$A$ particle moves with constant angular velocity in a circle. During the motion,its:

Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal forces is

The centripetal acceleration of a particle in uniform circular motion is $18 \,m/s^2$. If the radius of the circular path is $50 \,cm$, the change in velocity of the particle in a time of $\frac{\pi}{18} \,s$ is (in $\,m/s$)

$A$ particle,moving with uniform speed $v$,changes its direction by angle $\theta$ in time $t$. The magnitude of its average acceleration during this time is:

The real force $F$ acting on a particle of mass $m$ performing circular motion acts along the radius of a circle of radius $r$ and is directed towards the center of the circle. If the square root of the magnitude of such force is given by $\sqrt{F} = \frac{2 \pi}{T} \sqrt{m r}$,where $T$ is the periodic time,find the expression for the force $F$.

The position vector of a particle is given as $\vec{r} = (t^2 - 8t + 12)\hat{i} + t^2\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ........ $sec$.