I Why can centripetal force balance out gravity?

[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?

 

[lgmulti]

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?

 

[PeroK]

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?

 

[berkeman]

Welcome to PF.

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

 

[berkeman]

Diagram should look like this:

Looks upside-down?

 

[lgmulti]

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

 

[DaveC426913]

... 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?

 

[berkeman]

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?

 

[lgmulti]

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?

 

[berkeman]

What would the speed of the mass have to be in order for $\theta \rightarrow 0$?

 

[PeroK]

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.

 

[lgmulti]

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?

 

[lgmulti]

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?

 

[PeroK]

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.

 

[berkeman]

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.

More info on the equations for circular motion: http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html

 

[mpresic3]

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.

 

[Frabjous]

While it is beyond the scope of your question, if the object at the end of the rope generated lift, you could do it horizontally. https://en.wikipedia.org/wiki/Lift_(force)

 

[lgmulti]

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.

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

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

 

[pbuk]

While it is beyond the scope of your question, if the object at the end of the rope generated lift, you could do it horizontally. https://en.wikipedia.org/wiki/Lift_(force)

I think there is a mentor here who is/was a fan of RTP flying.

 

[mpresic3]

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.

 

[A.T.]

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.

 

[lgmulti]

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]

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
$$T \cos \theta = F_c = \frac{mv^2}{l\cos\theta}$$

 

[sophiecentaur]

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!!