A circular disc is rotating with an angular speed (in radian per sec)

$\mathrm{\omega }=2{\mathrm{t}}^2$

C P =2 m

Given, $\mathrm{CP}=2 \mathrm{m}$

In terms of $\hat{\mathrm{i}},\mathrm{j}$ and $\hat{\mathrm{k}},\mathrm{at} t=1 \mathrm{s}$

find,

 linear acceleration of the particle lying at P

$$\begin{gathered} \mathrm{\omega }=2{\mathrm{t}}^2\Rightarrow \mathrm{\alpha }=\frac{\mathrm{d\omega }}{\mathrm{dt}}=4\mathrm{t} \\ \text{At }\mathrm{t}=1 \mathrm{s}, \\ \text{ For the particle at }\mathrm{P}, \\ \mathrm{r}=\mathrm{CP}=2\mathrm{rad}/\mathrm{s}\text{ and } \mathrm{\alpha }=4\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Acceleration of the particle has two components

(i) ${\mathrm{a}}_{\mathrm{r}}={\mathrm{r\omega }}^2$

(radial component)

$=\left(2\right)(2{)}^2=8 \mathrm{m}/{\mathrm{s}}^2$

This components is towards centre C.

(ii) ${\mathrm{a}}_{\mathrm{t}}=\mathrm{r\alpha }$

(tangential component)

$=\left(2\right)\left(4\right)=8 \mathrm{m}/{\mathrm{s}}^2$

This component is in the direction of linear velocity, as $\mathrm{\omega }$ or v is increasing.

$\therefore \mathbf{a}=\left(8\cos 53°-8\cos 37°\right)\left(\hat{\mathbf{i}}\right)$$+\left(8\sin 53°+8\sin 37°\right)(-\hat{\mathrm{j}})$

or $\mathrm{a}=(-1.6\hat{\mathrm{i}}-11.2\hat{\mathrm{j}})\mathrm{m}/{\mathrm{s}}^2$