$A$ point mass $m$ is suspended from a light thread of length $l,$ fixed at $O,$ and is whirled in a horizontal circle at a constant speed as shown. From your point of view,stationary with respect to the mass,the forces acting on the mass are:

$A$ car goes around a uniform circular track of radius $R$ at a uniform speed $v$ once in every $T$ seconds. The magnitude of the centripetal acceleration is $a_c$. If the car now goes uniformly around a larger circular track of radius $2R$ and experiences a centripetal acceleration of magnitude $8a_c$,then its time period is:

$A$ merry-go-round starts from rest and accelerates at $0.4 \ rad \ s^{-2}$ for the first $5 \ s$. It then rotates at this constant angular velocity for $30 \ s$ and finally comes to rest with the same magnitude of angular deceleration. What is the total linear distance traveled by a child sitting $3 \ m$ from the center of the merry-go-round?

From the ground level,a ball is to be shot with a certain speed. The graph shows the range $(R)$ of the particle versus the angle of projection from the horizontal $(\theta)$. The values of $\theta_1$ and $\theta_2$ are:

Two cars $S_1$ and $S_2$ are moving in coplanar concentric circular tracks in the opposite sense with the periods of revolution $3 \, min$ and $24 \, min$,respectively. At time $t = 0$,the cars are farthest apart. Then,the two cars will be

The position vector of a particle is given as $\vec{r} = (t^2 - 4t + 6)\hat{i} + (t^2)\hat{j}$. The time after which the velocity vector and acceleration vector become perpendicular to each other is equal to ....... $sec$.