Pulling a yoyo over a surface with tension

[Justforthisquestion1]

Homework Statement

In the following homework a yoyo is pulled to the right. I have trouble understanding, why it would rotate clockwise. A similar problem has been posted in the past, but i did not understand its solution.

Relevant Equations

Angular momentum, Fixed axis rotation

My Work so far:
So there is a Force F applied to the right and since we have a no slipping condition we have a static friction force fs to the left. So the only thing that can move the center of mass (cm) of the yoyo is rotation. F would lead to a ccw rotation and fs to a clockwise rotation. From a yoyo in reality i know that i can observe a cw rotation. So fs * R > F * b. But since the yoyo is moving F should be larger than fs? I am misunderstanding something here. Thank you for your help!

 

[BvU]

Hi,
What axis of rotation are you considering?

 

[BvU]

(Was a rhetorical question). So you find a certain angular acceleration. What is the connection with the no-slipping condition?

 

[Justforthisquestion1]

Hi,
What axis of rotation are you considering?

I had considered the cm to be the axis of rotation but now that you mention it i am beginning to think that the contact point might be the axis of rotation. The contact point has always a speed of zero? So it stays fixed. then the torque through fs would be zero and the torque through F would be in the plane leading to a cw rotation. I dont understand it fully yet, but is my train of thoughts correct so far?

 

[BvU]

Yes, that clarifies the clockwise angular acceleration. Now the link to no-slip, where the linear acceleration comes in....

 

[Justforthisquestion1]

(Was a rhetorical question). So you find a certain angular acceleration. What is the connection with the no-slipping condition?

I dont know the connection between angular acceleration and non slipping. I know that when non slipping the relationsship between the speed of the cm and the angular speed: v(cm) = R *omega

 

[BvU]

And $a$ as a function of $\alpha$ ?

 

[Justforthisquestion1]

And $a$ as a function of $\alpha$ ?

Ah yes sorry $a$ = $\alpha$ * $R$
Okay now it gets confusing again. So i have an acceleration $a$ due to a force $F$. This force leads to a rotation with an acceleraton depending on R. Meaning i have an angular acceleration. But i am still a little bit struggeling to accept, that the axis of rotation leads to this kind of rotation. But i think i know now why i am having such a hard time. In preivious problems the point of contact would always be a fixed pivot. So no rotation of a yoyo, but a sphere as a whole would rotate around the pivot point. I think my real proplem right now is, that i am not fully understanding the meaning of the contact point having a velocity of zero

 

[BvU]

One way to progress is to consider an extreme case at the opposite end: what happens if the friction is zero ?

 

[Justforthisquestion1]

One way to progress is to consider an extreme case at the opposite end: what happens if the friction is zero ?

If the friction is zero, one could just pull the thread from the yoyo. The yoyo would rotate in the ccw direction and probably maintain its position? Thank you for being so patient!

 

[BvU]

So? What happened to $F=ma$ ?

 

[Justforthisquestion1]

So? What happened to $F=ma$ ?

Oh you are right so there would be no? rotating but the yoyo would translate in the direction of the force. So from rotating without sliding we now have moved to sliding without rotating?

 

[jbriggs444]

Oh you are right so there would be no? rotating but the yoyo would translate in the direction of the force. So from rotating without sliding we now have moved to sliding without rotating?

Why no rotation? We have an off-center force.

It is fairly easy to run the experiment with a pencil that is on a table. Flick the pencil perpendicular to its long axis. If you strike it off center, does it rotate only? Does it translate only? Does it do both?

 

[Justforthisquestion1]

Why no rotation? We have an off-center force.

It is fairly easy to run the experiment with a pencil that is on a table. Flick the pencil perpendicular to its long axis. If you strike it off center, does it rotate only? Does it translate only? Does it do both?

Looks like both? I have been sitting at this problem for too long now and i feel like there is a blockage in my head

 

[jbriggs444]

Looks like both? I have been sitting at this problem for too long now and i feel like there is a blockage in my head

Yes indeed. It is both.

Newton's second law tells us that it must translate. The object is subject to a net force. So its center of mass moves in response to that force. $F=ma$.

If we choose to consider an axis located at the center of mass, we have a net torque. So the object rotates in response to that force. $\tau = I \alpha$. (Torque = $\tau$, moment of inertia = $I$, angular acceleration = $\alpha$).

It is jarring to some students when they are told that the same force can have both effects at the same time. Both equations are correct. Both have full effect. The applied force is not diluted even though it has both effects. None of the linear impulse is "used up" for rotation. None of the rotational impulse is "used up" for linear acceleration.

 

[Justforthisquestion1]

So what is the difference between the case with friction and without friction? Do we have a slipping part without the friction?

 

[jbriggs444]

So what is the difference between the case with friction and without friction? Do we have a slipping part without the friction?

That is something that can be answered by writing down the equations and doing the math.

How rapidly would the yoyo accelerate linearly under a particular force. (Solve for $a$ of the center of mass).

How rapidly would its rotation accelerate under that same force? (Solve for angular acceleration $\alpha$).

What would the combination mean for the motion of the point on the yoyo that is momentarily touching the ground? If it is accelerating relative to the ground (and if friction can not hold it in place), then it will slip.

[There is an intuition that can be attached to this. It has to do with the sweet spot on a baseball bat]

 

[haruspex]

So what is the difference between the case with friction and without friction? Do we have a slipping part without the friction?

With friction there is a horizontal force to the left at the contact point.
This force is less than the applied force, so the yo-yo still moves right but more slowly.
But because it is further from the cm than the applied force is, it can have the greater magnitude torque about the cm, so the rotation can be clockwise instead of counterclockwise. And since there is no slipping, and motion is to the right, the rotation must indeed be clockwise.

 

[Justforthisquestion1]

Thank you all
I think i have a rough understanding of the problem now!

 

[BvU]

Care to demonstrate that by working out the exercise in post #1 ?

$\$

 

[Justforthisquestion1]

I already did :D But i already cross checked with the short solution that i have.
Rotation is kind of hard to imagine though. For some reason, i dont have any real intuition for it

 

[Justforthisquestion1]

I already did :D But i already cross checked with the short solution that i have.
Rotation is kind of hard to imagine though. For some reason, i dont have any real intuition for it

Care to demonstrate that by working out the exercise in post #1 ?

$\$

Okay i take everything back. I am trying to do the same thing now when pulling at an angle and all my solution attempts are wrong :D

 

[BvU]

Back to start ? Step zero: what is the problem statement ?
Step 1: make a drawing. Any proposals for the next step ?

 

[Justforthisquestion1]

Back to start ? Step zero: what is the problem statement ?
Step 1: make a drawing. Any proposals for the next step ?

View attachment 324694

I think i may be a step further...
Problem: What is max magnitude F so the yoyo is rolling without slipping?
Attempted solution: Since no angle is given i am assuming that i have to anticipate a cw rotation.
Again it is no slipping conditions.
For the max static friction i now anticipate:
F(s) = mu *(mg -F*sin(theta))
theta is the angle between the new F and the old one
Newton 2:
m * a = F * cos(theta) - F(s)
Torque
I(cm) * $alpha$ = F(s) * R - F * b

especially with F * b i am not sure. Is that correct`?

My next step would be to exchange a = $alpha$ * R in Newton 2 and then put that into the torque equation instead of $alpha$.

 

[BvU]

i am assuming that i have to anticipate a cw rotation

I like to play devil's advocate: also for $\theta > \pi/2$ ?

$\$

 

[Justforthisquestion1]

I like to play devil's advocate: also for $\theta > \pi/2$ ?

$\$

The angle where the direction of the rotation switches depens on b and R doesnt it? Or the ratio of those to be more precise. Walter Lewin has a video about that on youtube. I just hope i understood it correctly

 

[Justforthisquestion1]

I like to play devil's advocate: also for $\theta > \pi/2$ ?

$\$

So i finshed. Could you please tell me, if i have it correct?

F(max) = (3*mu*m*g*R) / (R*cos(theta)+3*mu*R*sin(theta)-2*b)
Please let that be correct :D

 

[BvU]

I would bet good money that $\theta = \pi/2$ causes ccw rotation

The angle where the direction of the rotation switches depens on b and R doesnt it? Or the ratio of those to be more precise.

I suppose so. Let's do the math ... you started out in #24 with cw and (I assume ) $\theta < \pi/2$. I agree with the b times F (but with one comment: clockwise rotation is considered to go in a negative angle direction, so I would write $I\alpha=b\times F - R \times F_s$ )

$\$

 

[BvU]

So i finshed. Could you please tell me, if i have it correct?

F(max) = (3*mu*m*g*R) / (R*cos(theta)+3*mu*R*sin(theta)-2*b)
Please let that be correct :D

How does this $F_{\text {max }}$ follow from #24 ?

#\ ##

 

[Justforthisquestion1]

I would bet good money that $\theta = \pi/2$ causes ccw rotation I suppose so. Let's do the math ... you started out in #24 with cw and (I assume ) $\theta < \pi/2$. I agree with the b times F (but with one comment: clockwise rotation is considered to go in a negative angle direction, so I would write $I\alpha=b\times F - R \times F_s$ )

$\$

I chose my coordinate system with i_hat to the right j_hat down. Doesnt that make cw positive? Otherwise i would also have to correct the other lines?

So i would now have
F(s) = mu *(- mg + F*sin(theta))

 

[Justforthisquestion1]

How does this $F_{\text {max }}$ follow from #24 ?

#\ ##

First of all i used $a$ = $alpha$ * $R$
So i switched the $a$ in m * $a$ in Newtons second law for $alpha$ * $R$
so now i have
$m * alpha * R$ = F * cos(theta) - F(s)
then i simply adjust that to
$alpha$ = (F * cos(theta) - F(s)) / (m * R)

Next i insert this $alpha$ in line 24 and solve for F

 

[BvU]

I chose my coordinate system with i_hat to the right j_hat down. Doesnt that make cw positive?

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

$\$

 

[Justforthisquestion1]

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

$\$

Almost!
The -2b are the problem!
it should be +2b!
Thank you! i didnt think of this easy yet effective method to test this solution
Do you have any idea where the mistake could come from? If not then i will write my solution on a paper and scan it!

 

[Justforthisquestion1]

Almost!
The -2b are the problem!
it should be +2b!
Thank you! i didnt think of this easy yet effective method to test this solution
Do you have any idea where the mistake could come from? If not then i will write my solution on a paper and scan it!

In my mind, the solution for $theta = 0$ should be
##Fmax= (mu * m * g * 3*R) / (2 * b +R)

 

[Justforthisquestion1]

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

$\$

Got it thank you so much
Now i can forget about yoyos forever!!!