Rotating ring on a rough surface- but with a twist

[erobz]

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.

 

[palaphys]

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

 

[erobz]

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.

 

[kuruman]

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?

 

[kuruman]

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.

 

[palaphys]

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.

 

[palaphys]

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

 

[kuruman]

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?

 

[palaphys]

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?

 

[palaphys]

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

 

[palaphys]

Here is my attempt for finding the frictional force. I have assumed that the center of mass is in circular motion with respect to the geometric center of the ring.

 

[kuruman]

Thank God. Why can't I take other point as geometric center of the ring?

Read the second paragraph in post #94. The center of the ring is NOT the &*%@# axis of rotation !!!!

I have assumed that the center of mass is in circular motion with respect to the geometric center of the ring.

Bad assumption.

The figure below shows the trajectory of a point on the ring (blue line), the trajectory of the CM (red line) and the trajectory of the center of the ring (dashed black line) assuming that the ring is rolling at constant velocity. This is not the case here because the ring will oscillate back and forth after it is released.

I am signing off too.

 

[palaphys]

I know that the center of mass is not rotating about the center, I just said it was in circular motion with respect to it.

Read the second paragraph in post #94. The center of the ring is NOT the &*%@# axis of rotation !!!!

Bad assumption.

The figure below shows the trajectory of a point on the ring (blue line), the trajectory of the CM (red line) and the trajectory of the center of the ring (dashed black line) assuming that the ring is rolling at constant velocity. This is not the case here because the ring will oscillate back and forth after it is released.

View attachment 357469
I am signing off too.

 

[palaphys]

I know that the center of mass is not rotating about the center, I just said it was in circular motion with respect to it.

Are you implying that my working is wrong once again? I completed the problem and am getting answers angular acceleration as g/10R ,frictional force as mg/2 and normal reaction as 49mg/10

 

[palaphys]

Also @kuruman I have confirmed with my teacher, who says that from the frame of the center of the ring, initially the center of mass would appear to trace a circular path, so it has an angular acceleration alpha (r/5) at that instant

 

[haruspex]

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.

@palaphys wrote that friction in general opposes a tendency to relative motion. Presumably that means the relative motion of the surfaces in contact, which is correct.

 

[haruspex]

Here is my attempt for finding the frictional force. I have assumed that the center of mass is in circular motion with respect to the geometric center of the ring.

Please don’t post working as images. It is contrary to forum rules for good reasons. Your writing is not completely clear, and it makes it hard for responders to refer to specific equations. If you must post working as images write very clearly and number the equations.
Also, you do not explain where your equations come from.

In your attachment to post #79, you seem to have an equation which has forces equal to $\alpha R/5$. Is there an m missing?
In the attachment to post #101, what is $\Sigma lp$? Or is it $\Sigma \tau p$? Or $\Sigma \tau_p$?

Wrt the answers, I do get something different.

 

[palaphys]

Please don’t post working as images. It is contrary to forum rules for good reasons. Your writing is not completely clear, and it makes it hard for responders to refer to specific equations. If you must post working as images write very clearly and number the equations.
Also, you do not explain where your equations come from.

In your attachment to post #79, you seem to have an equation which has forces equal to $\alpha R/5$. Is there an m missing?
In the attachment to post #101, what is $\Sigma lp$? Or is it $\Sigma \tau p$? Or $\Sigma \tau_p$?

Wrt the answers, I do get something different.

Yes indeed m is missing.
It is tau_p

 

[haruspex]

Yes indeed m is missing.
It is tau_p

Which means? It's hard to guess what your equations mean if you don’t define the variables.

 

[palaphys]

Please

Which means? It's hard to guess what your equations mean if you don’t define the variables.

Please ignore my poor handwriting. What I did was just taking the torques about the center due to friction and due to the weight from com and equated them to Ialpha

 

[Lnewqban]

... I humbly request someone to aid me in this problem and explain the mechanics here, thanks in advance. I'm only used to seeing rings roll where com coincides with geometric center.

Returning to your original post:
Have you had the same difficulty with the type of problems in which “rings roll where com coincides with geometric center”?
If not, have you have confusion regarding the direction and magnitude of the vector representing static friction between the ring and the surface?

 

[Lnewqban]

I don't understand this part please make it simple

Sorry, I will try.

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 center of the ring is forced to always keep the same r distance from the surface; therefore, it moves only parallel to it.
The many points of the periphery of the ring that sucesively work as a pivot are also forced to move along the surface.

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.

Both instantaneous points, center of ring and pivot, have the same direction of movement, as well as same horizontal velocity and tangential acceleration.

 

[erobz]

@palaphys wrote that friction in general opposes a tendency to relative motion. Presumably that means the relative motion of the surfaces in contact, which is correct.

I feel static friction is the interlocking of surfaces on a microscopic level in very non-obvious ways. Kinetic friction the rapid destruction of that interlock via shearing/deformation of the surfaces.

Maybe the definition covers it, but to me static and kinetic friction feel fundamentally different from each other and aren’t served well conceptually by a single definition. Just my opinion.

Friction is just the name we give for these complicated realities. Perhaps, it isn’t being covered adequately on a fundamental level because these definitions are attempting to be overly efficient. There needs some differentiation. They need to spend some more time on the topic in intro physics; Revisit the concepts as the problems evolve. This hang up is nothing new... I know firsthand.

 

[kuruman]

Also @kuruman I have confirmed with my teacher, who says that from the frame of the center of the ring, initially the center of mass would appear to trace a circular path, so it has an angular acceleration alpha (r/5) at that instant

Hang on to that thought because it is correct but is it useful? How is it going to help you find the linear acceleration relative to the ground and use it to find the force of static friction? Note that, relative to the center of the ring, the CM has tangential acceleration that points vertically straight down. That is useful for finding the vertical normal force, but not the force of friction that is horizontal.

 

[palaphys]

Hang on to that thought because it is correct but is it useful? How is it going to help you find the linear acceleration relative to the ground and use it to find the force of static friction? Note that, relative to the center of the ring, the CM has tangential acceleration that points vertically straight down. That is useful for finding the vertical normal force, but not the force of friction that is horizontal.

For force of friction I just took the torque about the center of the ring that's it

Also I SUDDENLY got another doubt...doesn't the bottom most point (here ICR) have an upward acceleration? Then why are we not considering the pseudo torque due to that..

 

[palaphys]

I'm so close to finishing this, just gotta clarify that major thing whether the IAR is an inertial frame

 

[palaphys]

I'm beginning to think that this problem is solvable only if the angular velocity just after rolling begins is small, otherwise this def isn't solvable.

 

[kuruman]

For force of friction I just took the torque about the center of the ring that's it

A torque is not a force and vice-versa.

 

[palaphys]

A torque is not a force and vice-versa.

I never said that I just said I used the center to find the frictional force, sorry for being unclear andsorry for annoying you for such a long time i promise if you answer my final question I posted above I will not ask anything else

 

[kuruman]

I never said that I just said I used the center to find the frictional force, sorry for being unclear andsorry for annoying you for such a long time i promise if you answer my final question I posted above I will not ask anything else

What question is this and in what post? Is it this

I'm so close to finishing this, just gotta clarify that major thing whether the IAR is an inertial frame

If so, you need to clarify to me what the IAR frame is.

You are not annoying me.