In the given figure,$a = 15 \, m s^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5 \, m$ at a given instant of time. The speed of the particle is ........ $m/s$.

What happens to the centripetal acceleration of a revolving body if you double the orbital speed $v$ and halve the angular velocity $\omega$?

$A$ point $P$ moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s = t^3 + 5$,where $s$ is in meters and $t$ is in seconds. The radius of the path is $20 \ m$. The acceleration of $P$ when $t = 2 \ s$ is nearly .......... $m/s^2$.

$A$ particle is moving in a circle of radius $50 \ cm$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4 \ m/s$,the time taken to complete the first revolution will be $\frac{1}{\alpha}[1-e^{-2 \pi}] \ s$,where $\alpha=$ . . . . . . .

The magnitude of the centripetal force acting on a body of mass $m$ executing uniform circular motion in a circle of radius $r$ with speed $v$ is:

$A$ particle of mass $5 \ kg$ is moving on a circle of radius $1 \ m$ with an angular velocity of $2 \ rad \ s^{-1}$. The centripetal force is (in $N$)