Question

The circular disc rotates about its centre O on the direction shown. At a certain instant point P on the rim has an acceleration $\overrightarrow{a}=-3\hat{i}-4\hat{j}m/s^2$. For this instant its angular velocity is?

Answer 1

$\overrightarrow{a}=-3\hat{i}-4\hat{j}m/s^2$

$⟹a=\sqrt{(-3{)}^2+(-4{)}^2}m/s^2$

$⟹a=5m/s^2$

This is the centripetal acceleration.

Now, we know that the centripetal acceleration is $a_r={\omega }^{2}r$

$⟹{\omega }^{2}r=5$

$⟹{\omega }^2=\frac{5}{0.5}=10$

$⟹\omega =3.16rad/s$