Yo-yo rolling without slipping

[PhizKid]

Homework Statement

Homework Equations

Angular momentum/fixed axis rotation

The Attempt at a Solution

Hi guys. I'm having a bit of trouble visualizing what will happen so if you could first indulge me with the mechanics. Imagine there was no friction to begin with; since the force is acting on the rope wound around the circular axle, will the yo - yo only slide to the right or will it slide to the right + roll backwards due to the pulling of the rope that is wound around the axle? It seems like it should be the second if I try to imagine it (don't have a yo - yo with me unfortunately :[). Now, regardless of that, the center of mass velocity will be directed to the right I think; was only assuming this based on physical grounds that the yo - yo will slide to the right but if the pulling of the rope around the axle causes it to roll backwards then idk if it will slide to the right or not anymore so if someone could clarify this that would be great. Going with that assumption, the friction force would be directed to the left. The equations of motion are $ma = F - f = F - \mu mg$ and, this is another point I would like someone to clarify for me because I assumed that since we want it to roll without slipping the rotation must be directed clockwise so that the tangential velocity at the contact point (left) is directed opposite to the assumed CM velocity direction (right) but I'm not sure if I'm supposed to assume this a priori or if its somehow supposed to come out of the equations at the end so if someone could clarify this as well, $-\frac{1}{2}mR^{2}\alpha = -\frac{1}{2}mRa = Fb - \mu mgR$ and putting all this together gives $F = \frac{3\mu mg}{(1 + \frac{2b}{R})}$. If you could clarify all the things I asked for that would be great. Thanks!

 

[SammyS]

I'll answer some of your questions.

Homework Statement

Homework Equations

Angular momentum/fixed axis rotation

The Attempt at a Solution

Hi guys. I'm having a bit of trouble visualizing what will happen so if you could first indulge me with the mechanics. Imagine there was no friction to begin with; since the force is acting on the rope wound around the circular axle, will the yo - yo only slide to the right or will it slide to the right + roll backwards due to the pulling of the rope that is wound around the axle? It seems like it should be the second if I try to imagine it (don't have a yo - yo with me unfortunately :[).

Yes, it is the second.

The Net force is to the right. (It's the only horizontal force.) The torque w.r.t the center of mass is counter-clockwise.

Now, regardless of that, the center of mass velocity will be directed to the right I think; was only assuming this based on physical grounds that the yo - yo will slide to the right but if the pulling of the rope around the axle causes it to roll backwards then idk if it will slide to the right or not anymore so if someone could clarify this that would be great. Going with that assumption, the friction force would be directed to the left.

Yes, the frictional force is to the left.

The equations of motion are $\ ma = F - f = F - \mu mg\$ and, this is another point I would like someone to clarify for me because I assumed that since we want it to roll without slipping the rotation must be directed clockwise so that the tangential velocity at the contact point (left) is directed opposite to the assumed CM velocity direction (right) but I'm not sure if I'm supposed to assume this a priori or if its somehow supposed to come out of the equations at the end so if someone could clarify this as well, $-\frac{1}{2}mR^{2}\alpha = -\frac{1}{2}mRa = Fb - \mu mgR$ and putting all this together gives $F = \frac{3\mu mg}{(1 + \frac{2b}{R})}$. If you could clarify all the things I asked for that would be great. Thanks!

(Can't you rig up something to emulate a yo-yo? Try to have b/R a fairly small fraction.)

If the frictional force is sufficient to keep the yo-yo from slipping then look at the torque w. r. t. the point of contact of the yo-yo with the table. It's fairly clear then that the yo-yo will rotate clockwise, and move to the right.

I know that doesn't answer all of your questions. Hopefully it's a useful start.

 

[PhizKid]

Hi Sammy, thanks for responding. So we agree the applied tangential force causes the yo - yo to roll backwards (counter clockwise) due to the pulling of the string around the axle. What I can't reconcile is if this means the center of mass velocity will also be directed backwards. Is it possible for the yo - yo to be rolling backwards but have the center of mass velocity simultaneously be directed along the direction of the applied tangential force (forward) or must the center of mass velocity be directed backwards since that is the direction the yo - yo will start rolling? I just can't picture something rolling and sliding at the same time. If you have a wheel rolling without slipping along a flat road for example then the center of mass velocity is directed in the direction of rolling (both are forward) and it seems here if the yo - yo rolls backward due to the pulling of the string around the axle, the center of mass velocity would have to be directed backwards as well (and this way it can cancel out with the tangential velocity at the contact point which will point forward if we want no slipping). If there is no friction then the yo - yo will simply slide forward with no rolling, so that the CM velocity is directed forward for sure, (right? I'm pretty sure of that) but if there IS friction then the applied force will cause it to roll backwards but I can't decide logically which direction the center of mass velocity will be directed in this case which I need to know in order to determine the direction of friction. This is where I need help. Thanks a ton!

 

[haruspex]

It might help if you mentally replace the entire yo-yo by a stick, initially upright, stretching from the point of contact on the ground to the yo-yo's centre. The string is attached to it at exactly the same point as it was to the yo-yo. Instantaneously, the stick will move just as the yo-yo would have done.
Another way to check it is to observe that the string will certainly move to the right. Since R > b, this means the yo-yo must roll to the right.

 

[SammyS]

Hi Sammy, thanks for responding. So we agree the applied tangential force causes the yo - yo to roll backwards (counter clockwise) due to the pulling of the string around the axle. What I can't reconcile is if this means the center of mass velocity will also be directed backwards. Is it possible for the yo - yo to be rolling backwards but have the center of mass velocity simultaneously be directed along the direction of the applied tangential force (forward) or must the center of mass velocity be directed backwards since that is the direction the yo - yo will start rolling? I just can't picture something rolling and sliding at the same time. If you have a wheel rolling without slipping along a flat road for example then the center of mass velocity is directed in the direction of rolling (both are forward) and it seems here if the yo - yo rolls backward due to the pulling of the string around the axle, the center of mass velocity would have to be directed backwards as well (and this way it can cancel out with the tangential velocity at the contact point which will point forward if we want no slipping). If there is no friction then the yo - yo will simply slide forward with no rolling, so that the CM velocity is directed forward for sure, (right? I'm pretty sure of that) but if there IS friction then the applied force will cause it to roll backwards but I can't decide logically which direction the center of mass velocity will be directed in this case which I need to know in order to determine the direction of friction. This is where I need help. Thanks a ton!

"Imagine there was no friction ..." This is the case for which I agreed the yo-yo rolled counter-clockwise, i.e., backwards.

I also said that if the friction was sufficient to keep the yo-yo from slipping on the surface, then the yo-yo would roll clockwise . --- The yo-yo rolls forward as it moves to the right.

By the way, if the frictional force is just right, compared to the applied force, then the yo-yo would slide along without rotating.

 

[PhizKid]

What I don't get is why the CM velocity must be directed in the same direction as the applied force. Assuming it moved to the right and if there was friction why would it roll forward instead of rolling backwards? Wouldn't that only be the case if the torque due to friction was greater than the torque due to the applied force?

 

[SammyS]

What I don't get is why the CM velocity must be directed in the same direction as the applied force. Assuming it moved to the right and if there was friction why would it roll forward instead of rolling backwards? Wouldn't that only be the case if the torque due to friction was greater than the torque due to the applied force?

Can the force of friction (which is to the left in this case) be larger than the applied force?

 

[PhizKid]

It has to be less if we want the yo - yo to start moving don't we?

 

[tms]

What I don't get is why the CM velocity must be directed in the same direction as the applied force. Assuming it moved to the right and if there was friction why would it roll forward instead of rolling backwards? Wouldn't that only be the case if the torque due to friction was greater than the torque due to the applied force?

When the yo-yo rolls, about which axis is it rolling, instantaneously? What torque does the frictional force exert around that axis?

 

[PhizKid]

Those don't answer my specific questions. The torque due to friction is -fR if friction is assumed to be directed to the left; it is rotating about the z axis.

 

[tms]

Those don't answer my specific questions.

I am trying to lead you to the answer by asking questions that will make you think in the right direction.

The torque due to friction is -fR if friction is assumed to be directed to the left; it is rotating about the z axis.

Where does friction act? The friction that is causing the rolling, that is. The friction between the yo-yo and the table.

By the z-axis I assume you mean an axis perpendicular to the page. The question is where does that axis intercept the page.

 

[PhizKid]

Ok thanks but I just don't get why the yo - yo must SLIDE to the right (i.e. why the CM velocity is directed to the right)? If I can clear that up then I get the rest of the problem because if it slides to the right friction is directed to the left and then it's just a matter of making the torque due to friction greater than the torque due to the applied force so that the yo - yo rolls clock - wise and allows for the CM velocity to cancel out with the tangential velocity at the contact point and we can then replace alpha with a/ R due to no slipping. That is all correct right? I just need that first point clarified.

 

[tms]

I'll ask again: Where does friction act? (The friction that is causing the rolling, that is. The friction between the yo-yo and the table.) What axis is the yo-yo rotating about (instantaneously)?

 

[PhizKid]

Friction acts at the point of contact of the yo - yo with the ground. The instantaneous axis of rotation goes through the center

 

[SammyS]

Friction acts at the point of contact of the yo - yo with the ground. The instantaneous axis of rotation goes through the center

No that's not the instantaneous axis of rotation if the friction is sufficient to keep the yo-yo from slipping.

I pretty much covered this back in post #2.:

...

If the frictional force is sufficient to keep the yo-yo from slipping then look at the torque w. r. t. the point of contact of the yo-yo with the table. It's fairly clear then that the yo-yo will rotate clockwise, and move to the right.

 

[tms]

Friction acts at the point of contact of the yo - yo with the ground.

Right.

The instantaneous axis of rotation goes through the center

Wrong. The yo-yo is rolling on the ground; rolling objects instantaneously rotate about the point in contact with the surface they are rolling on.

What, then, is the torque due to friction?

 

[PhizKid]

You are interchangeably using the word axis and the instantaneously co - moving reference frame. If you are using them synonymously then there is no absolute "axis of rotation" i.e. family of instantaneously co - moving reference frames. I can just as easily take my coordinate system's origin to be fixed to the center of the disk, for which there is only one element in the family of instantaneously co - moving reference frames, and call that, according to your terminology, the "axis of rotation". You can't state one "axis" is wrong over the other if you choose to use the word in that sense. Of course for pre - calculation intuitive purposes, going with Sammy's post, if I take an element of the family of ICRF's and attach one to each point the yo - yo and ground come into contact along the trajectory, it is easier to see the net torque is clock - wise in each such frame and hence the tangential velocity points to the left so we, in order to have no slipping, we would have to require the CM velocity to be directed to the right (Sammy could you clarify if this is indeed what you were saying because this is how I pictured it). Thank you all.

 

[rcgldr]

Friction acts at the point of contact of the yo - yo with the ground. The instantaneous axis of rotation goes through the center

The yo-yo is rolling on the ground; rolling objects instantaneously rotate about the point in contact with the surface they are rolling on.

Regardless if the yo-yo is sliding or rolling, the yo-yo's axis of rotation is at the center of the yo-yo.

 

[ehild]

You can take the yo-yo rolling either about the CM or about the instantaneous axis of rotation.
The yo-yo rolls clockwise, the CM moves to the right. There are two forces acting: F, and the force of static friction f at the contact point with the ground. Both F and the friction f have a torque with respect to the CM. Writing up the equations both for linear acceleration of the CM and angular acceleration of rotation about the CM and using the rolling condition acceleration = angular acceleration X Radius we can solve the equations for the force of friction which can not be greater than μMg.

ehild

 

[haruspex]

Regardless if the yo-yo is sliding or rolling, the yo-yo's axis of rotation is at the center of the yo-yo.

No, if it's rolling then the point which is instantaneously stationary is the point of contact with the ground. That is therefore the axis of rotation. Yes, you can represent the motion as the sum of a linear motion and a rotational one, but there are many ways to do that, and making the centre of the yo-yo the axis of that rotation is an arbitrary choice.

 

[ehild]

No, if it's rolling then the point which is instantaneously stationary is the point of contact with the ground. That is therefore the axis of rotation.

It is the axis of rotation if we consider the motion instantaneously as rotation about fixed axis. But it can be also treated as combined from translation of the CM and rotation about the CM.


ehild

 

[rcgldr]

Regardless if the yo-yo is sliding or rolling, the yo-yo's axis of rotation is at the center of the yo-yo.

No, if it's rolling then the point which is instantaneously stationary is the point of contact with the ground.

The contact point isn't a pivot point for the yo-yo. Even in the instantaneous case, the motion is a combination of linear and angular movement. Choosing the center of the yo-yo (it's center of mass) as the axis of rotation makes the most sense when calculating the torques.

putting all this together gives $F = \frac{3\mu mg}{(1 + \frac{2b}{R})}$.

This seems to be the correct answer.

For the alternate cases, if the yo-yo is resting on a pair of rails, with a hub between the rails and larger than the yo-yo so that b > R, then pulling the string to the right would result in the yo-yo rolling to the left. If b = R, the yo-yo either doesn't move at all or it slides to the right without rolling.

If 1/2 R < b < R, then the yo-yo moves faster to the right than the string. If I remember correctly, if R < b < 2 R, then the yo-yo moves faster to the left than the string moves to the right. The closer b is to R, the faster the yo-yo moves (for both cases, b < R and b > R).

 

[PhizKid]

Well just to put it to rest, I did the calculation by attaching an instantaneously co - moving origin to the contact point at every instant of time. There is no difference in answer for whichever method we choose: the single frame attached to the center or the family of instantaneously co - moving references frames attached to the contact point at every instant; you just can't claim there is some natural/absolute axis of rotation. I just think Sammy's original intention to use such a family of frames was it make it clearer why the CM velocity was directed to the left because that was my issue from the begging: why is the CM velocity directed to the left? How can we gather this from the starting information? Well it would seem if we chose such a family of ICRF's, we don't need to care about the torque due to friction of course and the net torque would be clock - wise with respect to this origin which would result in a leftwards tangential velocity so we would need to conclude the CM velocity is to the right in order to have no slipping. This seems like a reasonable argument for why CM velocity would be directed to the right (correct?) but I cannot see why CM velocity would be directed to the right had we stayed in the center of mass frame. In this frame it is not obvious to me why a priori the CM velocity would be directed as such. This is what I needed clarified.

As for the calculations in this frame: choose some arbitrary instant of time and attach an instantaneously co - moving frame to the contact point. The vector pointing from this origin to the point of application of the force is $\vec{r} = (b - R)\hat{y}$ and the force is $\vec{F} = F\hat{x}$ so $\tau = \vec{r}\times \vec{F} = -(b - R)F = \frac{3}{2}MR^{2}\alpha$ (I have used the parallel axis theorem). Going by the above argument, CM velocity would be directed to the right so friction to the left and thus the equation of motion is $F - f = Ma = MR\alpha$ after applying the ,then justifiable, no slipping condition. Combining the two we get that $F(3 + 2(\frac{b}{R} - 1)) = 3f = 3\mu Mg$ where I have replaced f with uMg since f <= uMg in general and we want the maximum force so we take the equality. This simplifies to $F = \frac{3\mu Mg}{1 + \frac{2b}{R}}$ as before but with more annoying algebra (which I guess is a trade off for the intuition).

 

[SammyS]

... I just think Sammy's original intention to use such a family of frames was it make it clearer why the CM velocity was directed to the left because that was my issue from the begging: why is the CM velocity directed to the left? How can we gather this from the starting information? Well it would seem if we chose such a family of ICRF's, we don't need to care about the torque due to friction of course and the net torque would be clock - wise with respect to this origin which would result in a leftwards tangential velocity so we would need to conclude the CM velocity is to the right in order to have no slipping. This seems like a reasonable argument for why CM velocity would be directed to the right (correct?) but I cannot see why CM velocity would be directed to the right had we stayed in the center of mass frame. In this frame it is not obvious to me why a priori the CM velocity would be directed as such. This is what I needed clarified.

...

My intent with considering the torque w.r.t. the point of contact was that the force of friction produces no torque about the contact point so only the applied force produces torque, and that torque will tend to produce a clockwise rotation.

As far as the motion of the center of mass (and I didn't address this): As long as the applied force , F, is greater than the force of friction (a reasonable assumption), the net force is to the right, thus the center of mass accelerates to the right. This is also consistent with clockwise rotation if there's no slipping.

 

[PhizKid]

Yes I get that but the thing is that the direction of acceleration is not necessarily the direction of CM velocity in general. This was my qualm from the start.

 

[haruspex]

The contact point isn't a pivot point for the yo-yo. Even in the instantaneous case, the motion is a combination of linear and angular movement.

http://en.wikipedia.org/wiki/Instan...nt_centre_of_a_wheel_rolling_without_slipping

 

[SammyS]

Yes I get that but the thing is that the direction of acceleration is not necessarily the direction of CM velocity in general. This was my qualm from the start.

Well, if the yo-yo is initially rolling to the left, then it will decelerate for a while -- until it comes to rest, then it will accelerate to the right.

 

[haruspex]

Yes I get that but the thing is that the direction of acceleration is not necessarily the direction of CM velocity in general. This was my qualm from the start.

I'm not sure what you mean. Direction of linear acceleration or of angular acceleration? CM velocity or CM linear acceleration?

 

[rcgldr]

The contact point isn't a pivot point for the yo-yo. Even in the instantaneous case, the motion is a combination of linear and angular movement.

http://en.wikipedia.org/wiki/Instan...nt_centre_of_a_wheel_rolling_without_slipping

An instant center is a point where an object with both angular and linear velocities can considered to be rotating about for an instant, but it only applies to an instant in time. An instant center may lie completely outside of an object as shown in this article:

http://www.real-world-physics-problems.com/instant-center.html

From that same article:

In Closing:

Note that the spatial location of the instant center can change with time. So in general its use only applies at the instant considered, which corresponds to the information given in the problem, at the instant considered. In general, if the velocities of points A and B change, and/or their directions change, a new instant center must be calculated to solve for the (new) unknown linear velocities, or angular velocites.

In general, the instant center approach should not be used for finding the acceleration of points in a rigid body. This is because the instant center (IC) does not generally have zero acceleration.

There may be cases where using an instant center is mathematically convenient, but it's not needed for this problem. It's easier (at least for me) to understand the relative torques based on radial distance from the yo-yo's center of mass.

 

[haruspex]

An instant center is a point where an object with both angular and linear velocities can considered to be rotating about for an instant, but it only applies to an instant in time. An instant center may lie completely outside of an object as shown in this article:

http://www.real-world-physics-problems.com/instant-center.html

From that same article:

In Closing:

Note that the spatial location of the instant center can change with time. So in general its use only applies at the instant considered, which corresponds to the information given in the problem, at the instant considered. In general, if the velocities of points A and B change, and/or their directions change, a new instant center must be calculated to solve for the (new) unknown linear velocities, or angular velocites.

In general, the instant center approach should not be used for finding the acceleration of points in a rigid body. This is because the instant center (IC) does not generally have zero acceleration.

There may be cases where using an instant center is mathematically convenient, but it's not needed for this problem. It's easier (at least for me) to understand the relative torques based on radial distance from the yo-yo's center of mass.

I only ever claimed it was an instantaneous centre of rotation. That's all that's needed to understand which way it will move. No, it's not needed to solve this problem, but for some (me included) it may be the easiest way to see this.

 

[PhizKid]

I'm not sure what you mean. Direction of linear acceleration or of angular acceleration? CM velocity or CM linear acceleration?

Hi haruspex. I was referring to the linear acceleration and the linear (center of mass) velocity.

 

[haruspex]

Hi haruspex. I was referring to the linear acceleration and the linear (center of mass) velocity.

Starting from rest, the initial acceleration must have the same sign as the first nonzero velocity and the first nonzero displacement. You can see this by considering a velocity-time graph. Thereafter, the acceleration may change sign, making it, for a while at least, opposite in sign to the velocity.