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For a wheel in pure rotational motion; Does the center rotate? Is there a point in wheel which do not rotate? What can we say about angular velocity of the center? From where we can start to use the term "center" for that area?

Thanks.

Ali Aslantürk
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  • At what scale do you mean, in particular, at the atomic level? –  Jul 12 '15 at 20:09
  • @AcidJazz I think, eventually It must be thought at the atomic level. So, yes at the atomic level. However, I'd like to learn other side too. – Ali Aslantürk Jul 12 '15 at 20:14

3 Answers3

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Just to clarify there are two different cases:- 1) Purely rotational motion of a wheel where there is no translation of the centre of mass, here assumed to be the centre of the wheel. In this case the centre remains fixed and doesn't rotate. Angular velocity is defined for the entire rigid body as a whole, so it is the same for the centre as for any other point.

2) The other case which you may possibly consider is the rolling motion of a wheel. This is the case when the centre of the wheel is also translating in space, just in the way a car or bullock cart moves. The rolling without slipping condition implies the stationary point on the wheel is actually the lowermost point, which is motionless with respect to the ground. The centre does rotate in this case

Pratik Rath
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  • So, centre is conceptual and from definition that says angular velocity is same for every part we say centre does have angular velocity. – Ali Aslantürk Jul 12 '15 at 20:22
  • The answer by sintetico is much more elaborate taking into consideration all possibilities. My answer is true for a rigid body which I assumed your wheel to be. – Pratik Rath Jul 14 '15 at 10:00
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You know the law of motion that states that the net forces applied on a body affect the linear motion of the center of mass only. Complementary to this is the law that the net torques on body about the center of mass affects the rotational motion of the body.

Rotation is common to all points on a body so if any part rotates so does the center of mass. To check for rotation of a body add up all the applied torques and equipollent forces (troque due to force at distance) and if they add up to zero then there is no change in the rotational momentum. Similary of the all the applied forces add up to zero then there is no change in linear momentum.

In 3D vector form the above laws are often stated together as:

$$\begin{aligned} \sum_i (\bf{F}_i)=\sum_i (\bf{F}_{cm}) &= \frac{{\rm d}} {{\rm d}t}(m {\bf v}_{cm})\\ \sum_i (\bf{\tau}_i + \bf{r}_i \times \bf{F}_i)= \sum_i (\bf{\tau}_{cm}) &= \frac{{\rm d}} {{\rm d}t}(I_{cm} {\bf \omega}) \end{aligned}$$

where $\bf \tau_i$ are applied torques, $F_i$ are applied forces located at $\bf r_i$ relative to the center of mass, and $m$ and $I_{cm}$ are the mass and mass moment of inertia of the body about the center of mass.

John Alexiou
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The angular velocity is not well defined for the center of a wheel, or for the center of mass of a rigid body, in the same sense that the number $0/0$ is indeterminate. This is a consequence of the fact that the center of mass does not move if the motion is purely rotational. To see this, let us start ab ovo...

The angular velocity is defined as

The angular velocity of a particle is measured around or relative to a point, called the origin.

(Wikipedia)

You see that the angular velocity of any particle or of any point in a rigid body is defined with respect with the origin, which is usually the center of mass, which is at rest in the case of a purely rotational motion. More precisely, one can define the angular momentum of a point of mass $m$ with respect to a chosen origin $O$ as $$ \boldsymbol{\omega}= \frac{\mathbf{r}\times \mathbf{v}}{r^2}= \frac{\mathbf{r}\times\mathbf{p}}{r m^2}= \frac{\mathbf{L}}{r^2 m} $$ where $\mathbf{r}$ is the position measured from the origin $O$. On the other hand $\mathbf{v}$, $\mathbf{p}=m\mathbf{v}$, $\mathbf{L}$ are the velocity, momentum, and angular momentum, which do not depend on the choice of the origin $O$ of the reference frame. Now, what happens if we consider another reference frame, with a different origin $O$? The angular velocity will change. To see this, imagine to calculate the angular velocity of a wheel in uniform and purely rotational motion with respect to its center, and respect to a point $O$ external to the wheel. The former is constant, the latter is not.

We can finally come back to your question. What is the angular velocity of the center $C$ of a wheel in a purely rotational motion? We again have to specify the origin $O$. Well if the origin is different from the center $O\neq C$, the angular velocity is zero, since $\mathbf{v}=0$ and $\mathbf{r}\neq0$. If instead we calculate the angular velocity of the center with respect to the center $O=C$, the angular velocity is simply not defined, since one has $\mathbf{r}= \mathbf{v}=0$, which gives the indeterminate expression $0/0$ in the equation above.

Another observation. The angular velocity is a local quantity: it can be intended as a global properties only in the case of rigid bodies, where the rotation is, by definition, rigid.

sintetico
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