What constant force must be applied tangentially at the equator to stop the Earth's rotation in one day?

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 of radius $0.2 \ m$ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $10 \ N$ as shown in the figure. The established torque produces an angular acceleration of $2 \ rad/s^2$. The moment of inertia of the wheel is . . . . . . $kg \ m^2$. (Acceleration due to gravity $= 10 \ m/s^2$)

$A$ thin uniform hemispherical bowl of mass $m$ and radius $R$ is lying on a smooth horizontal surface. $A$ horizontal force $F$ is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration of the bowl will be

$A$ rod of length $50\,cm$ is pivoted at one end. It is raised such that it makes an angle of $30^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$) will be $(g = 10\,ms^{-2})$.

$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is: