# Why can centripetal force balance out gravity?

• I
• lgmulti
In summary: If you observe an object moving in a circle, then the combined...If you observe an object moving in a circle, then the combined centripetal and gravitational force is the resultant (combined) force of all the other applied forces.
lgmulti
TL;DR Summary
When an object is doing uniform circular motion in a plane parallel to the ground, the centripetal force is parallel to the ground while the gravity is vertical to the ground, so the combination of the two force must be downwards, which means that it's impossible to maintain the motion, but why it's possible sometimes?
I can't upload image, so the problem is the same as the summary: When an object is doing uniform circular motion in a plane parallel to the ground, the centripetal force is parallel to the ground while the gravity is vertical to the ground, so the combination of the two force must be downwards, which means that it's impossible to maintain the motion, but why it's possible sometimes?

Here is the figure for my question. The object is doing uniform circular motion around the gray circle that is in a plane parallel to the ground, the pink force F_combined is the combination of the two red force G (gravity) and F_c (centripetal force), so the pink force will always point downwards, meaning that the object will always move down, but why can the uniform circular motion sometimes be maintained?

lgmulti said:
TL;DR Summary: When an object is doing uniform circular motion in a plane parallel to the ground, the centripetal force is parallel to the ground while the gravity is vertical to the ground, so the combination of the two force must be downwards, which means that it's impossible to maintain the motion, but why it's possible sometimes?

I can't upload image, so the problem is the same as the summary: When an object is doing uniform circular motion in a plane parallel to the ground, the centripetal force is parallel to the ground while the gravity is vertical to the ground, so the combination of the two force must be downwards, which means that it's impossible to maintain the motion, but why it's possible sometimes?
Do you mean why is it possible to run round in circles?

russ_watters
Welcome to PF.

lgmulti said:
Here is the figure for my question. The object is doing uniform circular motion around the gray circle that is in a plane parallel to the ground, the pink force F_combined is the combination of the two red force G (gravity) and F_c (centripetal force), so the pink force will always point downwards, meaning that the object will always move down, but why can the uniform circular motion sometimes be maintained?
You should redraw your figure to reflect reality. If the mass is undergoing uniform circular motion while experiencing the downward force from gravity, the point at the center of the circle will be above the plane of rotation. So the force ##F_c## will be angled up at some angle that depends on the velocity of the rotating mass on the end of the string.

Much like these rides at amusement parks:

https://www.alamy.com/stock-photo-amusement-park-marry-go-round-ride-52775486.html

PeroK
DaveC426913 said:
Diagram should look like this:
Looks upside-down?

berkeman said:
Looks upside-down?
So does it look like this when the point at the center of the circle is above the plane?

If so, what will happen if the point at the center of the circle is on the plane

lgmulti said:
... what will happen if the point at the center of the circle is on the plane
What do you think will happen?

Think about swinging a bucket on a piece of string around yourself. What must you do to keep it in the air?

lgmulti said:
So does it look like this when the point at the center of the circle is above the plane?
Pretty close, you just need the resultant force ##F_{combined}## to point toward the center of the circle (in the horizontal plane). Otherwise, there will be an acceleration of the mass up or down. Your goal was uniform circular motion in the horizontal plane, right?

Oh, so the object can keep the uniform circular motion as long as
$$F_c \sin \theta = mg.$$
But it seems impossible to happen when ##\theta = 0## because ##\sin 0 = 0##. Is it right?

Lnewqban and berkeman
What would the speed of the mass have to be in order for ##\theta \rightarrow 0##?

lgmulti said:
View attachment 328058
Oh, so the object can keep the uniform circular motion as long as
$$F_c \sin \theta = mg.$$
But it seems impossible to happen when ##\theta = 0## because ##\sin 0 = 0##. Is it right?
For circular motion, the centripetal force is the resultant (combined) force of all the other applied forces. It's not a separate, applied force. To achieve a resultant centripetal force in your case, there must be an applied force with an upward component to cancel out gravity.

PeroK said:
For circular motion, the centripetal force is the resultant (combined) force of all the other applied forces. It's not a separate, applied force. To achieve a resultant centripetal force in your case, there must be an applied force with an upward component to cancel out gravity.
Oh, so is ##F_\rm{combined}## in my figure actually the centripetal force?

berkeman said:
What would the speed of the mass have to be in order for ##\theta \rightarrow 0##?
I don't know how ##\theta## affects the speed of the mass. Is it infinity?

Lnewqban and berkeman
lgmulti said:
Oh, so is ##F_\rm{combined}## in my figure actually the centripetal force?
If you observe an object moving in a circle, then the combined force must be centripetal (towards the centre). So, yes, in your diagram the combined force is the centripetal force. You then need an additional applied force upwards at some angle ##\theta## which has an upward component to balance gravity and a horizontal component that must be the resultant centripetal force.

lgmulti said:
Is it infinity?
Yes, it would take higher and higher speeds to get smaller values of ##\theta##. But as @PeroK says, as a practical matter, the force from the string will always need a vertical component to cancel out the downward force of gravity on the mass.

Lnewqban
Sounds like you have solved this problem, But I have some illustrative comments. In the diagram, F combined is the centripetal force as you have already noted. Shankar in his video on youtube mentions students get into trouble with free body diagrams (FBD) by including forces that they should not. Depending on your instructor, if (s)he wants you to include a FBD, you might have to be careful. (If the student got everything else correct by legitimate means, I generally never took off for a including a force that was supplied from the combination of gravity, electromagnetic, or tension forces, etc., but others might.

What I would include as forces on the mass are gravity. Clearly gravity acts on the mass. The other force to include is tension T from the string that is holding the mass so that it travels in a circle. These two forces together supply the centripetal force. (I never know whether to include it in the FBD. Perhaps including it by a dotted line is appropriate).

The force along the supporting string is the tension. In your analysis of the problem, you can extend the problem and now find the tension in the supporting string. You might think, why "extend the problem", when I am done with it? Well, the instructor might ask for the tension for this example in a future problem on the exam. If (s)he does, then you will be ready for it.

pbuk
PeroK said:
You then need an additional applied force upwards at some angle θ which has an upward component to balance gravity and a horizontal component that must be the resultant centripetal force.
berkeman said:
as a practical matter, the force from the string will always need a vertical component to cancel out the downward force of gravity on the mass
mpresic3 said:
The other force to include is tension T from the string that is holding the mass so that it travels in a circle. These two forces together supply the centripetal force.
So the object always need an upward force to maintain the circular motion, and it's usually not drawn on the FBD. In my case it's the tension and the gravity together that creates centripetal force. I got it. Thanks for everyone

lgmulti said:
So the object always need an upward force to maintain the circular motion, and it's usually not drawn on the FBD. In my case it's the tension and the gravity together that creates centripetal force. I got it. Thanks for everyone

No, this is still a misunderstanding. The example shown demonstrates the horizontal component of the tension is the component of the force that maintains the circular motion, this is not the upward force. In addition, I am not sure whether you want to include centripetal force on a FBD. You may want to ask your instructor (if you are not self-learning).

If for example you did include the centripetal force in the diagram to support your equations, and you derived for the horizontal component of the three forces:

For horizontal components.

F gravity = mg cos 90 degrees = 0
Tension cosine theta = horizontal component of the tension
F centripetal = m v squared / r = the horizontal component of the centripetal force

and summed the forces to get F gravity ( = 0 ) +T cos(theta) + m v squared / r = mass * centripetal acceleration = m v squared / r.

In this situation, you would be "counting" the centripetal force twice, and get tension T = 0. This is wrong.

I think it's OK to include centripetal force in the FBD, but you have to be careful in writing the equations, not to double-count the forces.

berkeman
lgmulti said:
In my case it's the tension and the gravity together that creates centripetal force.
In your case gravity is perpendicular to the circle, so it doesn't contribute to the centripetal force. The tension force provides the full centripetal force and also balances gravity.

Lnewqban
A.T. said:
In your case gravity is perpendicular to the circle, so it doesn't contribute to the centripetal force. The tension force provides the full centripetal force and also balances gravity.
Oh, so the force ##T## must also satisfy both
$$T \cos \theta = F_c = \frac{mv^2}{l\sin\theta}$$
and
$$T \sin \theta = mg.$$

At this time if I want to express ##v## in terms of other variables, I can rewrite the first equation into
$$Tl\cos\theta\sin\theta = mv^2$$
and plug the second equation in to get
$$(T\sin\theta) l\cos\theta = mv^2 \implies mgl\cos\theta = mv^2.$$
Therefore,
$$v = \sqrt{gl\cos\theta},$$
so for a fixed ##l## and ##\theta##, the velocity must be the same. Is that right?

lgmulti said:
Oh, so the force ##T## must also satisfy both
$$T \cos \theta = F_c = \frac{mv^2}{l\sin\theta}$$
and
$$T \sin \theta = mg.$$

At this time if I want to express ##v## in terms of other variables, I can rewrite the first equation into
$$Tl\cos\theta\sin\theta = mv^2$$
and plug the second equation in to get
$$(T\sin\theta) l\cos\theta = mv^2 \implies mgl\cos\theta = mv^2.$$
Therefore,
$$v = \sqrt{gl\cos\theta},$$
so for a fixed ##l## and ##\theta##, the velocity must be the same. Is that right?

I made a mistake. It should be
$$T \cos \theta = F_c = \frac{mv^2}{l\cos\theta}$$

PeroK
lgmulti said:
If so, what will happen if the point at the center of the circle is on the plane
Another force is needed from somewhere to maintain height. You will slip down a rotating vertical cylinder if it is smooth. However, the Fairground ride (The ROTOR) keeps you from slipping down if the walls are slightly sticky / rough. All that's needed is for the rotational velocity to be high enough to produce a sufficient friction force to counteract your weight. I went on one of these in the 50's at Battersea Funfair. (brilliant) Twenty years later I was very nearly sick!!

## Related to Why can centripetal force balance out gravity?

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## Why does centripetal force balance out gravity in orbital motion?

Centripetal force balances out gravity in orbital motion because it provides the necessary inward force to keep an object moving in a circular path. Gravity acts as the centripetal force, pulling the object towards the center of the orbit, while the object's inertia keeps it moving forward, creating a stable orbit.

## What is the relationship between centripetal force and gravitational force?

The relationship between centripetal force and gravitational force in orbital motion is that gravitational force acts as the centripetal force. For an object in orbit, the gravitational pull of the larger body (e.g., Earth) provides the necessary centripetal force to keep the object in its curved path.

## Can centripetal force and gravity be equal in magnitude?

Yes, centripetal force and gravity can be equal in magnitude for an object in stable orbit. When an object is in a stable orbit, the gravitational force pulling it towards the center is exactly balanced by the centripetal force needed to keep it moving in its circular path. This balance allows the object to maintain a consistent orbit.

## How does an object maintain a stable orbit with centripetal force and gravity?

An object maintains a stable orbit when the centripetal force, provided by gravity, is exactly equal to the force required to keep the object moving in its circular path. This balance prevents the object from flying off into space or falling into the larger body it is orbiting. The object's velocity and the gravitational pull must be precisely aligned to maintain this balance.

## Why doesn't an object in orbit fall to the ground if gravity is acting on it?

An object in orbit doesn't fall to the ground because its forward velocity is sufficient to counteract the pull of gravity. As the object moves forward, gravity continuously pulls it towards the center of the orbit, creating a curved path. This balance between forward motion and gravitational pull keeps the object in a stable orbit rather than falling straight down.



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