# Circular Motion - Newton's Second Law: Bead on a Rotating Hoop

• AzimD
In summary, the bead is rotating around an axis that the hoop is rotating around, and there is an inward force towards the center of the hoop.
AzimD
Homework Statement
A bead lies on a frictionless hoop of radius R that rotates around a vertical
diameter with constant angular speed ω, as shown in the figure below.

(a) What should ω be so that the bead maintains the same position on the hoop, at
an angle θ with respect to the vertical? Express you answer in terms of some or
all of the following: θ, R and g.
(b) Analyzing the answer for Part A, you will find that there is a range of angular
speeds, 0 < ω < ωo for which the fixed angle θ = 0 (meaning that the only
balanced position is at the bottom of the hoop). Find the value of ωo. Express
you answer in terms of some or all of the following: R and g.
Relevant Equations
a_r=-rω^2 (I'm not sure if it's relevant but I think it is.)

For whatever reason, I'm having a hard time conceptualizing this problem. I understand that the tangential components of all forces involved need to cancel out in order for the bead to be stationary. I also understand that there is a mgsinθ in the negative θ-hat direction. What I don't understand is to find the force opposite of it.

The bead is not stationary when ##\theta## is constant. Can you describe its motion?

What can you say about the net force for such motion?

MatinSAR
PeroK said:
The bead is not stationary when ##\theta## is constant. Can you describe its motion?

What can you say about the net force for such motion?
The bead is rotating around the axis that the hoop is rotating around, which means that there is inward Force toward the center of the hoop. I believe the Force would be characterized by F=ma_r=m(-rω^2) since there is no acceleration on ω thus there is no a_θ component.

If that is right, where I'm getting stuck is how do I use this inward Force to figure out the tangential component of this Force along the hoop itself to find the net zero?

AzimD said:
the tangential components of all forces involved need to cancel out
Only if there is no component of the bead's acceleration in the tangential direction.

MatinSAR
are you allowed to use the lagrangian approach for this problem?

Show us your free body diagram of the bead please.

MatinSAR
haruspex said:
Only if there is no component of the bead's acceleration in the tangential direction.
I don't believe there is an acceleration given that ω is constant as the Hoop rotates around the axis! Am I right about that part? Do I just need to find the Force that is opposite the direction of mgsinθ?

PhDeezNutz said:
are you allowed to use the lagrangian approach for this problem?
No, I am not!

PeroK
Chestermiller said:
Show us your free body diagram of the bead please.
I believe the Force inward would be characterized by F=ma_r=m(-rω^2).

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AzimD said:
I believe the Force inward would be characterized by F=ma_r=m(-rω^2).
The trajectory (for a bead which remains at a fixed angle ##\theta##) is circular and has radius r. So yes, the net force on the bead would be inward with magnitude ##m\omega^2 r##.

Edit: However, as @Steve4Physics makes clear, we should be wary -- what is ##r##?

But let us look at your free body diagram. You have the force of gravity downward. You've split this into components for some reason and drawn both the gravitational force vector and its two component vectors. You've also drawn a force labelled "F". What is that?

Presumably you mean that ##F## is the normal force of the wire on the bead. It would be nice if we knew the magnitude of ##F##. There is a way to get that.

(Sorry, @AzimD, looked at the text first and the drawing second)

Last edited:
MatinSAR
jbriggs444 said:
The trajectory (for a bead which remains at a fixed angle ##\theta##) is circular and has radius r. So yes, the net force on the bead would be inward with magnitude ##m\omega^2##.

But you've been asked for a free body diagram.
I provided it with the picture!

AzimD said:
The bead is rotating around the axis that the hoop is rotating around
Yes.

AzimD said:
which means that there is inward Force toward the center of the hoop.
No. It does not mean that!

The bead is moving in a horizontal circle.:
Is the centre of this circle the centre of the hoop?
What can you say about the direction of the net force on the bead?

By the way, you should state what force 'F' represents in your FBD.

AzimD
AzimD said:
I believe the Force inward would be characterized by F=ma_r=m(-rω^2).
If, by F in your equation, you mean the force F in your figure, then your equation is incorrect. What is the direction of the net force vector on the bead, (a) horizontal, (b) vertical, or (c) in some other direction? What is the vertical component of the normal force vector N of the wire on the bead (in terms of the magnitude N)? What is the horizontal component of the normal force vector N of the wire on the bead (in terms of the magnitude N)?

What is your parameter r expressed in terms of the parameter R in the original figure?

AzimD
AzimD said:
I don't believe there is an acceleration given that ω is constant as the Hoop rotates around the axis!
The bead is moving in a horizontal circle at constant speed. What is the direction of its acceleration?

MatinSAR and AzimD
haruspex said:
The bead is moving in a horizontal circle at constant speed. What is the direction of its acceleration?
Okay so after reading all of the replies, I think I understand it a bit better. The acceleration isn't directed towards the center of the hoop but rather the axis itself, since that's what the bead is rotating about.

I think I did it correctly for the most part, except I know that my signage is wrong. There should not be a negative in the result of ω. That being said, is the acceleration not directed inward?

I've attached my work to this!

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AzimD said:
Okay so after reading all of the replies, I think I understand it a bit better. The acceleration isn't directed towards the center of the hoop but rather the axis itself, since that's what the bead is rotating about.

I think I did it correctly for the most part, except I know that my signage is wrong. There should not be a negative in the result of ω. That being said, is the acceleration not directed inward?

I've attached my work to this!
Your FBD is missing a force.

erobz
You haven’t answered my questions in post #13

AzimD
Chestermiller said:
If, by F in your equation, you mean the force F in your figure, then your equation is incorrect. What is the direction of the net force vector on the bead, (a) horizontal, (b) vertical, or (c) in some other direction? What is the vertical component of the normal force vector N of the wire on the bead (in terms of the magnitude N)? What is the horizontal component of the normal force vector N of the wire on the bead (in terms of the magnitude N)?

What is your parameter r expressed in terms of the parameter R in the original figure?
Since this question, I realized the force in my FBD towards the center of the hoop was incorrect and rather it should be into the axis. The Net Force Vector would be in some other direction correct? Particularly in the direction tangential to the hoop itself?

I'm not entirely sure about the Normal Force in this question. Does it matter though given that it's dealing with ω and I really just need to find the Force Component acting opposite of mgsinθ to figure out ω?

haruspex said:
Your FBD is missing a force.
Is the missing force the Normal force? @Chestermiller made a mention of this force, I don't see how to fit it in, also I don't see how it's relevant.
If so, would the Normal Force be directed toward the center of the Hoop as it rotates?

Edit: Yes the r is R, I fixed that in the next FBD.

The net force is horizontal, perpendicular to the axis. The normal force is directed toward the center of the hoop .

AzimD said:
Is the missing force the Normal force? @Chestermiller made a mention of this force, I don't see how to fit it in, also I don't see how it's relevant.
Since you are only concerned with the tangential force balance, it is not relevant, but you really should show it.
The mistake, though, is that you have treated the centripetal force as an applied force balancing the tangential component of gravity. That has led to a spurious minus sign. Centripetal force is that component of the resultant force which acts perpendicularly to the velocity.

haruspex said:
Since you are only concerned with the tangential force balance, it is not relevant, but you really should show it.
I disagree. In my judgment, the normal force is very relevant.

The vertical component of the normal force is ##N\cos{\theta}## and the horizontal component is ##N\sin{\theta}##. So the force balances on the bead in the vertical and horizontal directions read:$$N\cos{\theta}=mg$$and $$N\sin{\theta}=m\omega^2r=m\omega^2(R\sin{\theta})$$What do you get if you eliminate N between these two equations?

The realization that the net force on the bead is horizontal provides a very compelling reason to express the force balance in the horizontal and vertical directions, rather than the normal and tangential directions. The resolution of the gravitational force into the tangential and normal directions as the OP did is just a waste of time and effort in my judgment.

SammyS
Chestermiller said:
the normal force is very relevant.
It has no component tangential to the hoop. If the force balance equation in that direction is all that is needed to solve the problem then it can be made irrelevant.

haruspex said:
It has no component tangential to the hoop. If the force balance equation in that direction is all that is needed to solve the problem then it can be made irrelevant.
I finally see what you are saying. Dotting the vector force balance by a unit vector in the tangential direction eliminates the contribution of the normal force and immediately gives you the desired relationship.

haruspex said:
The mistake, though, is that you have treated the centripetal force as an applied force balancing the tangential component of gravity. That has led to a spurious minus sign. Centripetal force is that component of the resultant force which acts perpendicularly to the velocity.
Could you explain this further? Should I only be considering the magnitude of the force? As I mentioned, I was always under the impression that the Centripetal force is an inward force.

AzimD said:
I was always under the impression that the Centripetal force is an inward force.
It is inward, but something has to supply it.
If you whirl a mass around on the end of a string then the only forces acting on it are gravity and the tension in the string. The resultant of those is called the centripetal force. It is not an additional force.
More generally, if a body has velocity ##\vec v## and acceleration ##\vec a##, its centripetal acceleration is defined to be the component of ##\vec a## normal to ##\vec v##. The net force on the body is ##m\vec a##, and the component of that normal to ##\vec v## is defined to be the centripetal force.

However, if you use the reference frame of the body then there is, by definition, no net acceleration. To balance the forces experienced, one has to introduce the "fictional" centrifugal force. This is an applied force, not a resultant.

In terms of unit vectors, the force balance reads: $$N(\cos{\theta}\mathbf{i_z}-\sin{\theta}\mathbf{i_r})-mg\mathbf{i_z}=-m\omega^2R\sin{\theta}\mathbf{i_r}$$

The unit tangent vector to the ring is $$\boldsymbol{\tau}=\cos{\theta}\mathbf{i_r}+\sin{\theta}\mathbf{i_z}$$
What do you get for the dot-product of the unit tangent with the force balance?

haruspex and PhDeezNutz
haruspex said:
However, if you use the reference frame of the body then there is, by definition, no net acceleration. To balance the forces experienced, one has to introduce the "fictional" centrifugal force. This is an applied force, not a resultant.
Applied by what? There is no interaction responsible for the fictional force. Or you mean to say that is NOT an applied force? Is there a specific meaning of "applied force" that you have in mind?

nasu said:
Applied by what? There is no interaction responsible for the fictional force. Or you mean to say that is NOT an applied force? Is there a specific meaning of "applied force" that you have in mind?
I believe that the distinction is between an identifiable individual force on an object and the resultant (vector sum) of those individual forces. One might also count "component forces" (the projection of a force along a particular direction) as being distinct from an "applied force".

In that sense, inertial forces such as gravity, centrifugal force, coriolis force and the Euler force count as "applied" forces, even though they appear or disappear depending on a choice of coordinates rather than arising from an interaction with another object.

So the distinction being made is between "applied forces" and "resultant forces" rather than between "interaction forces" and "fictitious forces".

haruspex
How do you know that a given force is applied or resultant? If not looking at where does it come from, single interaction or multiple interactions? How is this applied to inertial forces (zero interactions)?

nasu said:
How do you know that a given force is applied or resultant? If not looking at where does it come from, single interaction or multiple interactions? How is this applied to inertial forces (zero interactions)?
It is a resultant if it is just something you got by adding up a bunch of other forces.

It is an applied force if it is either an interaction force or a fictitious force resulting from your choice of non-inertial coordinates.

If there is only one force then the distinction becomes pretty much irrelevant. For instance a satellite orbitting a primary under the force of gravity. The resultant "centripetal force" and the applied "gravitational force" are pretty much one and the same thing.

How do you know that the forces you add are not resultants too? Is the weight of a macroscopic body applied or resultant? You can think of it as the force between the body and the Earth or as the resultant of all individual attractions between the molecules in the two objects or other parts you can divide the two objects into. What about friction? Which kind is it? Is the contact force between two solids applied or resultant? I don't see how this concept can be useful or/and well defined.
And for the inertial forces, what if you have several components, like Coriolis term, centrifugal term, for example. Is the total inertial force applied or resultant? What about the centrifugal term when the Coriolis term is not zero? It is a component or is applied?
Or you mean that the same force can be applied or resultant, depending on the context? Then, yes it is clear but I just don't see much usefulness. Have you seen this distinction made in textbooks?

nasu said:
Or you mean that the same force can be applied or resultant, depending on the context? Then, yes it is clear
Thank you. Then we are in agreement.
nasu said:
but I just don't see much usefulness.
It was useful in this thread. If you write down both individual forces and the resultant which is their sum on the same free body diagram then you risk double-dipping.
nasu said:
Have you seen this distinction made in textbooks?
I've seen the term "resultant" used, yes. I'd not paid enough attention to worry about digging up a word to refer to forces that are added together to produce resultants. "Applied" is not an unreasonable adjective for the notion.

nasu said:
How do you know that a given force is applied or resultant? If not looking at where does it come from, single interaction or multiple interactions? How is this applied to inertial forces (zero interactions)?
As reckoned by an observer in an accelerating frame of reference, all masses behave as if they are acted upon by a (fictitious) body force equal to -ma.

nasu said:
How do you know that the forces you add are not resultants too?
They often are, but when you draw an FBD you have to decide which are the forces acting on it. E.g. for gravity, it is usual to consider it a single force acting on the mass centre, though in reality it acts on each atom, or gluon perhaps, separately. Having determined these "applied" forces, a summation of some of them can be considered a resultant, and the resultant of all is the net force.
The error in post #1, and I have witnessed this several times, is to think that the centripetal force is an applied force and write ##(\Sigma \{F_r\})+F_{centripetal}=0##, where the ##\{F_r\}## are the radial components of the (actual) applied force; the correct equation, of course, being ##\Sigma \{F_r\}=F_{centripetal}##.
In the non inertial frame, we have instead ##(\Sigma \{F_r\})+F_{centrifugal}=0##, so the centrifugal force behaves like an applied force.

jbriggs444 and PeroK

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