A rigid body preserves the distance between every pair of its material points. In planar motion, the body's motion at any instant can be decomposed into translation of one chosen reference point plus rotation about that point.

Let $A$ and $B$ be two points fixed in the same rigid body. Their positions satisfy

$$\mathbf r_B=\mathbf r_A+\mathbf r_{B/A},$$

where $\mathbf r_{B/A}$ has constant magnitude because the body is rigid, although its direction can rotate.

Velocity relation

Differentiate the position relation. For a vector fixed in a body rotating with angular velocity $\boldsymbol\omega$,

$$\frac{d\mathbf r_{B/A}}{dt}=\boldsymbol\omega\times\mathbf r_{B/A}.$$

Therefore

$$\boxed{\mathbf v_B=\mathbf v_A+\boldsymbol\omega\times\mathbf r_{B/A}}.$$

The first term is the translation shared through the chosen reference point. The second is the velocity of $B$ relative to $A$ caused by the body's rotation.

In planar motion, $\boldsymbol\omega$ is perpendicular to the plane, so the relative velocity

$$\boldsymbol\omega\times\mathbf r_{B/A}$$

is perpendicular to the line joining $A$ and $B$, with magnitude

$$\boxed{v_{B/A}=|\omega|r_{B/A}}.$$

Acceleration relation

Differentiate the velocity relation:

$$\mathbf a_B =\mathbf a_A +\frac{d\boldsymbol\omega}{dt}\times\mathbf r_{B/A} +\boldsymbol\omega\times\frac{d\mathbf r_{B/A}}{dt}.$$

With angular acceleration

$$\boldsymbol\alpha=\frac{d\boldsymbol\omega}{dt}$$

and

$$\frac{d\mathbf r_{B/A}}{dt} =\boldsymbol\omega\times\mathbf r_{B/A},$$

we obtain

$$\boxed{\mathbf a_B =\mathbf a_A +\boldsymbol\alpha\times\mathbf r_{B/A} +\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{B/A})}.$$

The rotational part contains two geometrically different terms.

The tangential relative acceleration is

$$\boldsymbol\alpha\times\mathbf r_{B/A},$$

with magnitude

$$|\alpha|r_{B/A}.$$

The normal or centripetal relative acceleration is

$$\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{B/A}),$$

which points from $B$ toward $A$ and has magnitude

$$\omega^2r_{B/A}.$$

Worked example: translating and rotating bar

A rigid bar moves in the $xy$-plane. At one instant point $A$ has velocity

$$\mathbf v_A=3\hat{\mathbf x},\mathrm{m/s}.$$

Point $B$ is located

$$\mathbf r_{B/A}=0.50\hat{\mathbf x},\mathrm m$$

from $A$, and the bar rotates counterclockwise with

$$\boldsymbol\omega=4\hat{\mathbf z},\mathrm{rad/s}.$$

Then

$$\boldsymbol\omega\times\mathbf r_{B/A} =(4\hat{\mathbf z})\times(0.50\hat{\mathbf x}) =2\hat{\mathbf y},\mathrm{m/s}.$$

Therefore

$$\boxed{\mathbf v_B =3\hat{\mathbf x}+2\hat{\mathbf y},\mathrm{m/s}}.$$

Point $B$ does not simply share the translational velocity of $A$: rotation adds a perpendicular component.

Suppose at the same instant

$$\mathbf a_A=1\hat{\mathbf x},\mathrm{m/s^2}$$

and

$$\boldsymbol\alpha=6\hat{\mathbf z},\mathrm{rad/s^2}.$$

The tangential rotational contribution is

$$\boldsymbol\alpha\times\mathbf r_{B/A} =(6\hat{\mathbf z})\times(0.50\hat{\mathbf x}) =3\hat{\mathbf y},\mathrm{m/s^2}.$$

The normal contribution is

$$\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{B/A}) =-\omega^2\mathbf r_{B/A} =-8\hat{\mathbf x},\mathrm{m/s^2}.$$

Thus

$$\boxed{\mathbf a_B=-7\hat{\mathbf x}+3\hat{\mathbf y},\mathrm{m/s^2}}.$$

Choosing the reference point

The relations are valid for any two material points of the same rigid body. A convenient reference point can simplify a problem:

• the center of mass is useful when translation and rotation are both important;
• a fixed pivot has $\mathbf v_A=\mathbf0$ and $\mathbf a_A=\mathbf0$;
• in pure rolling, the instantaneous contact point can simplify velocity calculations, although its acceleration is generally not zero.

The key idea is that the velocities and accelerations of different points of a rigid body are constrained by one common angular velocity and angular acceleration. Translation and rotation are not separate motions assigned arbitrarily to each point; together they determine the motion of the entire rigid body.