# Rigid bodies

A rigid body is an extended area of material that includes all the points inside it, and which moves so that the distances and angles between all its points remain constant. The location of a rigid body can be described by the position of one point $P$ inside it, together with the rotation angle of the body (one angle in 2D, three angles in 3D).

location description velocity description
point mass position vector $\vec{r}_P$ velocity vector $\vec{v}_P$
rigid body in 2D position vector $\vec{r}_P$
angle $\theta$
velocity vector $\vec{v}_P$
angular velocity $\omega$
rigid body in 3D position vector $\vec{r}_P$
angles $\theta,\phi,\psi$
velocity vector $\vec{v}_P$
angular velocity vector $\vec{\omega}$

Neither point masses nor rigid bodies can physically exist, as no body can really be a single point with no extent, and no extended body can be exactly rigid. Despite this, these are very useful models for mechanics and dynamics.

## Rotation angles for rigid bodies

All points on a rigid body have the same angular rotation angles, as we can see on the figure below. Because the angular velocity is the derivative of the rotation angles, this means that every point on a rigid body has the same angular velocity $\vec{\omega}$, and also the same angular acceleration $\vec{\alpha}$.

In 2D the angle $\theta$ of a rigid body the angle of rotation from a fixed reference (typically the $\hat\imath$ direction), measured positive counter-clockwise. The angular velocity is $\omega = \dot\theta$ and the angular acceleration is $\alpha = \dot\omega = \ddot\theta$. The vector versions of these are $\vec\omega = \omega\,\hat{k}$ and $\vec\alpha = \alpha\,\hat{k}$, where $\hat{k}$ is the out-of-plane direction.

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All points on a rigid body rotate at the same rate. Body $\mathcal{B}_1$ is rotated by angle $\theta_1$, so its angular velocity is $\omega_1 = \dot{\theta}_1$, and similarly for the second body.

Warning: Labeling angular quantities on rigid bodies

Angular quantities ($\theta$, $\omega$, $\alpha$) for rigid bodies should not be labeled with points. If $P$ and $Q$ are points on a rigid body, we do not write $\omega_P$ or $\omega_Q$ for the angular velocity about these points.

Instead, we label angular quantities according to the body. If we have two rigid bodies $\mathcal{B}_1$ and $\mathcal{B}_2$, then $\omega_1$ is the angular velocity of the first body (and all points on it) and $\omega_2$ is the angular velocity of the second body.

Rotations in 3D are significantly more complicated than rotations in 2D. Unlike positions, velocities, etc, which simply go from 2D vectors to 3D vectors, rotational quantities go from scalars in 2D to full 3D vectors in 3D. Angular velocity and angular acceleration are somewhat straightforward, so equations #rkg-er hold in both 2D and 3D, but understanding the rotations themselves is significantly more complicated. There are three main ways that 3D rotations can be represented:

Detailed study of rotations in 3D is necessary for a full understanding of topics ranging from satellite attitude control to articulated robot construction, and is usually covered in advanced dynamics courses.

## Points on rigid bodies

If we know how one point $P$ on a rigid body is moving and we also know how the body is rotating, then we can calculate the movement of any other point $Q$ on the same body. The formulas for this are given below.

Rigid body point relations.

\begin{aligned} \vec{r}_Q &= \vec{r}_P + \vec{r}_{PQ} \\ \vec{v}_Q &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PQ} \\ \vec{a}_Q &= \vec{a}_P + \vec{\alpha} \times \vec{r}_{PQ} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ}) \end{aligned}

Points $P$ and $Q$ are two locations on a rigid body. Vectors $\vec{\omega}$ and $\vec{\alpha}$ are the angular velocity and angular acceleration of the rigid body.

The first equation is simply the definition of the offset vector $\vec{r}_{PQ}$. Differentiating the first equation gives \begin{aligned} \vec{r}_Q &= \vec{r}_P + \vec{r}_{PQ} \\ \dot{\vec{r}}_Q &= \dot{\vec{r}}_P + \dot{\vec{r}}_{PQ} \\ \vec{v}_Q &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PQ}, \end{aligned} where the derivative of $\vec{r}_{PQ}$ comes from the rotation formula, given that this offset vector is simply rotating with the rigid body.

If we differentiate again then we obtain \begin{aligned} \vec{v}_Q &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PQ} \\ \dot{\vec{v}}_Q &= \dot{\vec{v}}_P + \dot{\vec{\omega}} \times \vec{r}_{PQ} + \vec{\omega} \times \dot{\vec{r}}_{PQ} \\ \vec{a}_Q &= \vec{a}_P + \vec{\alpha} \times \vec{r}_{PQ} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ}), \end{aligned} where we use the fact that $\vec{\alpha} = \dot{\vec{\omega}}$.

Warning: Cross product order

Because cross products are not associative, it is very important to compute the centripetal acceleration term with the parentheses as shown. That is, we must not compute $(\vec{\omega} \times \vec{\omega}) \times \vec{r}_{PQ}$, as this is always zero.

The term $\vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ})$ in the acceleration equation above is called the centripetal (center-seeking) term, as it is always in the direction $-\vec{r}_{PQ}$ and so acts from $Q$ towards the “center” $P$.

Show: none position velocity acceleration

Two points $P$ and $Q$ on a moving and accelerating rigid body.

Example Problem: Points on a rigid body.

A rectangular block is sliding along the ground so that the contact point $P$ is moving at a constant speed of $3\rm\ m/s$ to the left. The block is rotating clockwise and just before point $Q$ hits the ground the angular velocity is $2\rm\ rad/s$ clockwise and the angular acceleration is $3\rm\ rad/s^2$ clockwise. What is the velocity and acceleration of $Q$ just before impact?

Using the regular $\hat\imath,\hat\jmath$ basis, the known information just before impact is $\vec{r}_{PQ} = 2\,\hat\imath \qquad \vec{v}_P = -3\,\hat\imath \qquad \vec{a}_P = 0 \qquad \vec\omega = -2\,\hat{k} \qquad \vec\alpha = -3\,\hat{k}.$ The velocity and acceleration of $Q$ are given by \begin{aligned} \vec{v}_Q &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PQ} \\ &= -3\,\hat\imath - 2\,\hat{k} \times 2\,\hat\imath \\ &= -3\,\hat\imath - 4\,\hat\jmath \\ \vec{a}_Q &= \vec{a}_P + \vec{\alpha} \times \vec{r}_{PQ} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ}) \\ &= 0 - 3\,\hat{k} \times 2\,\hat\imath - 2\,\hat{k} \times (-2\,\hat{k} \times 2\,\hat\imath) \\ &= -6\,\hat\jmath - 2\,\hat{k} \times (-4\,\hat\jmath) \\ &= -8\,\hat\imath - 6\,\hat\jmath. \end{aligned}

Example Problem: Coupled rigid bodies.

Three rigid bodies (two rods and a block) are coupled by pins at $O$, $P$, and $Q$ as shown. Body $\mathcal{B}_3$ cannot rotate and must slide horizontally along the ground. The rod lengths are $OP = 3\sqrt{2}\rm\ m$ and $PQ = 5\rm\ m$, and points $O$ and $Q$ are the same height above the ground.

At the instant shown, $\theta_1 = 45^\circ$, $\omega_1 = 4\rm\ rad/s$. What are $\omega_2$ and $\vec{v}_Q$ at this instant?

Taking a regular $\hat\imath,\hat\jmath$ basis at $O$, we can use the rod lengths and angles to find the relative position vectors: \begin{aligned} \vec{r}_{OP} &= 3\hat\imath + 3\hat\jmath\rm\ m \\ \vec{r}_{PQ} &= 4\hat\imath - 3\hat\jmath\rm\ m. \end{aligned} Using the vector angular velocities $\vec{\omega}_1 = \omega_1\,\hat{k}$ and $\vec{\omega}_2 = \omega_2\,\hat{k}$, we can now use #rkg-er to find first $\vec{v}_P$ and then $\vec{v}_Q$ in terms of the unknown angular velocity $\omega_2$: \begin{aligned} \vec{v}_{P} &= \vec{v}_O + \vec{\omega}_1 \times \vec{r}_{OP} \\ &= 4\hat{k} \times (3\hat\imath + 3\hat\jmath) \\ &= -12\hat\imath + 12\hat\jmath{\rm\ m/s} \\ \vec{v}_{Q} &= \vec{v}_P + \vec{\omega}_2 \times \vec{r}_{PQ} \\ &= (-12\hat\imath + 12\hat\jmath{\rm\ m/s}) + \omega_2\,\hat{k} \times (4\hat\imath - 3\hat\jmath) \\ &= (-12 + 3\omega_2)\,\hat\imath + (12 + 4\omega_2)\,\hat\jmath{\rm\ m/s}. \end{aligned} The last piece of information that we need to use is the constraint on the motion of $\mathcal{B}_3$, namely that it can only slide horizontally. This can be expressed by saying that the vertical component of $\vec{v}_Q$ is zero, or more generally that the component in the normal direction $\hat{e}_n = \hat\jmath$ is zero. That is: \begin{aligned} 0 &= \vec{v}_Q \cdot \hat{e}_n \\ &= \Big((-12 + 3\omega_2)\,\hat\imath + (12 + 4\omega_2)\,\hat\jmath\Big) \cdot \hat\jmath \\ &= 12 + 4\omega_2 \\ \omega_2 &= -3{\rm\ rad/s}. \end{aligned} Substituting this back into the expression above for $\vec{v}_Q$ gives: $\vec{v}_Q = -21\hat\imath{\rm\ m/s},$ which does indeed have no vertical component, as expected.

Warning: Rigid body points may be accelerating even if $\vec{a}_P = 0$ and $\alpha = 0$.

If $\vec{\alpha} = 0$ and $\vec{a}_P = 0$ for a rigid body then we would normally say that the body itself is not accelerating. This does not mean, however, that the points on the bodies are not accelerating. If the body is rotating (so $\vec{\omega} \ne 0$) then the centripetal acceleration means that in fact most points on the body are accelerating at any time.

## Rigid bodies in 2D

The #rkg-er equations above are valid in both 2D and 3D. If we know that we are in the 2D $\hat\imath,\hat\jmath$ plane then the angular velocity vector is orthogonal to the plane in the $\hat{k}$ direction and the equations can be written in a simpler form with the perpendicular vector notation #rvv-en, as shown below.

Rigid body point relations in 2D.

\begin{aligned} \vec{r}_Q &= \vec{r}_P + \vec{r}_{PQ} \\ \vec{v}_Q &= \vec{v}_P + \omega \, \vec{r}_{PQ}^\perp \\ \vec{a}_Q &= \vec{a}_P + \alpha \, \vec{r}_{PQ}^\perp - \omega^2 \, \vec{r}_{PQ} \end{aligned}

Points $P$ and $Q$ are two locations on a rigid body. Scalars $\omega$ and $\alpha$ are the scalar angular velocity and angular acceleration of the rigid body (positive counter-clockwise).

The position vectors $\vec{r}_P$ and $\vec{r}_Q$ lie in the $\hat\imath,\hat\jmath$ plane, so $\vec{v}_P$, $\vec{v}_Q$, and $\vec{r}_{PQ}$ are also in this plane. The angular velocity and angular acceleration vectors are orthogonal and thus can be written: \begin{aligned} \vec{\omega} &= \omega\,\hat{k} \\ \vec{\alpha} &= \alpha\,\hat{k}. \end{aligned} Evaluating #rkg-er and using the cross-product expression #rvv-e9 now gives the velocity expression: \begin{aligned} \vec{v}_Q &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PQ} \\ &= \vec{v}_P + \omega\,\hat{k} \times \vec{r}_{PQ} \\ &= \vec{v}_P + \omega \,\vec{r}_{PQ}^\perp, \end{aligned} and the acceleration equation: \begin{aligned} \vec{a}_Q &= \vec{a}_P + \vec{\alpha} \times \vec{r}_{PQ} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ}) \\ &= \vec{a}_P + \alpha\,\hat{k} \times \vec{r}_{PQ} + \omega\,\hat{k} \times (\omega\,\hat{k} \times \vec{r}_{PQ}) \\ &= \vec{a}_P + \alpha\,\vec{r}_{PQ}^\perp + \omega\,\hat{k} \times (\omega\,\vec{r}_{PQ}^\perp) \\ &= \vec{a}_P + \alpha\,\vec{r}_{PQ}^\perp + \omega^2 (\vec{r}_{PQ}^\perp)^\perp \\ &= \vec{a}_P + \alpha\,\vec{r}_{PQ}^\perp - \omega^2 \vec{r}_{PQ}, \end{aligned} where we used the fact that $\vec{r}^{\perp\perp} = -\vec{r}$ (two $90^\circ$ rotations make a $180^\circ$ rotation).

If a point $M$ on a rigid body has zero velocity then it is called the instantaneous center of rotation, because the velocity of all points on the body will be given by simple rotation about $M$ with the angular velocity $\vec{\omega}$ of the body. In 2D we can always find the instantaneous center with the following equation, although it might be outside of the physical body.

Instantaneous center of rotation $M$ in 2D.

$\vec{r}_{PM} = \frac{1}{\omega^2} \vec{\omega} \times \vec{v}_P = \frac{1}{\omega} \vec{v}_P^\perp$

Point $P$ has velocity $\vec{v}_P$ and is attached to a rigid body rotating with angular velocity $\vec{\omega}$.

The instantaneous center $M$ has the property that the point on the rigid body at $M$ has zero velocity. We thus want: $0 = \vec{v}_M = \vec{v}_P + \vec{\omega} \times \vec{r}_{PM}.$ Taking another cross product by $\vec{\omega}$ gives: \begin{aligned} 0 &= \vec{\omega} \times \Big( \vec{v}_P + \vec{\omega} \times \vec{r}_{PM} \Big) \\ &= \vec{\omega} \times \vec{v}_P + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{PM}) \\ &= \vec{\omega} \times \vec{v}_P - \omega^2 \, \vec{r}_{PM} \text{ (because \vec\omega \perp \vec{r}_{PM})} \\ \vec{r}_{PM} &= \frac{1}{\omega^2} \vec{\omega} \times \vec{v}_P. \end{aligned} Here we used the fact that if $\vec{r}_{PM}$ lies in the $\hat\imath,\hat\jmath$ plane and $\vec{\omega}$ is in the $\hat{k}$ direction, then from #rvv-e9 we have $\vec{\omega} \times (\vec{\omega} \times \vec{r}_{PM}) = \omega^2 \,\vec{r}_{PM}^{\perp\perp} = -\omega^2\,\vec{r}_{PM}$, as in the derivation of #rkg-e2.

In 3D there will only be an instantaneous center if $\vec{v}_P$ is orthogonal to $\vec{\omega}$, in which case there will many choices for the instantaneous center, all lying on a line in the $\vec{\omega}$ direction.

If $M$ is an instantaneous center, so it has zero velocity, then the velocity of any other point on the rigid body is given by the following equation.

Velocity from the instantaneous center $M$.

$\vec{v}_{Q} = \vec{\omega} \times \vec{r}_{MQ}$

Point $M$ is the instantaneous center of rotation for a rigid body rotating with angular velocity $\vec{\omega}$, and $Q$ is any point on the body.

By definition the instantaneous center has $\vec{v}_M = 0$, so the velocity formula #rkg-er gives: \begin{aligned} \vec{v}_Q &= \vec{v}_M + \vec{\omega} \times \vec{r}_{MQ} \\ &= \vec{\omega} \times \vec{r}_{MQ}. \end{aligned}

 Movement: translation var. translation rotation var. rotation slide hinge circle spin reverse spin oscillation

Point $P$ Body $Q$ Velocity $Q$ Acceleration
none none
$\vec{v}_P$ (trans.) $\vec{a}_P$ (trans.)
$\vec{\omega} \times \vec{r}_{PQ}$ (rot.) $\vec{\alpha} \times \vec{r}_{PQ}$ (ang.)
$\vec{\omega} \times (\vec{\omega} \times \vec{r}_{PQ})$ (cent.)
$\vec{v}_Q$ (total) $\vec{a}_Q$ (total)

Velocity and acceleration of points on a rigid body undergoing different motions.