrollbar on the left that will change the circle to an ellipse.

Last edited: Jul 7, 2006
8. Jul 7, 2006

### Delzac

Then this force is from the tangential velocity(at an instant) in orbit i guess....... but still u can treat it like projectile motion( sort of?) then earth while moving horizontally can also move inwards. ( treating horizontal and vertical velocity as seprated things? as in projectile motion)

9. Jul 7, 2006

### Delzac

what do u mean? i dun quite understand how different reference frame affect what force u are feeling.

10. Jul 7, 2006

### nazzard

The important difference is: Is it an inertial reference frame or a non-inertial reference frame. In the latter case the reference frame is accelerated and fictitious forces are acting on you.

Edit: I've edited post #7 to include a link to an applet that might be helpful.

Last edited: Jul 7, 2006
11. Jul 7, 2006

### Staff: Mentor

It doesn't, of course. Different reference frames just change the way you describe what's going on; they don't change what's going on.

If you stick to an inertial frame, then you only need to deal with real forces, not pseudoforces.

12. Jul 7, 2006

### Hootenanny

Staff Emeritus
This is all good.
This is not so good. You have said it yourself, velocity. Just because the velocity is in that direction doesn't mean there is a force. Forces create accelerations. Is there any force acting tangentially to the eraser? Is the eraser travelling at a constant speed? Remember Newton's first law. Remember that the eraser is travelling at the same velocity as the disc.

Last edited: Jul 7, 2006
13. Jul 7, 2006

### Hootenanny

Staff Emeritus
Just add to what the Doc said. A good way to get your head round this is to imagine a child on a round-a-bout. When you are watching a child going around on a round-a-bout what you see is the child holding on, 'pulling towards the centre of the circle'; this is the centripetal force. You are viewing this from an inertial (non-accelerating) reference frame. Now imagine you yourself are on the round-a-bout, you feel yourself being 'pulled' off the round-a-bout; this is the centrifugal force because you are in a non-inertial reference frame. Do you follow?

Last edited: Jul 8, 2006
14. Jul 8, 2006

### Delzac

Thx i understand now :)

15. Aug 2, 2006

### Delzac

Hi again, sry to dig up an old post again but i wanna clarify something. :P

U guys mention that when the acceleration is perpendicular to velocity, the object will not move towards the direction of the accelration, but move in circular motion right?

Say i have a rock obiting around a planet(going in circles). in circular motion, the rock should not fall into the planet right? but sometimes, or often the case is that the rock do fall into the planet sooner of later isn't? y does this happen, shoudn't it be in circular motion(meaning it will circle the planet forvever)?

hope u understand what i am driving at :P

16. Aug 2, 2006

### HallsofIvy

Staff Emeritus
If the rock is close enough to the planet that it is affected by noticable friction from the atmosphere, then there will be a component of acceleration (due to the friction force) not perpendicular to the velocity vector. That will slow the rock and cause it to eventually fall to the planet. If, however, the rock is far enough away or the planet has no atmosphere so that there is no friction, the rock will circle the planet forever.

17. Aug 2, 2006

### civil_dude

Maybe try thinking about a swinging bucket of water in a vertical plane. There is a minimum velocity such that the water will stay in the bucket. If you spin it slower, the water will spill out. If you maintain the minimum velocity, the water stays in, and at the instance at the top of the motion, the only force is gravity, which provides the centrifical force normal to the velocity.

18. Aug 3, 2006

### Delzac

HallsofIvy said that it is due to friction of the atmosphere but civil_dude said that it is the minimum velocity so whcih is it, friction or velocity or both?

19. Aug 3, 2006

### Staff: Mentor

Both are correct, they are just talking about different aspects of the same thing. In either case it's true that as long as the force remains perpendicular to the velocity, the rock will remain in a circular orbit.

Of course to get that rock in orbit in the first place requires a certain speed. As you know, if you throw a rock it just falls to the ground following a parabolic trajectory. The force starts out perpendicular to the velocity, but doesn't stay perpendicular. It will only stay perpendicular if the rock is moving fast enough. (Of course if the rock is moving too fast, it will have an elliptical orbit around the earth's center--assuming it doesn't just crash into the surface at some point. Faster still and it will "escape" the earth's gravity, never to return.)

Once the rock is in a circular orbit, as long as no other forces act it will stay there. The gravitational force and the velocity are perfectly matched to remain perpendicular. But if you add in the effect of atmospheric friction, as Halls' explained, the net force is no longer perpendicular to the velocity.

20. Aug 5, 2006

### Delzac

thx for the explaination! :)