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Circular Motion - Physics

Hi! I had a few doubts in the differences between vertical and horizontal circular motion.

For horizontal circular motion (ex. a car around a race track) I understand that the speed, centripetal force, centripetal acceleration and angular velocity all remain constant, provided the radius is constant.

How does it differ in a vertical circular motion (ex. me spinning a stone with a thread round vertically or the classic loop the loop situation). Well, I know the tension in the string would change but the centripetal force remains constant.

But in vertical circular motion, does the speed and angular velocity remain constant?

(edited 4 years ago)

Reply 1

username5232810

If the centripetal force and radius remain the same then, by the definition of the centripetal force, the velocity and angular velocity have to stay the same. Unless you’re losing mass or something.

Reply 2

qwert7890

Correct me if I'm wrong though, aren't there g.p.e and k.e. changes taking place while you rotate a string attached to a ball vertically?
Since the ball moves away from gravity during the first half it should 'lose' velocity, and while it comes back down, it should 'gain' velocity

Original post by e^(i*pi) +

Reply 3

Stonebridge

It depends what the conditions are for the circular motion you are thinking of.
The most obvious would be friction. It's probably best to consider frictionless idealised motion at first.
In the case of vertical motion, then you also have to factor in the gravitational force on the object.
Just thinking in terms of conservation of energy, then the object would have to be travelling faster at the bottom of the motion than the top.
In which case, it would not have the same angular or linear velocity at those 2 places.
If it has the same radius then the centripetal force would also be different at those 2 places.
At the top, gravity would be providing some of that force. At the bottom gravity acts in the direction away from the centre so doesn't provide any centripetal force.
If you swing the object round then you are providing other forces on the object which complicate the motion.

Reply 4

Pangol

Original post by qwert7890

Hi! I had a few doubts in the differences between vertical and horizontal circular motion.

For horizontal circular motion (ex. a car around a race track) I understand that the velocity, centripetal force, centripetal acceleration and angular velocity all remain constant, provided the radius is constant.

How does it differ in a vertical circular motion (ex. me spinning a stone with a thread round vertically or the classic loop de loop situation). Well, I know the tension in the string would change but the centripetal force remains constant.

But in vertical circular motion, does the velocity and angular velocity remain constant?

As others have said, it very much depends on the situation. It is not quite as simple as you suggest.

Horizontal circular motion might happen with a changing angular velocity and with a changing centripetal force. Think of a car driving around a roundabout, but speeding up or slowing down.

Vertical circular motion might happen at a constant angular velocity and with a constant centripetal force. Think of a Ferris wheel.

At A Level, things are usually as you say, where you consider constant horizontal circular motion and non-constant vertical circular motion, but it doesn't have to be that way.

Reply 5

qwert7890

Original post by Pangol

Okay, but I'm slightly confused. Must the centripetal force always remain constant? Can it change?

In the case of the Ferris Wheel, when the ride is at the bottom, the centripetal force = Contact Force - Weight.
When the ride travels through 90 degrees anticlockwise, the centripetal force = Contact Force.
When the ride is at the top, the weight 'contributes' so the centripetal force = Weight + Contact Force

Clearly, the equation for centripetal force is changing. So in this case, is centripetal force constant? Of course, that is only possible if we assume the contact force is changing so as to keep the centripetal force constant?

Reply 6

Pangol

Original post by qwert7890

Think of the equations for centripetal force, either mv²/r or mrw². The only factors that affect this are the radius of the orbit (constant, or we wouldn't have circular motion), the mass of the object involved (constant again) and the linear or angular velocity (constant in this example). So yes, if you've got circular motion at a constant velocity, the centripetal force is constant.

In your example, the thing that changes is the contact force. When you're at the bottom of the circle, it is large, larger than your weight, which is why you feel quite heavy there. When you're at the top of the circle, it is small, which is why you feel lighter there, and can even leave your seat if the ride goes too fast.

Reply 7

qwert7890

Original post by Pangol

Okay, thank you very much! That clears up a bit of it.
While we're talking of circular motion,
does F vary inversely or directly with radius r? In the equation F = mv2/r , there seems to be an inverse relation but when you look at F = mrw2 it seems direct? This is a little confusing

Reply 8

Pangol

Original post by qwert7890

If you are talking about the relationship between F and r in those equations, you have to think about what it would mean for the other terms to stay constant.

If you are thinking about something turning at a constant rate - the turntable of a record player, for example - and you want to investigate how the centripetal force depends on how far you are away from the centre of the circular motion, then you need to consider F = mrw², which is better thought of as F = (mw²)r. So, in this case, F is proportional to r. The further away from the centre of the turntable you are, the larger the centripetal force. But note that although the turntable is turning at a constant rate, the speed of the object you are considering is not constant - it will be larger the further out it is, becasue v = rw.

If you are thinking about something moving at a constant speed - a car turning in a circle on a roundabout where it could be in the inner or outer lane, for example - and you want to investigate how the centripetal force depends on how far you are away from the centre of the circular motion, then you need to consider F = mv²/r, which is better thought of as F = (mv²)(1/r). So, in this case, F is proportional to 1/r. The further away from the centre of the motion you are, the smaller the centripetal force. But note that although the speed of the car is constant, the rate of rotation, the angular velocity, will be smaller when the radius of the motion is larger, becasue w = v/r.

The upshot is that you can't simply say that F is proprtional to r or 1/r, it depends if you are keeping the speed or the angular velocity constant.

(edited 4 years ago)

Reply 9

Eimmanuel

Study Forum Helper

Original post by Pangol

...
If you are thinking about something moving at a constant velocity - a car turning in a circle on a roundabout where it could be in the inner or outer lane, for example - and you want to investigate how the centripetal force depends on how far you are away from the centre of the circular motion, then you need to consider F = mv²/r, which is better thought of as F = (mv²)(1/r). So, in this case, F is proportional to 1/r. The further away from the centre of the motion you are, the smaller the centripetal force. But note that although the velocity of the car is constant, the rate of rotation, the angular velocity, will be smaller when the radius of the motion is larger, because w = v/r.
....

I doubt we can say an object undergoing any kind of circular motion can have a constant velocity. I believe you mean constant speed.