Physics - Circular Motion Question with Solution | TestHub

The total acceleration is $a_{total}=\sqrt{a_t^2+a_c^2}$

$1^2=(0.6{)}^2+a_c^2$

$1=0.36+a_c^2$

$a_c^2=0.64⟹a_c=0.8m/s^2$.

Also, $a_t=r{\alpha }^{\prime }$, where ${\alpha }^{\prime }$ is angular acceleration.

$\omega ={\alpha }^{\prime }t=\frac{a_t}{r}t$.

$a_c=r{\omega }^2=r{\left(\frac{a_t}{r}t\right)}^2=\frac{a_t^2}{r}{t}^2$.

So, $0.8=\frac{(0.6{)}^2}{r}{t}^2=\frac{0.36}{r}{t}^2$.

This implies $r=\frac{0.36}{0.8}{t}^2=0.45t^2$.

The angular displacement is $\theta =\frac{1}{2}{\alpha }^{\prime }{t}^2=\frac{1}{2}\frac{a_t}{r}{t}^2$.

Substitute $r=0.45t^2$:

$\theta =\frac{1}{2}\frac{0.6}{0.45t^2}{t}^2=\frac{1}{2}\frac{0.6}{0.45}=\frac{0.3}{0.45}=\frac{30}{45}=\frac{2}{3}$ radians.

Convert to degrees: $\theta =\frac{2}{3}\times \frac{180}{\pi }=\frac{120}{\pi }$ degrees.

Given $\theta =\frac{20\alpha }{\pi }$ degrees.

$\frac{120}{\pi }=\frac{20\alpha }{\pi }$

$120=20\alpha$

$\alpha =6$.

The final answer is $\boxed{6}$.