Question 2.SP.17: In the preceding problem determine the magnitude and acceler......
In the preceding problem determine the magnitude and acceleration of the particle using the same two methods.
(a) Polar coordinates. See Fig. 2-13. Using Eq. (2.35),
\mathbf{a}=\left(\ddot{r}-r \mathbf{ω}^{2}\right) \mathbf{e}_{r}+(2 \dot{r} \mathbf{ω}+r \mathbf{α}) \mathbf{e}_{\phi} \qquad \qquad (2.35)\\ \\ \mathbf{a}=\left(\ddot{r}-r \dot{\theta}^{2}\right) \mathbf{e}_{r}+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \mathbf{e}_{\theta}=-4.77 \mathbf{e}_{r}+20.94 \mathbf{e}_{\theta}
since \theta=t^{2}, \dot{\theta}=2 t, \ddot{\theta}=2 ; r=2 \theta=2 t^{2}, \dot{r}=4 t, \ddot{r}=4 ; t=1.023 \mathrm{~s} at \theta=\pi / 3. Thus,
a=\sqrt{(-4.77)^{2}+(20.94)^{2}}=\underline{21.5 \mathrm{~m} / \mathrm{s}^{2}}\quad \text { with } \quad \theta_{x}=30^{\circ}-12.8^{\circ}=\underline{17.2^{\circ}}
(b) Cartesian coordinates. See Fig. 2-14.
Continue the time derivatives, using x and y from Problem 2.16(b), and evaluate at t=1.023\left(t^{2}=\pi / 3\right) :
\begin{array}{l}\ddot{x}=4 \cos t^{2}+4 t\left(-\sin t^{2}\right)(2 t)+12 t^{2}\left(-\sin t^{2}\right)-4 t^{3}\left(\cos t^{2}\right)(2 t)=-20.52 \mathrm{~m} / \mathrm{s}^{2} \\\ddot{y}=4 \sin t^{2}+4 t\left(\cos t^{2}\right)(2 t)+12 t^{2}\left(\cos t^{2}\right)+4 t^{3}\left(-\sin t^{2}\right)(2 t)=6.34 \mathrm{~m} / \mathrm{s}^{2}\end{array}
Hence, \quad a=\sqrt{(-20.52)^{2}+(6.34)^{2}}=\underline{21.5 \mathrm{~m} / \mathrm{s}^{2}} \quad \text{ with } \quad \theta_{x}=\tan ^{-1} \frac{6.34}{20.52}=\underline{17.2^{\circ}}
