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According to the physics of a wheel rolling without slipping, the topmost point moves twice as fast as the wheel.

But I tried an experiment:

Take a wheel on a table and hold a ruler horizontally in contact with the top of the wheel. Roll the wheel without slipping by pushing the ruler forward, such that the center of the wheel moves forward by $10 \,\text{cm}$.

But the ruler was supposed to move $20 \,\text{cm}$, isn't it? But it didn't, it just seemed to move $10 \,\text{cm}$. Am I interpreting something wrong?

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    $\begingroup$ Velocity means "compared to something else". Just clarify all the velocities involved $\endgroup$
    – Fattie
    Commented May 7 at 18:49

3 Answers 3

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The ends of the ruler will indeed move forward by $20 \,\text{cm}$ with respect to the table. However, the point of contact between the wheel and the ruler will only move by $10 \,\text{cm}$ with respect to the ruler. In other words, if the wheel was initially touching the $0 \,\text{cm}$ mark on the ruler, it will be touching the $10 \,\text{cm}$ mark after the rolling. The other $10 \,\text{cm}$ is precisely because the wheel itself has rolled forward.

So the ruler moves forward $10 \,\text{cm}$ with respect to the wheel and the wheel moves forward $10 \,\text{cm}$ with respect to the table which gives $20 \,\text{cm}$ in total. There is no contradiction.

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Vincent Thacker has explained why your experiment got the results it did. I'm going to explain the actual movement of the wheel.

Dissecting the movement of the wheel

There are 2 things happening: the wheel is rotating, and it's moving forwards.

Rotation

Imagine a wheel hanging freely on a stationary axle. If we assume the wheel is not deforming, we can say that every part of the surface of the wheel is moving at the same speed, perpendicular to the direction to the center of the wheel. This always applies, but it's relative to the center of the wheel.

Forward movement

Now imagine a freely rolling wheel. If it's not slipping, the contact patch at bottom of the wheel has, by definition, 0 relative velocity to the surface. It's stationary. If the wheel is moving at speed $v$, then the bottom of the wheel must have a relative velocity to the center of the wheel of $-v$, or a relative speed of $v$ with the velocity vector pointing backwards.

Conclusion

If we put the two together, we determine that the top of the wheel must also have a relative speed of $v$, but at the top of the wheel, the velocity vector points forwards. (it must, or eventually there will be no rubber to take the place at the bottom as the wheel rotates)

This is a velocity relative to the center of the wheel, so to find the velocity relative to the surface, we have to add the velocity of the center of the wheel, and find a velocity of $2v$.

Addendum

Another way of looking at it is that if the wheel as a whole moves at $v$, then every part of it must, in the direction of travel, move at $v$ on average, or it won't stay with the rest of the wheel. At the bottom (again assuming no slip), it must come to a stop, so it must move at a higher speed at other points of the rotation. Given that a wheel's rotation is uniform, a reasonable non-rigorous assumption would be that the velocity varies symmetrically around the average speed of $v$ ($= 1*v$), so between $0 = 0*v$ and $2*v$.

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Not sure I understand what you mean be "pushing the ruler forward". You probably did not measure the effect you meant to measure.

First let me say that what you describe is only for wheels rolling without slipping. The latter statement means that the lowest point, that touches the floor actually has instantaneous velocity zero at that time. But only for that instance. Just before or just after touching the floor it will have finite velocity.

Realizing that the lowest point is at zero velocity, and the center moves with velocity v makes it perhaps more intuitive that the top point has to move with 2v.

Mathematically it is even simpler. In the reference frame pinned to the axle of the wheel the top point moves with $v= \omega R$, the center is at velocity zero by definition, and the bottom point has velocity $-v$. Now you do a Galilei transformation to a frame of reference fixed to the floor. In that frame the center moves with $v = 2\pi R/T = \omega R$ where $T$ is the period of rotation and the axle needs to cover a distance equal to the circumference. Since velocities are additive in Galilei transformations the velocities at the bottom and top are now $-v+v =0$ and $v+v=2v$.

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