Rotating ring on a rough surface- but with a twist

AI Thread Summary
The discussion revolves around the mechanics of a ring with two fixed masses rolling on a rough surface. The original poster is confused about the direction of friction and how to calculate the angular acceleration, frictional force, and normal reaction. Participants suggest using the parallel axis theorem and drawing free body diagrams (FBD) to analyze the forces and torques acting on the system. The conversation emphasizes the importance of identifying the instantaneous axis of rotation and considering the entire assembly's moment of inertia to solve the problem effectively. Overall, the thread highlights the complexities of rotational dynamics in systems with varying mass distributions.
  • #51
kuruman said:
About what point? About point O it's zero.

Right. In what direction does it point?
About center of mass
Static friction points forward? Yes? No? I think so because the center of mass is probably accelerating forward.. i cannot think about how the center of mass is moving here.
 
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  • #52
palaphys said:
Static friction points forward? Yes? No?
Are you asking me or are you telling me? Reason it out.
palaphys said:
i cannot think about how the center of mass is moving here
Have you seen bicycles with reflectors on the spokes of their wheels? Imagine a reflector at distance ##\frac{1}{5}R## from the axis of the wheel. The path that it describes as the bicycle moves forward is a cycloid. That's the path of the CM here.
 
  • #53
May I throw-in a couple of questions for @palaphys ?

a) You are in your new Porsche , stationary, on level ground. . The ground is super slippery – negligible friction between the ground and the wheels. What happens when you try to accelerate?

b) What happens if there is friction (e.g. if the ground is dry concrete)?

The centre of mass of an object will only accelerate when a net force acts on the object. What can you deduce about the type and direction of the force which makes your Porsche accelerate?
 
  • #54
palaphys said:
About center of mass
Static friction points forward? Yes? No? I think so because the center of mass is probably accelerating forward.. i cannot think about how the center of mass is moving here.
Please, stop and think about the excellent reasoning led by Kuruman. :smile:

You always have a pair of forces, action and reaction.
When you ask about friction force and its direction, it is important to be clear about what body the force of interest is acting upon.

You have the force with which the ring pushes on the rough surface, and you have the force with which the rough surface pushes on the ring.
Since at that always changing point of contact neither ring nor surface moves respect to the other, something else yields and moves forward, which is the geometrical center of the ring.

That forward movement forces the points of contact to separate, while new points of contact will come to establish a new pivot of zero velocity and no relative movement.

Please, see:
https://www.astro.ucla.edu/~malkan/astro8/physics1a/rolling.htm


roll2.gif



roll1.gif


rollmot.gif
 
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  • #55
To @palaphys :
You have not provided so far the final answer for the angular acceleration in part (a). Please do before worrying about the force of friction. That's because your plan of attack should be to find the magnitude of the linear acceleration of the CM and from it the horizontal component of the acceleration of the CM.

I have to sign off for a while now.
 
  • #56
kuruman said:
Are you asking me or are you telling me? Reason it out.

Have you seen bicycles with reflectors on the spokes of their wheels? Imagine a reflector at distance ##\frac{1}{5}R## from the axis of the wheel. The path that it describes as the bicycle moves forward is a cycloid. That's the path of the CM here.
Ok. I assume cycloid is some kind of curvilinear motion. Now in GENERAL rolling cases, where the center of mass is the geometric center, it just goes in a straight line right?. So I feel that the center of mass here may have two accelerations, namely linear and tangential. Am I right?
 
  • #57
Steve4Physics said:
May I throw-in a couple of questions for @palaphys ?

a) You are in your new Porsche , stationary, on level ground. . The ground is super slippery – negligible friction between the ground and the wheels. What happens when you try to accelerate?

b) What happens if there is friction (e.g. if the ground is dry concrete)?

The centre of mass of an object will only accelerate when a net force acts on the object. What can you deduce about the type and direction of the force which makes your Porsche accelerate?
a)the weels will slide and not sure if I would move
B) pure rolling would take place and I would go forward
Oh so you are saying that here as center of mass moves forward, it is due to friction?
 
  • #58
palaphys said:
Oh so you are saying that here as center of mass moves forward, it is due to friction?
If you would draw a FBD you should be able to see that if the wheel is accelerating to the right, there is some force acting to the right. What force can act to the right? Which way is normal force. Which, direction are the weight forces, is anyone pushing the wheel etc... are questions you should be asking yourself. At the very least by process of elimination you should see that there must be a horizontal force, and where it must act.
 
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  • #59
palaphys said:
... as center of mass moves forward, it is due to friction?
... and to the imbalance of masses around the ring: that is the driving torque.

If you take your point of view as the center of the ring, you will see the ring pushing the surface rearward.
In order to do that, the ring needs "to be able to grab" onto the surface.
 
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  • #60
erobz said:
If you would draw a FBD you should be able to see that if the wheel is accelerating to the right, there is some force acting to the right. What force can act to the right? Which way is normal force. Which, direction are the weight forces, is anyone pushing the wheel etc... are questions you should be asking yourself. At the very least by process of elimination you should see that there must be a horizontal force, and where it must act.
But will friction not cause angular retardation if it acts towards the right? Then how will the no slip condition be maintained?
 
  • #61
Lnewqban said:
... and to the imbalance of masses around the ring: that is the driving torque.

If you take your point of view as the center of the ring, you will see the ring pushing the surface rearward.
In order to do that, the ring needs "to be able to grab" onto the surface.
I have a question: if I look at ANY point on the ring relative to some other point, will it be in circular motion? Or is this true only relative to the geometric center of the ring and IAOR?
 
  • #62
palaphys said:
. . . the center of mass is the geometric center, it just goes in a straight line right?.
Wrong. The center of the ring follows a straight line because it is always at distance ##R## above ground. The distance of the CM above ground varies from ##R+\frac{1}{5}R## at its highest to ##R-\frac{1}{5}R## at its lowest.
 
  • #63
kuruman said:
Wrong. The center of the ring follows a straight line because it is always at distance ##R## above ground. The distance of the CM above ground varies from ##R+\frac{1}{5}## at its highest to ##R-\frac{1}{5}## at its lowest.
I meant in a general case, not this case. Here the center of mass follows a circular path relative to the geometric center?
 
  • #64
erobz said:
If you would draw a FBD you should be able to see that if the wheel is accelerating to the right, there is some force acting to the right. What force can act to the right? Which way is normal force. Which, direction are the weight forces, is anyone pushing the wheel etc... are questions you should be asking yourself. At the very least by process of elimination you should see that there must be a horizontal force, and where it must act.
Only the center of the wheel is accelerating right. Correct? The center of mass of the wheel has a tangential accn, centripetal acceleration, and the acceleration of the center of the ring. Please correct me if wrong
 

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  • #65
palaphys said:
But will friction not cause angular retardation if it acts towards the right? Then how will the no slip condition be maintained?
Is there a moment arm of ##f## about the instantaneous center of rotation?
 
  • #66

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  • #67
erobz said:
Is there a moment arm of ##f## about the instantaneous center of rotation?
No, but there is a moment arm about the center of the ring right.. the IAOR is always at rest momentarily?
 
  • #68
palaphys said:
Only the center of the wheel is accelerating right. Correct? The center of mass of the wheel has a tangential accn, centripetal acceleration, and the acceleration of the center of the ring. Please correct me if wrong
Mark the center of mass on a diagram. Next let system it roll without slipping to the point where the ##2m## mass is directly below the center of the ring and the ##m## mass directly above it. Can you conclude whether the center of mass has changed position?
 
  • #69
erobz said:
Mark the center of mass on a diagram. Next let system it roll without slipping to the point where the ##2m## mass is directly below the center of the ring and the ##m## mass directly above it. Can you conclude whether the center of mass has changed position?
Yes I can. It is going in circular motion right.. with radius r÷5 if u see from center of ring
 
  • #70
palaphys said:
Yes I can. It is going in circular motion right.. with radius r÷5 if u see from center of ring
Well how did it change position if it didn't accelerate?
 
  • #71
erobz said:
Well how did it change position if it didn't accelerate?
It has tangential and centripetal accl? Combined with accl of geometric center like this maybe?
 

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  • #72
palaphys said:
Only the center of the wheel is accelerating right. Correct?
You said this. It makes me believe that you believe the center of mass is not accelerating to the right initially as well?
 
  • #73
erobz said:
You said this. It makes me believe that you believe the center of mass is not accelerating to the right initially as well?
I wanted to say that the geometric center of the wheel is just accelerating in a straight line sorry for confusion
 
  • #74
erobz said:
You said this. It makes me believe that you believe the center of mass is not accelerating to the right initially as well?
Please tell me how the tendency to slip is towards the left.. I still can't believe it.. please
 
  • #75
palaphys said:
I wanted to say that the geometric center of the wheel is just accelerating in a straight line sorry for confusion
Well, I'm trying to get you to see that if the center of mass (not explicitly the center of the wheel) there must be a force acting on the body ( via Newton's Second) that accelerates the center of mass as it moves in this case.
 
  • #76
erobz said:
Well, I'm trying to get you to see that if the center of mass (not explicitly the center of the wheel) there must be a force acting on the body ( via Newton's Second) that accelerates the center of mass as it moves in this case.
I see that .. just can't predict what direction friction is going to be. It seems intuitive that it would be towards right, but I want a physics proof why it is so.
 
  • #77
palaphys said:
I see that .. just can't predict what direction friction is going to be. It seems intuitive that it would be towards right, but I want a physics proof why it is so.
In this case the wheel is initially pushing on the ground to the left, and the ground is pushing the wheel to the right...via friction. This thing is just a pendulum whose pivot moves back and forth. The wheel will have some oscillatory motion, first it goes to the right then it turns back and heads to the left. friction flips with acceleration. Its observer to be "accelerationally" dependent. If you assume it, and correctly carry out the mathematics, the math will tell you. Here its observationally obvious, in other problems it might be challenging to get it out of the gate.
 
  • #78
palaphys said:
I have a question: if I look at ANY point on the ring relative to some other point, will it be in circular motion? Or is this true only relative to the geometric center of the ring and IAOR?
All points of the ring instantaneously rotate about the instantaneous ring-surface point of contact, which is constantly disappearing while another contact point, located just a little further, is constantly assuming the function of a new instantaneous pivot, around which all the points of the ring rotate again, and so on.

The succession of those points of contact moves horizontally at the same velocity of the geometrical center of the rolling ring, keeping always apart from each other the same vertical distance, which equals the radius of the ring.
 
  • #79
I think I've figured out the normal reaction part. Please check my work someone.
 

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  • #80
Lnewqban said:
The succession of those points of contact moves horizontally at the same velocity of the geometrical center of the rolling ring, keeping always apart from each other the same vertical distance, which equals the radius of the ring.
I don't understand this part please make it simple
 
  • #81
erobz said:
In this case the wheel is initially pushing on the ground to the left, and the ground is pushing the wheel to the right...via friction. This thing is just a pendulum whose pivot moves back and forth. The wheel will have some oscillatory motion, first it goes to the right then it turns back and heads to the left. friction flips
I am extremely confused with friction in this case. What i was taught was that friction opposes relative tendency of motion. Here all of you are saying something else. Generally how I found direction of friction in rolling type of questions was by assuming none at all, and observing tendency of slipping. Please explain in simpler way pls
 
  • #82
Could someone please verify this
 

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  • #83
palaphys said:
I am extremely confused with friction in this case. What i was taught was that friction opposes relative tendency of motion. Here all of you are saying something else. Generally how I found direction of friction in rolling type of questions was by assuming none at all, and observing tendency of slipping. Please explain in simpler way pls
You are confusing kinetic friction with static friction. Kinetic friction is molecules sliding past each other, breaking structure permanently - converting mechanical energy to internal energy, and static friction is armwrestling with a wall...you might be warping it a bit, but it's springing back. Thats as far as I can see in plain terms.
 
  • #84
palaphys said:
@kuruman hope this is right
It is correct.
 
  • #85
erobz said:
You are confusing kinetic friction with static friction. Kinetic friction is molecules sliding past each other, breaking structure, and static friction is armwrestling with a wall...you might be warping it a bit, but it's springing back.
That's not my confusion. I know that it's static friction here. The direction of friction is spoiling my day. I'm not able to figure it out.
 
  • #86
kuruman said:
It is correct.
How about this
 

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  • #87
palaphys said:
I see that .. just can't predict what direction friction is going to be. It seems intuitive that it would be towards right, but I want a physics proof why it is so.
You don't need a proof, just logic. For the CM to accelerate to the right there must be a horizontal force acting on it. How many forces are there that act in the horizontal direction on any of the masses here?
 
  • #88
kuruman said:
You don't need a proof, just logic. For the CM to accelerate to the right there must be a horizontal force acting on it. How many forces are there that act in the horizontal direction on any of the masses here?
How is the CM accelerating to the right? You said it was a cycloid path right? Also check the screenshot I posted.how can I conclude that it is horizontal?
 
  • #89
palaphys said:
That's not my confusion. I know that it's static friction here. The direction of friction is spoiling my day. I'm not able to figure it out.
But you are confused about it. You said you were taught friction always opposes motion. It doesn't, I'm trying to clarify that concept for you. There are two distinct names, two concepts to grapple with. One of them (static friction) is a fickle mistress.
 
  • #90
erobz said:
But you are confused about it. You said you were taught friction always opposes motion. It doesn't, I'm trying to clarify that concept for you. There are two distinct names, two concepts to grapple with.
friction does not oppose relative motion?
 
  • #91
palaphys said:
friction does not oppose relative motion?
Static friction is a fickle mistress. Here its opposing that an atom on the wheel is not passing through an atom on the road.
 
  • #92
erobz said:
Static friction is a fickle mistress. Here its opposing that an atom on the wheel is not passing through an atom on the road.
Ok but how do u find the frictional force here. I thought about it for a while and now I accept that friction is towards the right. Pls tell me how to find it.I will try 5ma_cm,x =f_s
 
  • #93
palaphys said:
Ok but how do u find the frictional force here. I thought about it for a while and now I accept that friction is towards the right. Pls tell me how to find it.I will try 5ma_cm,x =f_s
Its 12 am here (I'm probably already going to have trouble relaxing now). I'm signing off. You'll get there, it takes time.
 
  • #94
palaphys said:
How is the CM accelerating to the right? You said it was a cycloid path right? Also check the screenshot I posted.how can I conclude that it is horizontal?
The acceleration of the CM has both horizontal and vertical components. The vertical component of the acceleration is provided by the vertical component of the net force. The horizontal component is provided by the horizontal component of the net force. If the CM has zero horizontal component it will have to fly off the surface or sink into it. Is that what's happening here?

As for the diagram that you drew, you assume that the CM is rotating about the center of the ring. No, no, no. We have already established that it rotates about the point of contact O and calculated the torque and moment of inertia about that point. When you write the tangential acceleration ##a_T=\alpha r##, what does ##r## represent?
 
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  • #95
palaphys said:
Ok but how do u find the frictional force here. I thought about it for a while and now I accept that friction is towards the right. Pls tell me how to find it.I will try 5ma_cm,x =f_s
You need to find the horizontal component of the linear acceleration of the CM.
 
  • #96
kuruman said:
The acceleration of the CM has both horizontal and vertical components. The vertical component of the acceleration is provided by the vertical component of the net force. The horizontal component is provided by the horizontal component of the net force. If the CM has zero horizontal component it will have to fly off the surface or sink into it. Is that what's happening here?

As for the diagram that you drew, you assume that the CM is rotating about the center of the ring. No, no, no. We have already established that it rotates about the point of contact O and calculate the torque and moment of inertia about that point. When you write the tangential acceleration ##a_T=\alpha r##, what does ##r## represent?
r represents R/5, and I was taught that in a rigid body in rotation every point goes in a circular motion relative to some other point.
 
  • #97
kuruman said:
You need to find the horizontal component of the linear acceleration of the CM.
How do I find that, seem to be blank on that
 
  • #98
palaphys said:
r represents R/5, and I was taught that in a rigid body in rotation every point goes in a circular motion relative to some other point.
You taught correctly. What do you think this "other" point is here?
 
  • #99
kuruman said:
You taught correctly. What do you think this "other" point is here?
Thank God. Why can't I take other point as geometric center of the ring?
 
  • #100
kuruman said:
You taught correctly. What do you think this "other" point is here?
This is how my mind is visualising the motion. According to this diagram the center of mass is indeed in circular motion with respect to the center
 

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