Read the second paragraph in post #94. The center of the ring is NOT the &*%@# axis of rotation !!!!palaphys said:Thank God. Why can't I take other point as geometric center of the ring?
Bad assumption.palaphys said:I have assumed that the center of mass is in circular motion with respect to the geometric center of the ring.
kuruman said: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.
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/10palaphys said: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.
@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.erobz said: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.
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.palaphys said: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.
Yes indeed m is missing.haruspex said: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.
Which means? It's hard to guess what your equations mean if you don’t define the variables.palaphys said:Yes indeed m is missing.
It is tau_p
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 Ialphaharuspex said:Which means? It's hard to guess what your equations mean if you don’t define the variables.
Returning to your original post:palaphys said:... 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.
Sorry, I will try.palaphys said:I don't understand this part please make it simple
Lnewqban said:All points of the ring instantaneously rotate about the instantaneous ring-surface point of contact,
The center of the ring is forced to always keep the same r distance from the surface; therefore, it moves only parallel to it.Lnewqban said: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.
Both instantaneous points, center of ring and pivot, have the same direction of movement, as well as same horizontal velocity and tangential acceleration.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 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.haruspex said:@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.
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 said: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
For force of friction I just took the torque about the center of the ring that's itkuruman said: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.
A torque is not a force and vice-versa.palaphys said:For force of friction I just took the torque about the center of the ring that's it
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 elsekuruman said:A torque is not a force and vice-versa.
What question is this and in what post? Is it thispalaphys said: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
If so, you need to clarify to me what the IAR frame is.palaphys said:I'm so close to finishing this, just gotta clarify that major thing whether the IAR is an inertial frame
A frame is more than just a point (IAR=instantaneous axis of rotation, I assume). There are also the matters of directions and acceleration.palaphys said:I'm so close to finishing this, just gotta clarify that major thing whether the IAR is an inertial frame
If you are taking torques about an axis that lies on a vertical line through the centre then the vertical acceleration of the contact point is irrelevant.palaphys said: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..
It wasn’t a definition. It was a factual statement about the direction in which the forces we call static friction and kinetic friction both act. Whether that is pure coincidence or arises because of a common underlying cause may be a research topic but is not important here.erobz said:Maybe the definition covers it
See attachment in post #21 wherein OP derives the equation ##\alpha = \dfrac{g}{10R}.##haruspex said:I assume this term came from an equation ##a=g/10##, but I can’t find where that comes from. Is the derivation of that in another attachment?
I think that this is can be modeled as a (physical) pendulum that starts from rest with the CM horizontally in-line with the point of support. If it is given non-zero initial angular velocity, then it becomes an inverted pendulum.TSny said:The equation ##\large \sum \tau_{A} = I_{A} \alpha## yields the correct answer for ##\alpha## for this particular problem. But, in general, this equation is not valid. For example, if the ring has angular velocity ##\omega## at the instant shown below, then ##\sum \tau_{A} = I_{A} \alpha## will not work for finding ##\alpha## at this instant.
View attachment 357496
However, taking torques about the center of mass of the system will work: ##\large \sum \tau_{cm} = I_{cm} \alpha##.
This is already a long thread and I hesitated to post this. But I want to caution students that ## \sum \tau_{A} = I_{A} \alpha## is not generally correct.
If you want to use point ##A## as the origin for torques, then the valid equation is ##\sum \tau_{A} = \dfrac {dL_A}{dt}##, where ##L_A## is the angular momentum of the system about ##A##. But, generally, ##\dfrac {dL_A}{dt} \neq I_A \alpha##.
You can put THIS sudden doubt to rest. Consider the downward (component of) acceleration of the top most point of the ring.palaphys said: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..
SammyS said:- - ICR - IAR - M-O-U-S-E ...
I'm not sure. The physical pendulum always rotates about a fixed axis in the lab frame. The rolling ring does not rotate about a fixed axis.kuruman said:I think that this is can be modeled as a (physical) pendulum that starts from rest with the CM horizontally in-line with the point of support.
I'm not familiar with defining different ##\alpha##'s this way. So, I don't know.haruspex said:How about if we distinguish the angular acceleration about A (as a fixed point on the ground) from that about the centre, ##\alpha_A=\alpha_C+a/R##? Is ##\tau_A=I_A\alpha_A## ok?
It's not rolling initallyTSny said:I'm not familiar with defining different ##\alpha##'s this way. So, I don't know.
For the problem of post #123 where the ring is already rolling with angular speed ##\omega## in the figure, I find the angular acceleration at that instant to be ##\alpha = \frac 1 {10}(\frac {g}{R} + \omega^2)##. If you have time, you could try your idea and see what you get.
Just communicated with my teacher. How says all my answers were correct and as per the problem the ring starts rolling just after rest (weird) and we can ignore the angular velocity at that instant just after resterobz said:This problem seems to have some devils in the details. I just want to work it generally from the coordinates of the mass center.
View attachment 357498
Does anyone think this is problematic? I'm using ##\sum \boldsymbol{F} = M \boldsymbol{a}##( 2 linear coordinate equations ##x,y##), ##\sum \boldsymbol{\tau_G} = I_G \boldsymbol{ \ddot \theta}##( angular coordinate ##\theta##) as @TSny suggests, to get everything in terms of the angular coordinate, then sub in ##\theta, \dot \theta = 0 ## to get ##f## at that instant. It sure seems icky.
In that case please post a summary of your approved answers for the benefit of everyone who reads this thread. Thanks.palaphys said:Just communicated with my teacher. How says all my answers were correct and as per the problem the ring starts rolling just after rest (weird) and we can ignore the angular velocity at that instant just after rest
Yes, I understand. For the case where the ring is initially at rest, the formula ##\sum \tau_A = I_A \, \alpha## happens to give the correct answer for the initial value of ##\alpha##. However, this equation does not hold at instants of time when the ring is in motion. So, I don't think it's a good idea to promote the use of this formula.palaphys said:It's not rolling initally
Yes, will do by the end of the day. Thank you for tolerating me and clearing many of my questionskuruman said:In that case please post a summary of your approved answers for the benefit of everyone who reads this thread. Thanks.
OK gotitTSny said:Yes, I understand. For the case where the ring is initially at rest, the formula ##\sum \tau_A = I_A \, \alpha## happens to give the correct answer for the initial value of ##\alpha##. However, this equation does not hold at instants of time when the ring is in motion. So, I don't think it's a good idea to promote the use of this formula.
However, it can be shown that the formula ##\sum \tau_{cm} = I_{cm \,} \alpha## holds at all instants of time during the motion of the ring. You are safer with this formula.
I withdraw that.TSny said:I'm not familiar with defining different ##\alpha##'s this way. So, I don't know.
TSny said:I'm not sure. The physical pendulum always rotates about a fixed axis in the lab frame. The rolling ring does not rotate about a fixed axis.
The third term on the right should not have the factor of ##\frac 1 2 ## for a ring. Otherwise, it looks good.erobz said:$$ I_G = 2m \left( \frac{4}{5}R \right)^2+ m \left( \frac{6}{5}R \right)^2 + \frac{1}{2} 2m R^2 + 2m \left( \frac{1}{5}R \right)^2 = \,\, ...$$
I could have sworn it was a disk! Thanks for finding that and your time.TSny said:The third term on the right should not have the factor of ##\frac 1 2 ## for a ring. Otherwise, it looks good.
TSny said:The third term on the right should not have the factor of ##\frac 1 2 ## for a ring. Otherwise, it looks good.
Check the fraction on the right side.erobz said:$$ mR^2 \left( \frac{32}{25} + \frac{36}{25}+\frac{50}{25}+\frac{2}{25} \right)= \frac{24}{25}MR^2 $$
I get ## \frac{120}{25} = \frac{24}{5} ##TSny said:Check the fraction on the right side.
Oops, my oversight. I didn't notice the switch from ##m## to ##M##. I agree with your result for ##I_G##.erobz said:I get ## \frac{120}{25} = \frac{24}{5} ##
Then I sub in that ## m = \frac{1}{5}M ## I get ##I_G = \frac{24}{25}MR^2##
I'm not seeing it.
Is the lever arm correct for ##N##? Did you mean to use ##r## here instead of ##R##?erobz said:## \circlearrowright^{+} \sum \tau_G##
$$ N R \cos \theta - f R = I_G \ddot \theta $$
It was supposed to be ##r##... Another blunder! I'll recalculate.TSny said:Is the lever arm correct for ##N##? Did you mean to use ##r## here instead of ##R##?
The CM moves on a circle that slips on the ##\frac 4 5 R ## line as the ring rolls. The motion of the CM is a curtate cycloid. (I just learned this term!kuruman said:After additional consideration, I think that the model for the motion is the cycloidal pendulum. The simulation on the right is taken from the cited reference. In this current problem, the CM is a point on a circle of radius ##\frac{1}{5}R## that rolls without slipping on a line parallel to the surface at distance ##\frac{4}{5}R## above it. Therefore, it describes a cycloid.