91Ó°ÊÓ

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

Circular trajectory = $\stackrel{→}{r}=c\cos \left(k{t}^2\right)\widehat{\mathrm{Ä\pm }}+c\sin \left(k{t}^2\right)\widehat{\mathrm{È·}}$

c= 5 m

k= 8 x 10⁴rad/sec²

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

$\stackrel{→}{r}=c\cos \left(k{t}^2\right)\widehat{\mathrm{Ä\pm }}+c\sin \left(k{t}^2\right)\widehat{\mathrm{È·}}$

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 $\stackrel{→}{r}=c\cos \left(k{t}^2\right)\widehat{\mathrm{Ä\pm }}+c\sin \left(k{t}^2\right)\widehat{\mathrm{È·}}$

c= 5 m

k= 8 x 10⁴rad/sec²

Velocity is calculated as

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

Substitute the value we get

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

Part(c) Step 1 : given information

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

Circular trajectory = $\stackrel{→}{r}=c\cos \left(k{t}^2\right)\widehat{\mathrm{Ä\pm }}+c\sin \left(k{t}^2\right)\widehat{\mathrm{È·}}$

c= 5 m

k= 8 x 10⁴rad/sec²

Force is calculated as

$F⃗=ma⃗$

Lets find acceleration

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

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

Now substitute values we get

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