9.6: Integration and differentiation over angles
The vector equations from the previous sections are very useful for determining the kinematics of points in the rigid body at a certain time instant. However, to predict the motion at all times, the coordinate functions \(x_{A}(t), y_{A}(t)\) and
\(\phi_{B / A}(t)\) need to be determined. Using Equation 9.5 we can then describe the planar time-dependent motion of a point \(B\) as follows:
\[\overrightarrow{\boldsymbol{r}}_{B, 2 D}(t)=\overrightarrow{\boldsymbol{r}}_{A}(t)+\left|\overrightarrow{\boldsymbol{r}}_{B / A}\right| \cos \phi_{B / A}(t) \hat{\boldsymbol{\imath}}+\left|\overrightarrow{\boldsymbol{r}}_{B / A}\right| \sin \phi_{B / A}(t) \hat{\boldsymbol{\jmath}} \tag{9.66} \label{9.66}\]
The kinematics to describe the motion \(\overrightarrow{\boldsymbol{r}}_{A}(t)\) of point \(A\) is exactly the same as for the kinematics of point masses that were described in Ch. 5, using different coordinate systems and path curves. The new aspect is that for rigid bodies we also have to describe the time dependent angle \(\phi(t) \equiv \phi_{B / A}(t)\).
Luckily, for planar kinematics in 2D the required mathematics is completely the same as that for the path coordinate \(s_{A}\), such that the methods we discussed in Sec. 5.7 and Sec. 5.8 can be directly applied to the angle \(\phi(t)\). This thus just requires replacing \(s\) by \(\phi, v\) by \(\omega\) and \(a\) by \(\alpha\). For completeness we provide all the resulting equations for angles here. For time differentiation we get:
\[\begin{align} \phi(t) & \rightarrow \tag{9.67} \label{9.67}\\[4pt] \omega(t) & =\dot{\phi}(t) \rightarrow \tag{9.68} \label{9.68}\\[4pt] \alpha(t) & =\dot{\omega}(t)=\ddot{\phi}(t) \tag{9.69} \label{9.69}\end{align}\]
For time integration we obtain:
\[\begin{align} \alpha(t) & \rightarrow \tag{9.70} \label{9.70}\\[4pt] \omega\left(t_{2}\right) & =\omega\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \alpha(t) \mathrm{d} t \rightarrow \tag{9.71} \label{9.71}\\[4pt] \phi\left(t_{2}\right) & =\phi\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \omega(t) \mathrm{d} t \tag{9.72} \label{9.72}\end{align}\]
Similar to having functions along the path curve like \(a_{s}(s)\) and \(v_{s}(s)\), we can have functions for angle dependent angular velocity \(\omega_{\phi}(\phi)\) and angular acceleration \(\alpha_{\phi}(\phi)\). Differentiation and integration over these functions is similar to that over the path coordinate \(s\) :
\[\begin{align} \alpha_{\phi}(\phi)=\frac{\mathrm{d} \omega_{\phi}}{\mathrm{d} \phi} \omega_{\phi}(\phi) \tag{9.73} \label{9.73}\\[4pt] \omega_{\phi}^{2}\left(\phi_{2}\right)=\omega_{\phi}^{2}\left(\phi_{1}\right)+2 \int_{\phi_{1}}^{\phi_{2}} \alpha_{\phi}(\phi) \mathrm{d} \phi \tag{9.74} \label{9.74}\end{align}\]
Gearwheel \(A\) in Figure 9.8 has a constant acceleration \(\alpha_{A}\) and \(\omega_{A}=0, \phi_{A}=0\) at \(t=0\). Determine the time dependent angle \(\phi_{A}(t)\).
Solution
It is straightforward to obtain the solution by integration:
\[\begin{align} \omega_{A}(t) & =\omega_{A}(0)+\int_{0}^{t} \alpha_{A} \mathrm{~d} t=\alpha_{A} t \tag{9.75} \label{9.75}\\[4pt] \phi_{A}(t) & =\phi_{A}(0)+\int_{0}^{t} \alpha_{A} t \mathrm{~d} t=\frac{1}{2} \alpha_{A} t^{2} \tag{9.76} \label{9.76}\end{align}\]