Scientists design a new particle accelerator in which protons (mass 1.7 x 10^-27kg ) follow a circular trajectory given by $\overrightarrow{r}=c\cos \left(k{t}^2\right)\hat{ı}+c\sin \left(k{t}^2\right)\hat{ȷ}$ where c=5.0 m and k=8.0 x 10⁴ rad/s² are constants and t is the elapsed time.
a. What is the radius of the circle?
b. What is the proton's speed at t=3.0s ?
c. What is the force on the proton at t=3.0 s ? Give your answer in component form.

Mass of the particle m=1.7 x 10^-27 kg

Circular trajectory = $\overrightarrow{r}=c\cos \left(k{t}^2\right)\hat{ı}+c\sin \left(k{t}^2\right)\hat{ȷ}$

c= 5 m

k= 8 x 10⁴rad/sec²

We can write an equation of particle motion is given as,

$\overrightarrow{r}=c\cos \left(k{t}^2\right)\hat{ı}+c\sin \left(k{t}^2\right)\hat{ȷ}$

Where,
c =radius of the circular path

So we can say radius of the path is 5 m

Mass of the particle m=1.7 x 10^-27 kg

Circular trajectory $\overrightarrow{r}=c\cos \left(k{t}^2\right)\hat{ı}+c\sin \left(k{t}^2\right)\hat{ȷ}$

c= 5 m

k= 8 x 10⁴rad/sec²

Velocity is calculated as

$$\begin{gathered} \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\mid \\ =\frac{d}{dt}\left(c\cos \left(k{t}^2\right)\hat{i}+c\sin \left(k{t}^2\right)\hat{j}\right) \\ =2kct\left(-\sin \left(k{t}^2\right)\hat{i}+\cos \left(k{t}^2\right)\hat{j}\right) \end{gathered}$$

Substitute the value we get

$$\begin{gathered} v=2\times (8\times {10}^{4}rad/s^2)\times \left(5m\right)\times \left(3s\right)[-sin(8\times {10}^{4}rad/s^2)\times {\left(3s\right)}^2)i+cos(8\times {10}^{4}rad/s^2)\times {\left(3s\right)}^2\left)j\right] \\ v=2.4\times {10}^{6}m/s \end{gathered}$$

Part(c) Step 1 : given information

Mass of the particle m=1.7 x 10^-27 kg

Circular trajectory = $\overrightarrow{r}=c\cos \left(k{t}^2\right)\hat{ı}+c\sin \left(k{t}^2\right)\hat{ȷ}$

c= 5 m

k= 8 x 10⁴rad/sec²

Force is calculated as

$F⃗=ma⃗$

Lets find acceleration

$$\begin{gathered} \overrightarrow{a}=\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d\overrightarrow{v}}{dt}=\frac{d}{dt}\left(2kct\left(-\sin \left(k{t}^2\right)\hat{i}+\cos \left(k{t}^2\right)\hat{j}\right)\right) \\ =2kc\frac{d}{dt}\left(t\left(-\sin \left(k{t}^2\right)\hat{i}+\cos \left(k{t}^2\right)\hat{j}\right)\right) \\ =2kc\left[\left(-\sin \left(k{t}^2\right)\hat{i}+\cos \left(k{t}^2\right)\hat{j}\right)+2k{t}^2\left(-\cos \left(k{t}^2\right)\hat{i}-\sin \left(k{t}^2\right)\hat{j}\right)\right] \\ k{t}^2=8\times {10}^4\times 3^2=7.2\times {10}^5 \\ \sin \left(k{t}^2\right)=-0.36\text{and}\cos \left(k{t}^2\right)=-0.93 \end{gathered}$$

$$\begin{gathered} \overrightarrow{a}=2\times (8\times {10}^{4}rad/se{c}^2)\times \left(5m\right)\left[(-0.36\hat{i}-0.93\hat{j})+2\times 7.2\times {10}^5(0.93\hat{i}+0.36\hat{j})\right] \\ =8\times {10}^5\times 2\times 7.2\times {10}^5(0.93\hat{i}+0.36\hat{j}) \\ =1.15\times {10}^{12}(0.93\hat{i}+0.36\hat{j})m/s^2 \end{gathered}$$

Now substitute values we get

$$\begin{gathered} \overrightarrow{F}=m\overrightarrow{a}=(1.7\times {10}^{-27}kg)\times (1.15\times {10}^{12}(0.93\hat{i}+0.36\hat{j})m/se{c}^2) \\ =1.96\times {10}^{-15}(0.93\hat{i}+0.36\hat{j})N \\ =\left(1.82\times {10}^{-15}\hat{i}+0.71\times {10}^{-15}\hat{j}\right)\mathrm{N} \end{gathered}$$