# Why doesn't a yo-yo come back to its original height?

1. Nov 19, 2013

### geoduck

Does a yo-yo lose all its translational kinetic energy when it reaches the bottom and turns around?

It seems at the bottom the string should give the axle of the yo-yo a violent tug, so maybe some of the translational kinetic energy is converted to rotational kinetic energy and is not lost as tension waves in the string?

Or is all translational kinetic energy lost, and what causes the yo-yo to climb is a conversion of its remaining energy (rotational) to gravitational potential energy?

2. Nov 19, 2013

### gabriel.dac

Friction. The yo-yo will be attached to a string, won't it?

3. Nov 19, 2013

### geoduck

If you drop the yo-yo without winding it, then the kinetic energy is lost when the yo-yo reaches its lowest point.

If you wind it, the yo-yo drops slower, but it starts spinning. It spins and moves downward at the same time.

I suspect that the downward motion is completely lost when the yo-yo reaches its lowest point. But the spinning remains. This spinning allows the yo-yo to climb back up.

So maybe the reason the yo-yo doesn't reach the height it started is because it lost energy at its lowest point, rather than along the way?

It might be the case that no energy is lost as the lowest point, and friction as you say is responsible. I just want to make sure.

4. Nov 19, 2013

### CWatters

At the bottom there are TWO possibilities...

If you want the yo-yo to stay down there then you allow the axel to spin in the loop at the end of the string. Most of the PE it had at the top will be converted to rotational kinetic energy (0.5ω2). The string stops it falling further so any linear KE it get will be absorbed by the string stretching or the persons arm etc.

If you want the yo-yo to come back up you let the string trap itself so that the yo-yo winds itself back up the string (PE at the top is converted to rotational KE at the bottom and then back again to PE as it climbs back up)

In both cases some energy will be lost to friction between the string and the inside of the slot and the axel of the yo-yo. Plus whatever is absorbed by the string/arm.

5. Nov 19, 2013

### Staff: Mentor

Ideally, a yoyo with a negligible diameter axle will convert all of its GPE to rotational KE at the bottom, losing nothing. So a real yoyo loses energy to the jerk (inelasticity of the string) and friction.

6. Nov 19, 2013

### phinds

No, friction is why it CAN come back up (most of the way). It doesn't come all the way back up because as Russ pointed out, it loses energy at the jerk at the bottom of its travel. If there were no friction between the string and the axle then it wouldn't come back up at all, just lose some energy to the jerk and then convert the rest to rotational energy.

7. Nov 19, 2013

### Staff: Mentor

Let's not forget air viscosity, it slows down the rotation. You can put it into the "friction" drawer.

It would be interesting to compare how high back yo-yo goes in the air and in the vacuum.

8. Nov 19, 2013

### CWatters

I think we (or at least I) was talking about friction between the string and the sides of the slot as a source of loss.

9. Nov 19, 2013

### phinds

Ah. Good point. I wasn't thinking of that.

10. Nov 19, 2013

### phinds

I certainly agree that air friction is a factor but I would expect it to be nearly negligible compared to the jerk at the bottom.

EDIT: OK, wait ... I'm thinking of a yo-yo that has been given a significant downward force but the hand/wrist but of course what we're talking about here is one that is just released. I don't have a good feel for whether or not that would have a jerk that would be strong enough to make air friction negligible by comparison so maybe I've got that wrong.

11. Nov 20, 2013

### geoduck

Assuming the rope does not slip, isn't there a linear relationship between the rotational KE and the translational KE of the yo-yo on its descent?

v=rω
v2=r2ω2
.5 m v2=.5m r2ω2
KEtrans=KErot/(Imr^2)

where I is the moment of inertia?

Both the rotational and translational KE should be max just before it reaches the bottom, and somehow when it reaches the bottom, the translational KE is gone. If it gets converted entirely to rotational KE, what is the mechanism behind this?

12. Nov 20, 2013

### jbriggs444

if the string is tied firmly to the yoyo's axle [which, in a real yoyo, it will not be] then the jerk at the bottom of the yoyo's stroke involves a torque. As the yoyo decellarates to zero linear momentum with the attachment point on the one side of the axle and rotating toward the middle, this torque will be accellerating the yoyo's rotation. As the yoyo accelerates to non-zero upward linear momentum with the attachment point rotating on toward the opposite side, this will be decellerating the yoyo's rotation.

Instead of a sudden jerk that dissipates energy, there is a smooth transition that, at least in the ideal case of zero friction and massless strings, trades linear kinetic energy for rotational kinetic energy with no losses.

Except for the one instant where the string exactly lines up with the axis of rotation there is no problem sourcing or sinking angular momentum into the string, of course.

13. Nov 20, 2013

### 256bits

I would agree. I was going to say the yo-yo acts as a pendulum at the very bottom with a movable pivot. I suppose the length of string would have an effect upon whether the horizontal translation of the pivot augments the transition since a wave would be set up in the string.

14. Nov 20, 2013

### A.T.

You can fix the string to the axle without relying on friction.