# How can different points on a rigid body move with different speeds but also be relatively at rest?

For a rigid body rotating with a constant angular speed, the points near the axis must have lower linear velocity than the points farther away. If they have different linear velocities, they must have a non-zero relative velocity.

If they have a non-zero relative velocity, the distance between them would change over time. But the distance between any two particles in a rigid body has to remain fixed while it moves. How is this possible?

EDI- Instead of the scalar distance, let's talk about the position vector of a ball that I rotate using a string. If I also rotate my body along with it with the same angular velocity, I'd find the ball to be at rest in my point of view. If there was a relative velocity $$\vec{v}$$, wouldn't the position vector of the ball change given by $$\vec{r(t+dt)}=\vec{r_0}+\vec{v}dt$$?

• Could you say more about both how a non-zero relative velocity makes it inevitable the distance must change and how, specifically, time affects that? The distance between particles in a rigid body does remain constant, but the particles clearly have different speeds and velocities. Aug 14, 2020 at 23:14
• What does the word "linear" in your phrase "linear velocity" mean to you? Aug 16, 2020 at 14:02

Having a non-zero relative velocity is fine as long as the distance between the points isn't changing. This certainly holds for a rotating rigid body. As another example, take a ball on a string and rotate it in a horizontal circle. Is the ball moving relative to you? Yes. Is it moving towards or away from you? No.

Therefore this part

If they have a non-zero relative velocity, the distance between them would change over time.

is the invalid step. This is not necessarily true, and it isn't true for rigid bodies.

See Mike Stone's answer for a simple geometric "proof" of this.

EDIT- Instead of the scalar distance, let's talk about the position vector of a ball that I rotate using a string. If I also rotate my body along with it with the same angular velocity, I'd find the ball to be at rest in my point of view. If there was a relative velocity $$\vec v$$ , wouldn't the position vector of the ball change given by $$\vec r(t+dt)=\vec r_0+\vec v\,\text dt$$?

Yes, if you are rotating with the ball then you would observe the ball to be at rest. You will be in what is called a non-inertial frame of reference. It is non-inertial because it is rotating (accelerating). In this frame of reference you would see a constant position vector for the ball and a $$0$$ velocity vector.

• Comments are not for extended discussion; this conversation has been moved to chat. Aug 14, 2020 at 21:15

If the velocity of point B relative to point A is always at right angles to the line AB joining them, then the distance does not change.

• Two particlesA&B, A revolving about B . Will A see that B is revolving around it? Aug 14, 2020 at 14:59

It seems the sticking point is your notion that non-zero relative velocity implies changing distance. To see that this is not the case, consider a car. When you turn in a car, the outer tire moves faster with respect to the road than the inner tire, i.e. the two tires have non-zero relative velocities. Yet, the car does not fall apart.

The reason is that the relative velocity of the tires is perpendicular to the separation vector.

To prove this, let $$\vec{r}_{AB} \equiv \vec{r}_B - \vec{r}_A$$ be the separation vector from object A to object B. We compute \begin{align} \frac{d}{dt} ||\vec{r}_{AB} || &= \frac{d}{dt} \sqrt{\vec{r}_{AB} \cdot \vec{r}_{AB}} = \frac{1}{2\sqrt{\vec{r}_{AB}\cdot\vec{r}_{AB}}} (2 \dot{\vec{r}}_{AB} \cdot \vec{r}_{AB}) = \frac{\dot{\vec{r}}_{AB} \cdot \vec{r}_{AB}}{||\vec{r}_{AB}||} \end{align} From which it follows $$\frac{d}{dt} ||\vec{r}_{AB}|| = 0 \iff \dot{\vec{r}}_{AB} \cdot \vec{r}_{AB} = 0$$ Which is to say that two objects with a fixed distance can have a relative velocity. In fact, this is possible precisely when the relative velocity is orthogonal to the separation vector.

To see that the relative velocity really is perpendicular to the separation vector in a rigid body, note that in a rigid body (in a frame comoving with the center of mass) $$\dot{\vec{r}}_A = \omega \times \vec{r}_A$$ and $$\dot{\vec{r}}_B = \omega \times \vec{r}_B$$. Thus $$\dot{\vec{r}}_{AB} = \dot{\vec{r}}_B - \dot{\vec{r}}_A = \omega \times \vec{r}_{B} - \omega \times \vec{r}_{A} = \omega \times (\vec{r}_{B} - \vec{r}_{A}) = \omega \times \vec{r}_{AB}$$ so $$\dot{\vec{r}}_{AB} \cdot \vec{r}_{AB} = (\omega \times \vec{r}_{AB}) \cdot \vec{r}_{AB} = 0$$

To summarize: 1) It is possible for two objects with a fixed distance to have a relative velocity; they need only to have a relative velocity perpendicular to their separation vector. 2) Points on a rigid body move with a relative velocity that is perpendicular to their separation vector.

you can see it also with these equations:

assuming all vectors are perpendicular thus:

$$\omega=\frac{r_1\,v_1}{r_1^2}=\frac{r_2\,v_2}{r_2^2}\tag 1$$

$$\Rightarrow$$

$$r_1=\frac{v_1}{v_2}\,r_2$$ and $$r_1-r_2=\frac{v_1}{v_2}\,r_2-r_2=r_2\frac{v_1-v_2}{v_2}=\text{constant}\tag 2$$

thus $$\frac{v_1-v_2}{v_2}$$ must be constant.

with:

$$v_1=\omega\,r_1~,v_2=\omega\,r_2$$

$$\Rightarrow$$ $$\frac{v_1-v_2}{v_2}=\frac{r_1-r_2}{r_2}=\text{constant}$$

I haven't seen it described, so I'll add: it all depends on the "observer", or the frame in which you are describing the motion.

Imagine a fly sitting on a frisbee traversing through the sky. If we strap the fly to the frisbee and it keeps its eyes open, every point on the frisbee will remain stationary from its point of view. However, different points on the frisbee will move at different speeds relative to the ground, basically depending on the speed of rotation and their location on the frisbee.

In more mathy terms, the velocity of a point on the frisbee relative to any other point on the frisbee, expressed in a frame fixed on the frisbee is zero. The velocities of those two different points expressed in a "ground" frame would be different.

Remember that in vector kinematics, the way you take a derivative is by using the transport theorem, $$^A\frac{d}{dt}(\bar{r}) = ^B\frac{d}{dt}(\bar{r}) + \bar{\omega}_{B/A} \times \bar{r}$$ where the super-indices reflect the frame you're taking the derivative in - or where the "observer" is sitting.