At time $t = 0$,a $2\, kg$ particle has position vector $\vec{r} = (4\hat{i} - 2\hat{j})\, m$ relative to the origin. Its velocity is given by $\vec{v} = 2t^2\hat{i}\, m/s$. The torque acting on the particle about the origin at $t = 2\, s$ is ........ $\hat{k}\, N\cdot m$.

$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

$A$ billiard ball is hit by a cue at a point distance $h$ above the centre. It acquires a linear velocity $v_0$. Let $m$ be the mass and $r$ be the radius of the ball. The angular velocity acquired by the ball is

$A$ wheel is rotating at a rate of $900$ revolutions per minute about its axis. When the power source is cut off,it comes to rest in $1$ minute. The angular retardation of the wheel in $rad/s^2$ is:

$A$ solid sphere of radius $4\text{ cm}$ and mass $5\text{ kg}$ is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of $1200\text{ rpm}$. It is brought to rest in $10\text{ s}$ by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are . . . . . . and . . . . . . respectively.