A uniform rod of length $'l'$ is pivoted at one of its ends on a vertical shaft of negligible radius When the shaft rotates at angular speed $\omega$ the rod makes an angle $\theta$ with it (see figure). To find $\theta$ equate the rate of change of angular momentum (direction going into the paper ) $\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the CM. The value of $\theta$ is then such that:

A disc of radius $5\, m$ is rotating with angular frequency $10\, rad / sec .$ A block of mass $2\, kg$ to be put on the disc friction coefficient between disc and block is $\mu_{ k }=0.4,$ then find the maximum distance from axis where the block can be placed without sliding (in $cm$)

As shown in Figure the two sides of a step ladder $BA$ and $CA$ are $1.6 m$ long and hinged at $A$. A rope $DE, 0.5 \;m$ is tied half way up. A weight $40\;kg$ is suspended from a point $F , 1.2\; m$ from $B$ along the ladder $BA$. Assuming the floor to be frictionless and neglecting the wetght of the ladder. find the tension in the rope and forces exerted by the floor on the ladder. (Take $g=9.8 \;m / s ^{2}$ )

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 

$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2 $and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero then the magnitude of  $\vec F_3$ is