Solve Unwinding Yo-Yo Homework: No Slipping Allowed

[rcgldr]

To clarify the issue here, the yo-yo outer surface has to be spinning faster than it's moving down the incline. The kinetic friction would have to be relatively low and/or the angle relatively steep in order for the yo-yo to move at all. So if there is kinetic friction, what is the direction of friction force exerted by the incline onto the yo-yo?

 

[ChrisBrandsborg]

To clarify the issue here, the yo-yo outer surface has to be spinning faster than it's moving down the incline. The kinetic friction would have to be relatively low and/or the angle relatively steep in order for the yo-yo to move at all. So if there is kinetic friction, what is the direction of friction force exerted by the incline onto the yo-yo?

F_fr = μk⋅Mgsinθ ? (Perpendicular to the normal force and in the direction opposite to the motion)

 

[TomHart]

Okay, I went out today and bought a yo-yo - $3 Duncan. I was kind of afraid to try it because, if I was wrong . . .

What I found was that, on a surface with friction, the yo-yo would not roll down an incline when I held the string as shown in the original figure for this problem. (And yes, I made sure that the yo-yo was positioned such that the string was on the lower side of the small (center) disk, and not on the upper side, as that would be a much different situation.) As a matter of fact, and it was somewhat surprising (although it shouldn't have been), I could roll (without slipping) the yo-yo uphill with the string as shown in the figure - the string winding itself up on the yo-yo as it went. Of course, if the angle was too steep, or if I pulled too hard, the yo-yo would slip.

 

[haruspex]

For a), it might be easier to think in terms of the instantaneous centre of rotation. When a wheel rolls, at any given instant, the wheel is rotating about the point of contact with the ground. So for the purposes of part a) you can replace the yo-yo with a stick standing normally to the slope. At the bottom it is in contact with the slope, at the top tied to the string. Can it move without slipping?

For b and c, as you note, there is no friction. So yes, there will certainly be angular acceleration. What will be the instantaneous centre of rotation now, i.e. what part of the yo-yo is, for the instant, unable to move?

 

[ChrisBrandsborg]

The Torque comes from the string force, right? But how do we calculate the angular velocity when it rolls and slides? Because then energy isn't conserved?

 

[rcgldr]

The Torque comes from the string force, right? But how do we calculate the angular velocity when it rolls and slides? Because then energy isn't conserved?

Friction force = kinetic coefficient of friction times normal force, and yes it consumes energy (so energy conservation approach won't help here). What helps here is you can assume the kinetic coefficient of friction remains the same regardless of how fast the yo-yo is spinning relative to the incline. To follow up on my prior comment / question, the direction of the kinetic friction force exerted by the incline onto the yo-yo is down the incline. I haven't done the math yet, but I suspect that the friction decreases angular acceleration of the yo-yo while increasing the tension in the string, and despite the fact that the kinetic friction force is down the incline, I suspect the overall result is that the yo-yo ends up moving slower with kinetic friction than without.

To solve the problem, I would focus on the angular and linear accelerations. Angular acceleration = the torque exerted by the string which is opposed by the kinetic friction force from the incline, and the net torque divided by the angular inertia. The linear acceleration of the center of mass of the yo-yo is related to the rate of increase of the distance of the unwinding string, and is related to the net downwards force (= m g sin(θ) + friction force - tension in string) divided by the linear momentum. Have you figured out the case for when the kinetic friction force is zero?

 

[ChrisBrandsborg]

Friction force = kinetic coefficient of friction times normal force, and yes it consumes energy (so energy conservation approach won't help here). What helps here is you can assume the kinetic coefficient of friction remains the same regardless of how fast the yo-yo is spinning relative to the incline. To follow up on my prior comment / question, the direction of the kinetic friction force exerted by the incline onto the yo-yo is down the incline. I haven't done the math yet, but I suspect that the friction decreases angular acceleration of the yo-yo while increasing the tension in the string, and despite the fact that the kinetic friction force is down the incline, I suspect the overall result is that the yo-yo ends up moving slower with kinetic friction than without.

To solve the problem, I would focus on the angular and linear accelerations. Angular acceleration = the torque exerted by the string which is opposed by the kinetic friction force from the incline, and the net torque divided by the angular inertia. The linear acceleration of the center of mass of the yo-yo is related to the rate of increase of the distance of the unwinding string, and is related to the net downwards force (= m g sin(θ) + friction force - tension in string) divided by the linear momentum. Have you figured out the case for when the kinetic friction force is zero?

But the in problem a-c we assume that there are no friction, which makes the yoyo- roll with sliding.
So we cannot use energy-conservation to calculate ω. How can we then do it?

 

[jbriggs444]

But the in problem a-c we assume that there are no friction, which makes the yoyo- roll with sliding.
So we cannot use energy-conservation to calculate ω. How can we then do it?

One can still use energy conservation to calculate ω. One must account for the energy of at least three types. Which types can you think of?

 

[rcgldr]

But the in problem a-c we assume that there are no friction, which makes the yoyo- roll with sliding. So we cannot use energy-conservation to calculate ω.

For no friction case, you can use conservation of energy. I was referring to the case where there is friction, where you will need to use a different approach.

 

[jbriggs444]

For no friction case, you can use conservation of energy. I was referring to the case where there is friction, where you will need to use a different approach.

Unless I am missing something, an accounting based on energy still works. There's just another bucket to account for.

 

[ChrisBrandsborg]

One can still use energy conservation to calculate ω. One must account for the energy of at least three types. Which types can you think of?

Potential at the top, Kinetic at the bottom (rotational and linear) and thermal energy.

 

[jbriggs444]

Potential at the top, Kinetic at the bottom (rotational and linear) and thermal energy.

Yes, that should do it. You should have enough information to quantify each of these.

 

[rcgldr]

Potential at the top, Kinetic at the bottom (rotational and linear) and thermal energy.

Yes, that should do it. You should have enough information to quantify each of these.

I'm not sure how you quantify the thermal energy component.

The forces and torques can be defined with the given data, angular acceleration is proportional to linear acceleration, so the equation for angular acceleration can be combined with the equation for linear acceleration to produce a single equation to solve for.

 

[jbriggs444]

I'm not sure how you quantify the thermal energy component.

The forces and torques can be defined with the given data, angular acceleration is proportional to linear acceleration, so the equation for angular acceleration can be combined with the equation for linear acceleration to produce a single equation to solve for.

I had in mind figuring the normal force, multiplying by the coefficient of friction to get sliding force and computing work done from there.

 

[rcgldr]

I'm not sure how you quantify the thermal energy component.

I had in mind figuring the normal force, multiplying by the coefficient of friction to get sliding force and computing work done from there.

You'd also need to calculate the "distance", which I'll leave up to Chris to figure out. So either an energy approach or a force / torque / acceleration approach should work. The energy approach may be simpler.

 

[ChrisBrandsborg]

You'd also need to calculate the "distance", which I'll leave up to Chris to figure out. So either an energy approach or a force / torque / acceleration approach should work. The energy approach may be simpler.

Distance is d, but do you mean the height?
h = dsinθ

 

[jbriggs444]

Distance is d, but do you mean the height?
h = dsinθ

No, not the height.

What @rcgldr is hinting at is that in order to calculate work, you have to know the distance that the two surfaces slip past one another. If this were a simple case of a block sliding down a ramp, that distance would simply be the length of the slope. But the yo-yo is rotating as it slides. That makes a difference.

 

[ChrisBrandsborg]

No, not the height.

What @rcgldr is hinting at is that in order to calculate work, you have to know the distance that the two surfaces slip past one another. If this were a simple case of a block sliding down a ramp, that distance would simply be the length of the slope. But the yo-yo is rotating as it slides. That makes a difference.

Yeah, but how do we calculate the sliding length? I don't think I have calculated that before. Most of our problems have been rolling without sliding..

 

[jbriggs444]

Yeah, but how do we calculate the sliding length? I don't think I have calculated that before. Most of our problems have been rolling without sliding..

Hint: If you wanted to calculate how fast a point on the surface of the yo-yo was sliding past a point on the ground where the two surfaces touched, how would you go about it?

[Or you could use the @rcgldr approach of calculating accelerations based on forces and torques, thereby sidestepping the need to calculate that distance]

 

[rcgldr]

Yeah, but how do we calculate the sliding length?

I'm not sure if the motion of the yo-yo is understood at this point. Given the variables r and R, how does angular velocity ω relate to the linear velocity of the yo-yo? It may be easier to consider to understand this assuming a frictionless incline.