For a particle in circular motion,$a_r = 3 \ m/s^2$ and $a_T = 4 \ m/s^2$. If $\theta$ is the angle between the resultant acceleration $a$ and the radial acceleration $a_r$,then .......

$A$ 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 $\alpha=$ . . . . . . .

For a particle in a non-uniform accelerated circular motion:

$A$ body is moving along a circular track of radius $100 \ m$ with velocity $20 \ m/s$. Its tangential acceleration is $3 \ m/s^{2}$,then its resultant acceleration will be (in $m/s^{2}$)

The centripetal acceleration is given by

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: