$A$ small ball of mass $m$ is suspended from the ceiling by a string of length $L$. The ball moves along a horizontal circle with constant angular velocity $\omega$, as shown in the figure. The torque about the centre $(O)$ of the horizontal circle is

$A$ $V$-shaped rigid body has two identical uniform arms. What must be the angle between the two arms,so that when the body is hung from one end,the other arm is horizontal?

An object of mass $8\,kg$ is hanging from one end of a uniform rod $CD$ of mass $2\,kg$ and length $1\,m$ pivoted at its end $C$ on a vertical wall as shown in the figure. It is supported by a cable $AB$ such that the system is in equilibrium. The tension in the cable is $............\,N$ (Take $g=10\,m/s^2$)

$A$ uniform rod of length $200 \, cm$ and mass $500 \, g$ is balanced on a wedge placed at $40 \, cm$ mark. $A$ mass of $2 \, kg$ is suspended from the rod at $20 \, cm$ and another unknown mass $'m'$ is suspended from the rod at $160 \, cm$ mark as shown in the figure. Find the value of $'m'$ such that the rod is in equilibrium. $(g = 10 \, m/s^2)$

For a system to be in equilibrium,the torques acting on it must balance. This is true only if the torques are taken about

$A$ wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However,the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium?
How would you set a half wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?