A particle is moving along a circular path having a

Given:
r=4 in, ${\displaystyle \theta =\cos \left(2t\right)}$,
Acceleration in this case is equal to
${\displaystyle \overrightarrow{a}=a_r{\overrightarrow{u}}_r+a_{\theta }{\overrightarrow{u}}_{\theta }}$
${\displaystyle a_r=\ddot{r}-r\dot{{\theta }^2}}$
${\displaystyle =a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }}$
Since the radius in circular motion is constant: ${\displaystyle \dot{r}=\ddot{r}=0}$, and time-derivatives of position are equal to: ${\displaystyle \dot{\theta }=-2\sin \left(2t\right),\ddot{\theta }=-4\cos \left(2t\right)}$
If we want to find out in what time the particle will have ${\displaystyle \theta =30=\frac{\pi }{6}}$, we use following formula:
${\displaystyle \theta =\cos \left(2t\right)}$
${\displaystyle \frac{\pi }{6}=\cos \left(2t\right)}$
${\displaystyle 2t=1.0197}$
${\displaystyle t=0.51s}$
Acceleration is finally equal to:
${\displaystyle \overrightarrow{a}=a_r{\overrightarrow{u}}_r+a_{\theta }{\overrightarrow{u}}_{\theta }}$
${\displaystyle =-r{\dot{\theta }}^2{\overrightarrow{u}}_r+r\ddot{\theta }{\overrightarrow{u}}_{\theta }}$
${\displaystyle =-4\cdot {(-2\sin (2\cdot 0.51))}^2{\overrightarrow{u}}_r+4\cdot -4\cos (2\cdot 0.51){\overrightarrow{u}}_{\theta }}$
${\displaystyle =-11.62{\overrightarrow{u}}_r-8.37{\overrightarrow{u}}_{\theta }}$
Magnitude of the acceleration is:
${\displaystyle a=\sqrt{a_r^2+a_{\theta }^2}}$
${\displaystyle =\sqrt{{\left(11.62\right)}^2+{\left(8.37\right)}^2}}$
${\displaystyle a=14.32\frac{in}{s^2}}$

Step 1
Given data:
The radius of the circular path, ${\displaystyle r=4}$ in.
Position of a particle, $\theta =\cos 2t$
Let,
${\displaystyle a_r=}$ radial acceleration
${\displaystyle u_r=}$ unit vector along radial direction
${\displaystyle a_{\theta }=}$ acceleration along ${\displaystyle \theta }$ direction
${\displaystyle u_{\theta }=}$ unit vector along ${\displaystyle \theta }$ direction
Step 2
Now calculating the time (t) when ${\displaystyle \theta =30°}$
$\theta =\cos 2t$
$\frac{\pi }{6}=\cos 2t$
${\displaystyle 2t=1.0197}$
${\displaystyle t=0.51s}$
Waiting cclerton in veto form
${\displaystyle \overrightarrow{a}=a_r{\bar{u}}_r+a_{\theta }{\bar{u}}_{\theta }}$
${\displaystyle =-r{\dot{\theta }}^2{\bar{u}}_r+r\ddot{\theta }{\bar{u}}_{\theta }}$
${\displaystyle =-4\cdot {[-2\sin (2\cdot 0.51)]}^2{\bar{u}}_r+4[-4\cos (2\cdot 0.51)]{\bar{u}}_{\theta }}$
${\displaystyle =-11.62{\bar{u}}_r-8.37{\bar{u}}_{\theta }}$
Step 3
Finally the magnitude of the acceleration of the particle (a) when ${\displaystyle \theta =30°}$ is calculated as:

${\displaystyle a=\sqrt{a_r^2+a_{\theta }^2}}$
${\displaystyle =\sqrt{{11.62}^2+{8.37}^2}}$
${\displaystyle =14.32\frac{in}{s^2}}$
Hence the required time (t) when ${\displaystyle \theta ={30}^{\circ }}$ is 0.51 seconds and the magnitude of the acceleration of the particle when ${\displaystyle \theta =30°}$ is ${\displaystyle 14.32\frac{in}{s^2}}$.

When

$\begin{array}{}\theta =\frac{\pi }{6}rad,\frac{\pi }{6}=2\cos 2tt=0.5099s\\ \dot{\theta }\frac{d\theta }{dt}=-2\sin 2t\mid t=0.5099s=-1.7039\frac{rad}{s}\\ \ddot{\theta }\frac{d^2\theta }{d{t}^2}=-4\cos 2t\mid t=0.5099s=-2.0944\frac{rad}{s^2}\\ r=4,\dot{r}=0,\ddot{r}=0\\ a_r=\ddot{r}-r{\dot{\theta }}^2=0-4(-1.7039{)}^2=-11.6135\frac{in.}{s^2}\\ a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }=4(-2.0944)+0=-8.3776\frac{in.}{s^2}\\ a=\sqrt{a_r^2+a_{\theta }^2}=\sqrt{(-11.6135{)}^2+(-8.3776{)}^2}=14.3\frac{in.}{s^2}\end{array}$