9.5: Special types of motion
Let us summarise the \(3 \mathrm{D}\) equations for the velocity and acceleration of point \(B\) in a rigid body \(F\) for the three types of motion:
Pure translation
\[\begin{align} \overrightarrow{\boldsymbol{\omega}} & =\overrightarrow{\mathbf{0}} \tag{9.30} \label{9.30}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{v}}_{A} \tag{9.31} \label{9.31}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B} & =\overrightarrow{\boldsymbol{a}}_{A} \tag{9.32} \label{9.32}\end{align}\]
Pure rotation around a fixed axis in space
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A} & =\overrightarrow{\mathbf{0}} \text { and } \overrightarrow{\boldsymbol{a}}_{A}=\overrightarrow{\mathbf{0}} \tag{9.33} \label{9.33}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / A} \tag{9.34} \label{9.34}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B} & =\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / A}+\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / A}\right) \tag{9.35} \label{9.35}\end{align}\]
General motion, translation and rotation
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{v}}_{A}+\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / A} \tag{9.36} \label{9.36}\\[4pt] \overrightarrow{\boldsymbol{a}}_{B} & =\overrightarrow{\boldsymbol{a}}_{A}+\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{B / A}+\overrightarrow{\boldsymbol{\omega}} \times\left(\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / A}\right) \tag{9.37} \label{9.37}\end{align}\]
As an example of the kinematic methods for rigid bodies, consider the square \(F\) in Figure 9.6. The CoM of the square falls with a velocity \(\overrightarrow{\boldsymbol{v}}_{G}=-1 \hat{\boldsymbol{\jmath}} \mathrm{m} / \mathrm{s}\) and acceleration \(\overrightarrow{\boldsymbol{a}}_{G}=-1 \hat{\boldsymbol{\jmath}} \mathrm{m} / \mathrm{s}^{2}\). At the same time the angular velocity and acceleration of the square are \(\overrightarrow{\boldsymbol{\omega}}=1 \hat{\boldsymbol{k}} \mathrm{rad} / \mathrm{s}\) and \(\overrightarrow{\boldsymbol{\alpha}}=1 \hat{\boldsymbol{k}} \mathrm{rad} / \mathrm{s}^{2}\). The question is: Determine the velocity and acceleration vectors of point \(B\).
Solution
To solve this problem, we can use Eqs. (9.36) and (9.37). Instead of point \(A\) we choose point \(G\) as reference point, because we have a lot of information on \(G\). Then we determine the vector \(\overrightarrow{\boldsymbol{r}}_{B / G}=(3.5 \hat{\boldsymbol{\imath}}-0.5 \hat{\boldsymbol{\jmath}}) \mathrm{m}\) from the figure. Now we use Equation 9.36 to obtain:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{B} & =[-1 \hat{\boldsymbol{\jmath}}+1 \hat{\boldsymbol{k}} \times(3.5 \hat{\boldsymbol{\imath}}-0.5 \hat{\boldsymbol{\jmath}})] \mathrm{m} / \mathrm{s} \tag{9.38} \label{9.38}\\[4pt] & =(0.5 \hat{\boldsymbol{\imath}}+2.5 \hat{\boldsymbol{\jmath}}) \mathrm{m} / \mathrm{s} \tag{9.39} \label{9.39}\end{align}\]
And using Equation 9.37 we obtain:\[\begin{align} \overrightarrow{\boldsymbol{a}}_{B} & =\left[-1 \hat{\boldsymbol{\jmath}}+1 \hat{\boldsymbol{k}} \times(3.5 \hat{\boldsymbol{\imath}}-0.5 \hat{\boldsymbol{\jmath}})-1^{2}(3.5 \hat{\boldsymbol{\imath}}-0.5 \hat{\boldsymbol{\jmath}})\right] \mathrm{m} / \mathrm{s}^{2} \tag{9.40} \label{9.40}\\[4pt] & =[(0.5 \hat{\boldsymbol{\imath}}+2.5 \hat{\boldsymbol{\jmath}})-1(3.5 \hat{\boldsymbol{\imath}}-0.5 \hat{\boldsymbol{\jmath}})] \mathrm{m} / \mathrm{s}^{2} \tag{9.41} \label{9.41}\\[4pt] & =(-3 \hat{\boldsymbol{\imath}}+3 \hat{\boldsymbol{\jmath}}) \mathrm{m} / \mathrm{s}^{2} \tag{9.42} \label{9.42}\end{align}\]
9.5.1 Instantaneous centre of rotation
In \(2 D\) kinematics it can be shown that at every instance there is always a single point in space, the instantaneous centre of rotation IC, around which all points in the rigid body move in a circular path (pure rotation). The velocity of the rigid body at \(\overrightarrow{\boldsymbol{r}}_{I} C\) is zero.
Using the position vector \(\overrightarrow{\boldsymbol{r}}_{I C}\), the velocity of all points \(i\) in the rigid body can be determined using the equation for pure rotation:
\[\overrightarrow{\boldsymbol{v}}_{i}=\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{i / I C} \tag{9.43} \label{9.43}\]
Let us illustrate the use of the \(I C\) with an example.
In Figure 9.7 a wheel \(F\) with radius \(R\) rolls at an angular velocity \(\overrightarrow{\boldsymbol{\omega}}\) without slip over a horizontal surface. Find the expressions for the velocity vectors of its CoM \(G\) and point \(B\).
Solution
This problem can be solved by realising that the point at which the wheel touches the ground has zero velocity and is therefore the instantaneous centre of rotation \(I C\). Now Equation 9.43 is used to determine the velocity vector of point \(G\) and \(B\). We have \(\overrightarrow{\boldsymbol{r}}_{G / I C}=R \hat{\boldsymbol{\jmath}}\) and \(\overrightarrow{\boldsymbol{r}}_{B / I C}=R(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})\) and \(\overrightarrow{\boldsymbol{\omega}}=\omega \hat{\boldsymbol{k}}\). Then we have:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{G} & =\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{G / I C} \tag{9.44} \label{9.44}\\[4pt] & =\omega R(\hat{\boldsymbol{k}} \times \hat{\boldsymbol{\jmath}}) \tag{9.45} \label{9.45}\\[4pt] & =-\omega R \hat{\boldsymbol{\imath}} \tag{9.46} \label{9.46}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B} & =\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\boldsymbol{r}}_{B / I C} \tag{9.47} \label{9.47}\\[4pt] & =\omega R(\hat{\boldsymbol{k}} \times[\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}]) \tag{9.48} \label{9.48}\\[4pt] & =\omega R(\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{\imath}}) \tag{9.49} \label{9.49}\end{align}\]So we see that the centre \(G\) of a wheel with a positive angular velocity \(\omega\) rolls in the negative \(x\) direction.Note that the \(I C\) is not a fixed point but can move in time, like the point of contact of the wheel in the previous example. We also remark that in 3D the \(I C\) is an axis of rotation instead of a point. Let us now describe how to find the \(I C\).
9.5.2 Finding the instantaneous centre of rotation
If a point with zero velocity is not known or present in the rigid body one can try to find the \(I C\) by utilising that all points in the rigid body move along concentric circles around point \(\overrightarrow{\boldsymbol{r}}_{I C}\). This allows three methods to find the IC for planar kinematics:
- If the angular velocity, and the velocity of a point \(i\) are known, determine the distance of that point, using the equation \(\left|\overrightarrow{\boldsymbol{r}}_{i / I C}\right|=\left|\overrightarrow{\boldsymbol{v}}_{i}\right| /|\omega|\). Project this distance perpendicular to the vector \(\overrightarrow{\boldsymbol{v}}_{i}\) to determine \(\overrightarrow{\boldsymbol{r}}_{I C}\).
- Use two known velocity vectors \(\overrightarrow{\boldsymbol{v}}_{i}\) and \(\overrightarrow{\boldsymbol{v}}_{j}\) of different points \(i\) and \(j\) on the rigid body. Determine the location \(\overrightarrow{\boldsymbol{r}}_{I C}\) by drawing the lines perpendicular to these two vectors, which are parallel to the radii of the two concentric circles along which \(i\) and \(j\) move. These two lines intersect at the common centre of rotation \(\overrightarrow{\boldsymbol{r}}_{I C}\).
- If two known velocity vectors \(\overrightarrow{\boldsymbol{v}}_{i}\) and \(\overrightarrow{\boldsymbol{v}}_{j}\) are parallel, determine the angular velocity by using the rate at which the speed increases at larger distance from the IC: \(|\omega|=\left(|| \overrightarrow{\boldsymbol{v}}_{i}|-| \overrightarrow{\boldsymbol{v}}_{j}||\right) /\left|\overrightarrow{\boldsymbol{r}}_{i / j}\right|\) and continue like in point 1.
9.5.3 Choosing the reference point for rotation
As we have seen from the previous examples, multiple choices are possible for the point \(A\) that is used to analyse the kinematics of a rigid body. Ideally the expressions for the translation and rotation of the rigid body are as simple as possible.
Depending on the situation several choices for the reference point \(\overrightarrow{\boldsymbol{r}}_{A}\) on the rigid body are possible. Often a point that is fixed in space, or does not accelerate is a good choice. Otherwise a point with zero velocity, like an \(I C\) can make the analysis simpler. Later we will see that a point where the effects of forces and moments can be easily calculated can also be good choice. In all cases, it is important to carefully consider the reference point of the rigid body, before starting to analyse its dynamics.
As illustrated in the next example, the kinematic equations discussed in this chapter can also be used to analyse the kinematics of multiple rigid bodies with relative motion that is described by constraint equations, similar to the relatively constrained kinematics for point masses discussed in Sec. 5.2.4.
Problem: Figure 9.8 shows three gearwheels \(A, B\) and \(C\), that rotate without slip. The angular velocity \(\overrightarrow{\boldsymbol{\omega}}_{A}=\omega_{A} \hat{\boldsymbol{k}}\) and angular acceleration \(\overrightarrow{\boldsymbol{\alpha}}_{A}=\alpha_{A} \hat{\boldsymbol{k}}\) of gearwheel \(A\) are known. Determine \(\omega_{B}, \omega_{C}, \alpha_{B}\) and \(\alpha_{C}\).
Solution
To solve this problem first determine the constraint equations. Since the gearwheels rotate without slip, there is static friction and we know from Equation 6.28 that for the no slip condition, the tangential components of acceleration and velocity of the surfaces at the points where they touch are equal \(\left(a_{P_{A B}, t}=a_{P_{B A}, t}\right)\). It is important to realise that at a contact point there are points of two objects that have the same position, and one should label and distinguish those separately, like \(P_{A B}\) and \(P_{B A}\) in Figure 9.8, to properly write the constraint equations:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{P_{A B}} & =\overrightarrow{\boldsymbol{v}}_{P_{B A}} \tag{9.50} \label{9.50}\\[4pt] \boldsymbol{v}_{P_{B C}} & =\overrightarrow{\boldsymbol{v}}_{P_{C B}} \tag{9.51} \label{9.51}\\[4pt] a_{P_{A B}, t} & =a_{P_{B A}, t} \tag{9.52} \label{9.52}\\[4pt] a_{P_{B C}, t} & =a_{P_{C B}, t} \tag{9.53} \label{9.53}\end{align}\]
We have 4 constraint equations and 4 unknown scalars that need to be determined, so the problem is solvable. There is pure rotation, since the gearwheels rotate around fixed axes, so we have the following kinematic equations for gearwheel \(A\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{P_{A B}} & =\overrightarrow{\boldsymbol{\omega}}_{A} \times \overrightarrow{\boldsymbol{r}}_{P_{A B} / A} \tag{9.54} \label{9.54}\\[4pt] & =\omega_{A} \hat{\boldsymbol{k}} \times R_{A} \hat{\boldsymbol{\imath}} \tag{9.55} \label{9.55}\\[4pt] & =\omega_{A} R_{A} \hat{\boldsymbol{\jmath}} \tag{9.56} \label{9.56}\\[4pt] \overrightarrow{\boldsymbol{a}}_{P_{A B}} & =\overrightarrow{\boldsymbol{\alpha}}_{A} \times \overrightarrow{\boldsymbol{r}}_{P_{A B} / A}-\omega_{A}^{2} \overrightarrow{\boldsymbol{r}}_{P_{A B} / A} \tag{9.57} \label{9.57}\\[4pt] & =\alpha_{A} \hat{\boldsymbol{k}} \times R_{A} \hat{\boldsymbol{\imath}}-\omega_{A}^{2} R_{A} \hat{\boldsymbol{\imath}} \tag{9.58} \label{9.58}\\[4pt] a_{P_{A B}, t} & =\alpha_{A} R_{A} \hat{\boldsymbol{\jmath}} \tag{9.59} \label{9.59}\end{align}\]
In the last step we separated the component of acceleration tangential to the contact surface, since that is the only direction in which the friction constrains the acceleration.
Since we know the velocity and acceleration of point \(P_{A B}\), we can combine it with the constraint and kinematic equations to determine the angular velocity and acceleration of gearwheel \(B\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{P_{A B}} & =\overrightarrow{\boldsymbol{v}}_{P_{B A}} \tag{9.60} \label{9.60}\\[4pt] \omega_{A} R_{A} \hat{\boldsymbol{\jmath}} & =\overrightarrow{\boldsymbol{\omega}}_{B} \times \overrightarrow{\boldsymbol{r}}_{P_{B A} / B}=-\omega_{B} R_{B} \hat{\boldsymbol{\jmath}} \tag{9.61} \label{9.61}\\[4pt] \omega_{B} & =-\omega_{A} \frac{R_{A}}{R_{B}} \tag{9.62} \label{9.62}\\[4pt] a_{P_{A B}, t} & =a_{P_{B A}, t} \tag{9.63} \label{9.63}\\[4pt] \alpha_{A} R_{A} \hat{\boldsymbol{\jmath}} & =\overrightarrow{\boldsymbol{\alpha}}_{B} \times \overrightarrow{\boldsymbol{r}}_{P_{B A} / B}=-\alpha_{B} R_{B} \hat{\boldsymbol{\jmath}} \tag{9.64} \label{9.64}\\[4pt] \alpha_{B} & =-\alpha_{A} \frac{R_{A}}{R_{B}} \tag{9.65} \label{9.65}\end{align}\]
After having obtained the angular velocity and acceleration of gearwheel \(B\), we can follow the same procedure for wheel \(C\).