Question

The total acceleration of a particle in circular motion is $5m/s^2$. If the speed of the particle decreases at a rate of $3m/s^2$, assuming $r=2m$, the angular speed $\omega$ of the particle ( in $rad/s$) relative to the centre of the circle is

Answer 1

Let the tangential acceleration be $a_t$
and centripetal acceleration be $a_c$
Given, $a=5m/s^2,a_t=3m/s^2,r=2ma=\sqrt{a_t^2+a_c^2}\Rightarrow a^2=a_t^2+a_c^2\Rightarrow 5^2=3^2+a_c^2$
$\therefore a_c=4m/s^2$

We know $a_c=\frac{v^2}{r}$
i.e$\frac{v^2}{r}=\frac{{\omega }^2{r}^2}{r}=4\Rightarrow {\omega }^2=\frac{4}{2}=2$
$\therefore \omega =\sqrt{2}rad/s$