problem 4\nthe two-dimensional motion of a particle is defined by the relations $r = 2a \\cos\\theta$ and…

planation:

Step1: Find $\dot{r}$ and $\dot{\theta}$

First, differentiate $r = 2a\cos\theta$ and $\theta = \frac{bt^2}{2}$ with respect to $t$: $\dot{\theta} = \frac{d}{dt}\left(\frac{bt^2}{2}\right) = bt$ $\dot{r} = \frac{dr}{dt} = \frac{dr}{d\theta}\cdot\frac{d\theta}{dt} = -2a\sin\theta \cdot bt = -2abt\sin\theta$

Step2: Calculate velocity magnitude

In polar coordinates, velocity magnitude $v = \sqrt{\dot{r}^2 + (r\dot{\theta})^2}$: Substitute $\dot{r}$, $r$, and $\dot{\theta}$: $$ \begin{align*} v &= \sqrt{(-2abt\sin\theta)^2 + (2a\cos\theta \cdot bt)^2}\ &= \sqrt{4a^2b^2t^2\sin^2\theta + 4a^2b^2t^2\cos^2\theta}\ &= \sqrt{4a^2b^2t^2(\sin^2\theta+\cos^2\theta)}\ &= 2abt \end{align*} $$ (using $\sin^2\theta+\cos^2\theta=1$)

Step3: Find $\ddot{r}$ and $\ddot{\theta}$

Differentiate $\dot{r}$ and $\dot{\theta}$: $\ddot{\theta} = \frac{d}{dt}(bt) = b$ $$ \begin{align*} \ddot{r} &= \frac{d}{dt}(-2abt\sin\theta)\ &= -2ab\sin\theta - 2abt\cos\theta \cdot \dot{\theta}\ &= -2ab\sin\theta - 2ab^2t^2\cos\theta \end{align*} $$

Step4: Calculate acceleration magnitude

In polar coordinates, acceleration components are $a_r = \ddot{r} - r\dot{\theta}^2$, $a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta}$. First compute each component: $$ \begin{align*} a_r &= (-2ab\sin\theta - 2ab^2t^2\cos\theta) - (2a\cos\theta)(bt)^2\ &= -2ab\sin\theta - 2ab^2t^2\cos\theta - 2ab^2t^2\cos\theta\ &= -2ab\sin\theta - 4ab^2t^2\cos\theta \end{align*} $$ $$ \begin{align*} a_\theta &= (2a\cos\theta)(b) + 2(-2abt\sin\theta)(bt)\ &= 2ab\cos\theta - 4ab^2t^2\sin\theta \end{align*} $$ Acceleration magnitude $a = \sqrt{a_r^2 + a_\theta^2}$: $$ \begin{align*} a &= \sqrt{(-2ab\sin\theta - 4ab^2t^2\cos\theta)^2 + (2ab\cos\theta - 4ab^2t^2\sin\theta)^2}\ &= \sqrt{4a^2b^2\sin^2\theta + 16a^2b^3t^2\sin\theta\cos\theta + 16a^2b^4t^4\cos^2\theta + 4a^2b^2\cos^2\theta - 16a^2b^3t^2\sin\theta\cos\theta + 16a^2b^4t^4\sin^2\theta}\ &= \sqrt{4a^2b^2(\sin^2\theta+\cos^2\theta) + 16a^2b^4t^4(\sin^2\theta+\cos^2\theta)}\ &= \sqrt{4a^2b^2 + 16a^2b^4t^4}\ &= 2ab\sqrt{1 + 4b^2t^4} \end{align*} $$

Step5: Find radius of curvature $\rho$

Use the formula $\rho = \frac{v^2}{a_n}$, where $a_n$ is the normal acceleration. First find $a_n$: We know $a^2 = a_t^2 + a_n^2$, and $a_t = \frac{dv}{dt} = \frac{d}{dt}(2abt) = 2ab$. $$ \begin{align*} a_n &= \sqrt{a^2 - a_t^2}\ &= \sqrt{4a^2b^2(1+4b^2t^4) - 4a^2b^2}\ &= \sqrt{16a^2b^4t^4}\ &= 4ab^2t^2 \end{align*} $$ Then $\rho = \frac{v^2}{a_n} = \frac{(2abt)^2}{4ab^2t^2} = \frac{4a^2b^2t^2}{4ab^2t^2} = a$

Answer:

(a) Velocity magnitude: $\boldsymbol{2abt}$; Acceleration magnitude: $\boldsymbol{2ab\sqrt{1 + 4b^2t^4}}$ (b) Radius of curvature: $\boldsymbol{a}$