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This question already has an answer here:

This has been a bit of an awkward question that's been plaguing me ever since I started watching space documentaries on discovery about 10 years ago. I was saving this for the day I would ever meet Professor Brian Cox where I can point and say "HA!" see you can't explain that one (he probably can and so could someone here).

It's a simple question: when you get on a merry-go-round or something that rotates fast, then the force pushes you outward, but when they talk about things in space that rotate they refer to the force that pulls them inwards.

So which is it. If I spun something like a big beach ball really really fast then I'd still be thrown off it. If I took that into space should that suddenly keep me glued to it?

I apologise for my ignorance but I've never had this explained to me.

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marked as duplicate by Brandon Enright, BMS, Ali, John Rennie, Kyle Kanos Aug 28 '14 at 13:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ This has got to be a duplicate... $\endgroup$ – BMS Aug 28 '14 at 2:02
  • $\begingroup$ It's possibly a duplicate, but that question was never answered to the OP's satisfaction. My opinion: Invoking centrifugal force more often than not makes things more confusing to someone who doesn't yet have a good grasp on the basics. It's better to leave those fictitious forces out of the answers. $\endgroup$ – David Hammen Aug 28 '14 at 2:38
  • $\begingroup$ I'm starting to think this question is much more subtle/advanced than the fictitious forces concept. Are you essentially asking what creates the gravitational force that we all feel? $\endgroup$ – tpg2114 Aug 28 '14 at 3:18
  • $\begingroup$ I'm going to take a pretty big stab at this... but I think the question that's actually here isn't really what OP means to ask (based on the comments on the answers)... I think the question is "what makes gravity work" which is mixed up with the notion that rotation is required for a body to have a gravitational field. If I'm right, then this question should help out and you can ask more questions based on what you understand/don't understand: physics.stackexchange.com/questions/116608/… $\endgroup$ – tpg2114 Aug 28 '14 at 3:23
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There is no outward force on the merry-go-round. The force is inward.

I'll start with linear motion, then go to rotational motion. If you have a powerful car, you have felt yourself sink back into the seat and then felt that push on your back when you step on the accelerator. You sink back into the seat because of Newton's first law. When you first step on the accelerator, your velocity is not quite the same as the car's. Newton's first law says you'll keep going with that velocity. You sink into the seat. The car can push against you once the seat cushion is compressed. The car is accelerating forwards, and yet you feel a force acting from behind you.

The same thing happens when you step on the brakes, hard. Your velocity takes you forward relative to the car until you seatbelt locks. Now you will feel pressure on your chest pushing you rearward. Once again the force comes from a point opposite the direction of the acceleration.

The exact same thing happens once again when you take a hard turn toward the passenger seat (that's a hard right turn in the US, a hard left turn in Europe). Just as before, you keep your current velocity until a force acts on you to change your velocity. When you take that corner, the difference between your velocity and the car's moves you toward the car door. Now the door pushes on you. Just as before, the push comes from the opposite side of the direction of the acceleration. You are accelerating toward the passenger seat. That's toward the center of the curvature of the turn. There is no outward force.


I suspect the issue that gets people so confused about merry-go-rounds and the like is a misunderstanding of what acceleration is. Any change in velocity involves an acceleration. Velocity has a magnitude (speed) and a direction. Changing the speed of a car requires some force. Changing the direction in which a car is moving also requires a force. The force is always in the direction of the change in the velocity vector. In the case of a turning car, or a person sitting on a merry-go-round, the change in velocity is toward the center of curvature, or inward.

Comments have suggested I talk about centrifugal force. I don't think that's a good idea. Understanding Newton's laws is a prerequisite for understanding those fictitious forces.

Last Try

So let's go back to the car. I assume you have no problem understanding that inertia is the reason you sink into the seat a bit after stepping on the accelerator, or that inertia is the reason you move forward a bit with respect to the car after hitting the brakes. Your velocity is slightly different from the car's velocity for a short time after you step on the accelerator or hit the brakes. Your position with respect to the car is the difference between your velocity and the car's, multiplied by the small amount of time.

Suppose you're traveling at 70 kph (43.5 mph) and you hit the accelerator. Assuming a decent but not great acceleration, the car will be going 71 kph a twentieth of a second later. (That corresponds to a 0-60 mph time of 4.8 seconds.) The car's going 71 kph, but you will still be moving at about 70 kph. Multiplying the car's average speed over that interval by that 1/20 second time interval yields 6.9 millimeters (0.27 inches), which is how much you've moved with respect to the car. That's just enough to sink into the seat and feel the acceleration. The same happens when you hit the brakes, only now you move forward initially instead of sinking into the seat. There's no force this forward or backward motion happen. It's just inertia. Your body keeps that 70 kph velocity until a force does act on your body to change that velocity.

Now let's suppose that instead of changing speed, you go around a corner at a constant speed of 70 kph. Suppose you're going straight north at 70 kph at the start of the turn. In that 1/20 second, the car's velocity changes from 70 kph due north to 69.998 kph north plus 0.99997 kph east. Almost all of the change in velocity is orthogonal to your velocity. This means that 1/20 second interval, you have slid sideways in the car by 6.9 millimeters. No force was involved in this sideways slide. It's just momentum, no different than sinking back into the seat when you hit the accelerator or leaning forward when you hit the brakes.

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  • $\begingroup$ People certainly do talk about centrifugal forces. Could you address that in your question? I think you touch on it, but not explicitly. $\endgroup$ – BMS Aug 28 '14 at 2:04
  • $\begingroup$ Thanks David. I understand the explanation of the linear force. It's quite easy to get my head around that but when I apply the same thought to a merry-go-round I'm drawing straight force lines outward from your explanation which is exactly what I've experienced in real life. I also understand also that gravity is a very weak force but we always say that gravity has a pull rather than a push which is opposite to the way that planet earth vs a merry-go-round works. If I had a big merry-go-round in space what would happen? Basically how is gravity created? $\endgroup$ – Dal Aug 28 '14 at 2:04
  • $\begingroup$ Thanks for chipping in BMS. Maybe the question I had wasn't addressing the right topic. Centrifugal forces sound like what I'm trying to compare to gravity, are they not similar at all? $\endgroup$ – Dal Aug 28 '14 at 2:09
  • $\begingroup$ The centrifugal force is a fictitious force. It is not a real force. The only real force acting on you when you are on a merry-go-round or in a turning car is inward. $\endgroup$ – David Hammen Aug 28 '14 at 2:13
  • $\begingroup$ I feel like I just swallowed the red pill...oh look a rabbit! 😲 $\endgroup$ – Dal Aug 28 '14 at 2:19
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Well from the way it was explained here I realised that my question was not correct. I should have been asking what makes gravity or how do I create gravity which lead me to a google search (didn't need to put anything about a merry-go-round in there)

This then lead me to a site which said that no-one knows how gravity is created and how it is communicated which means I'd be pretty surprised if the answer came from this site :)

For all that are interested I found a good podcast http://www.astronomycast.com/2008/08/ep-102-gravity/

Thanks for everyone's help.

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If you are moving in a curved path, there will be a tendency to fly off in a straight line. This is just Newton's First Law of Motion. And it holds everywhere.

We can get bogged down in semantics about what a force really is, or whose frame of reference we're in, but the general tendency to "fly off in an outward direction" will always be there.

Spinning in space doesn't pull you toward the center.

I think the confusion is the following: if you are moving in a circle without flying off in a straight line, as Newton says you are wont to do, and if you are in space so there is no material thing holding you to that motion, then there must be some other force (that is, gravity) pulling on you to counter that "flying outward" tendency.

So going in circles always induces an "outward push" (again, ignoring semantics and precision terminology here). Gravity always attracts. The two effects are often balancing each other (as in planets going around a star).

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  • $\begingroup$ Thanks for you're answer Chris. I'm still a little confused. Maybe if you could answer this one. I'm hoping that I'm right in saying that it is gravity that is keeping me glued to the earth. If so, is it partly to do with the size and speed of the object which is rotating? If we were talking about a perfect sphere maybe 10 miles diameter in a dead part of space where no other planets could have influence and we started spinning this thing. Would I have some sort of gravity? $\endgroup$ – Dal Aug 28 '14 at 1:39
  • $\begingroup$ @Dal No - spinning only pushes you off the surface. The faster the surface is spinning, the greater this effect. $\endgroup$ – user10851 Aug 28 '14 at 3:22
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The external force is the centrifugal force, which is a pseudo-force (or virtual force), it arises due to the fact that the motion is taking place in an accelerating force of reference.

The force pulling you inwards when in an orbit is actually Gravity. A pseudo-force is called that because it isn't actually exerted.

When we experience a pseudo force, it is because our frame of reference (the point from which the measurements are being made) is itself accelerating, like a car turning.

Classical mechanics doesn't distinguish between rest and uniform motion (at constant velocity). Acceleration however, changes the forces.

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