Rotating ring on a rough surface- but with a twist

[haruspex]

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.
The point of the ring that is instantaneously in contact with the ground is accelerating, but the fixed point on the ground it is currently in contact with is not.
You could also mean the contact point as a dynamic entity, neither fixed on the ground nor fixed to the ring. In that case it would be inertial if the ring were rolling at constant velocity, but in the present case it is accelerating.

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..

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.

In the attachment to post #79, the penultimate equation has a term at the end I had read as $\frac R5\frac 9{10R}$. You said there was an m missing, but I now think you meant $M=5m$. And what looked like a 9 is actually a g.
Fixing those gives $5m\frac R5\frac g{10R}$, which reduces to $m\frac g{10}$.
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?

 

[haruspex]

Maybe the definition covers it

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.

 

[TSny]

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.

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, $\dfrac {dL_A}{dt} \neq I_A \alpha$.

 

[kuruman]

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?

See attachment in post #21 wherein OP derives the equation $\alpha = \dfrac{g}{10R}.$

 

[kuruman]

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, $\dfrac {dL_A}{dt} \neq I_A \alpha$.

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.

 

[SammyS]

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..

You can put THIS sudden doubt to rest. Consider the downward (component of) acceleration of the top most point of the ring.

Please, if you're going to use abbreviations, at the very least, spell them out the first time you use them. - Better yet, avoid their use altogether. - - ICR - IAR - M-O-U-S-E ...

 

[TSny]

- - ICR - IAR - M-O-U-S-E ...

(giving away my age bracket)

 

[TSny]

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 sure. The physical pendulum always rotates about a fixed axis in the lab frame. The rolling ring does not rotate about a fixed axis.

 

[TSny]

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?

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.

 

[palaphys]

So I just went through the official question once more and it says that the ring starts from rest. All the values which have been asked to he computed are at the instant just after it begins to roll. So here I don't think there is much of a problem whether the bottom most point is inertial

 

[palaphys]

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.

It's not rolling initally

 

[erobz]

This problem seems to have some devils in the details. I just want to work it generally from the coordinates of the mass center.

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.

 

[palaphys]

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.

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

 

[kuruman]

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

In that case please post a summary of your approved answers for the benefit of everyone who reads this thread. Thanks.

 

[TSny]

It's not rolling initally

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.

 

[palaphys]

In that case please post a summary of your approved answers for the benefit of everyone who reads this thread. Thanks.

Yes, will do by the end of the day. Thank you for tolerating me and clearing many of my questions

 

[palaphys]

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.

OK gotit

 

[haruspex]

I'm not familiar with defining different $\alpha$'s this way. So, I don't know.

I withdraw that.
But as I wrote in post #121, one has to be careful what one means by "rotating about A”.
Is that the fixed point on the ground which was the initial point of contact, the point fixed to the ring which was the initial point of contact, or, dynamically, the point on the ground currently in contact?
Only the first of the three can be the origin of an inertial frame. The other two are accelerating.
Generally, it is safe to use the mass centre for the axis or any inertial frame. I believe it is also ok to use a point that is accelerating directly towards or away from the mass centre since that does not alter the angular acceleration about the point.
E.g. consider masses m, 2m at opposite ends of a seesaw length 2R. If we use a point below the middle but accelerating up at rate a, the apparent rate of change of angular momentum is $2m(aR+\alpha R^2)+m(-aR+\alpha R^2)=maR+3m\alpha R^2$, but the torque is still $mgR$.

 

[kuruman]

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.

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.

Parameter $a$ in the reference model can be found by matching the overall horizontal displacement in the model $\Delta x_{\text{mod}}=2\pi a$ to the overall horizontal displacement of the CM here, $\Delta x_{\text{cm}}=\frac{2}{5}R$ to get $$a=\frac{R}{5\pi}.$$ Then, one can use equations 19.9.5 and 19.9.6 to find the coordinates of the CM as a function of $\theta$ which is the angle that the straight part of the "string" forms w.r.t. the vertical, $$\begin{align} & x= \frac{R}{5\pi}(2\theta+\sin2\theta) \nonumber \\ & y=-\frac{2R}{5\pi}\cos^2\!\theta. \nonumber \end{align}$$ Finally, one can take derivatives to find $\ddot x$ and $\ddot y$, apply the initial condition $\dot {\theta}(0)=0$, find $\ddot{\theta}(0)$ from the torque equation and apply Newton's second law at $\theta=0$ to find expressions for the initial Cartesian components of the net force.

 

[erobz]

Well, I got a different answer for the initial angular acceleration... So what goes wrong here.

I find $\alpha = \frac{25}{51}\frac{g}{R}$

Here is what I did:
$$ x = R \theta + (\cos \theta - 1) r $$
$$ y = -r \sin \theta$$

The derivates follow as:

$$ \ddot x = R \ddot \theta - r \left( \dot \theta^2 \cos \theta+ \ddot \theta \sin \theta \right)$$

$$ \ddot y = r \left( \dot \theta^2 \sin \theta - \ddot \theta \cos \theta \right) $$

Where $r = R/5$, but leaving it general for now.

I get three equations of motion

$\sum F_x$

$$ f = M \ddot x $$

$\sum F_y$

$$ N - Mg = M \ddot y $$

$\circlearrowright^{+} \sum \tau_G$

$$ N R \cos \theta - f R = I_G \ddot \theta $$

Where $M = 5m$

Now get all the equations into functions of $\theta$ and its derivatives.

$$ \implies N = M r \left( \dot \theta^2 \sin \theta - \ddot \theta \cos \theta \right) + Mg $$

$$ \implies f = M \left(R \ddot \theta - r \left( \dot \theta^2 \cos \theta+ \ddot \theta \sin \theta \right) \right) $$

At this point I just make the substitutions for the initial conditions $\theta = 0 , \dot \theta = 0 \implies \cos \theta = 1, \sin \theta = 0$

$$ \implies N = Mg - Mr \ddot \theta $$

$$ \implies f = MR \ddot \theta $$

Rearrange the last:

$$ \ddot \theta = \frac{f}{MR} $$

And then sub into the sum of torques about $G$:

$$ f R = MgR - \cancel{MR}r\frac{f}{\cancel{MR}} - I_G \frac{f}{MR}$$

$$ \implies f = \frac{M^2 R^2 g }{MR^2 + MRr + I_G } $$

I find the mass moment of inertia about $g$ to be:

$$ I_G = 2m \left( \frac{4}{5}R \right)^2+ m \left( \frac{6}{5}R \right)^2 + \frac{1}{2} 2m R^2 + 2m \left( \frac{1}{5}R \right)^2 = \frac{21}{25}MR^2 $$

Finally

$$ f = \frac{25}{51}Mg $$

$$ \alpha = \frac{25}{51}\frac{g}{R} $$

Can someone see where I've gone off the rails? Thanks.

EDIT: Errors found-
1) Used wrong moment arm for torque of normal force, should be $r$.
2) Error in using the wrong model. I used a "disk" when the OP is a "ring"

Fixing the errors verifies solution for $\alpha = \frac{1}{10} \frac{g}{R}$ that has been previously shown.

 

[TSny]

$$ I_G = 2m \left( \frac{4}{5}R \right)^2+ m \left( \frac{6}{5}R \right)^2 + \frac{1}{2} 2m R^2 + 2m \left( \frac{1}{5}R \right)^2 = \,\, ...$$

The third term on the right should not have the factor of $\frac 1 2$ for a ring. Otherwise, it looks good.

 

[erobz]

The third term on the right should not have the factor of $\frac 1 2$ for a ring. Otherwise, it looks good.

I could have sworn it was a disk! Thanks for finding that and your time.

 

[erobz]

The third term on the right should not have the factor of $\frac 1 2$ for a ring. Otherwise, it looks good.

Unfortunately, still no dice...

$$ I_G = 2m \left( \frac{4}{5}R \right)^2+ m \left( \frac{6}{5}R \right)^2 + 2m R^2 + 2m \left( \frac{1}{5}R \right)^2 = mR^2 \left( \frac{32}{25} + \frac{36}{25}+\frac{50}{25}+\frac{2}{25} \right)= \frac{24}{25}MR^2 $$

$$ \alpha = \frac{25}{54} \frac{g}{R} $$

 

[TSny]

$$ mR^2 \left( \frac{32}{25} + \frac{36}{25}+\frac{50}{25}+\frac{2}{25} \right)= \frac{24}{25}MR^2 $$

Check the fraction on the right side.

 

[erobz]

Check the fraction on the right side.

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.

 

[TSny]

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.

Oops, my oversight. I didn't notice the switch from $m$ to $M$. I agree with your result for $I_G$.

 

[TSny]

$\circlearrowright^{+} \sum \tau_G$
$$ N R \cos \theta - f R = I_G \ddot \theta $$

Is the lever arm correct for $N$? Did you mean to use $r$ here instead of $R$?

 

[erobz]

Is the lever arm correct for $N$? Did you mean to use $r$ here instead of $R$?

It was supposed to be $r$... Another blunder! I'll recalculate.

 

[erobz]

Yep. It agrees now.

$$ \alpha = \frac{1}{10} \frac{g}{R} $$

Thanks again @TSny for verifying.

 

[TSny]

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.

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! ).