Point masses ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$ are placed at the opposite ends of a rigid of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ${\mathrm{\omega }}_0$ is minimum is given by

$1.<mspace\; width="1em"></mspace>\mathrm{x}<mspace\; width="1em"></mspace>=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\mathrm{L}$
$2.<mspace\; width="1em"></mspace>\mathrm{x}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\mathrm{L}$
$3.<mspace\; width="1em"></mspace>\mathrm{x}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{m}}_2\mathrm{L}}{{\mathrm{m}}_1+<mspace\; width="1em"></mspace>{\mathrm{m}}_2}$
$4.<mspace\; width="1em"></mspace>\mathrm{x}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{m}}_1\mathrm{L}}{{\mathrm{m}}_1+<mspace\; width="1em"></mspace>{\mathrm{m}}_2}$

A wheel is rotating about an axis through its centre at \(720~\text{rpm}.\) It is acted upon by a constant torque opposing its motion for \(8\) seconds to bring it to rest finally. The value of torque in \((\text{N-m })\) is:
(given \(I=\frac{24}{\pi}~\text{kg.m}^2)\) 
1. \(48\)
2. \(72\)
3. \(96\)
4. \(120\)

The centre of the mass of \(3\) particles, \(10~\text{kg},\)  \(20~\text{kg},\) and \(30~\text{kg},\) is at \((0,0,0).\) Where should a particle with a mass of \(40~\text{kg}\) be placed so that its combined centre of mass is \((3,3,3)?\)
1. \((0,0,0)\)
2. \((7.5, 7.5, 7.5)\)
3. \((1,2,3)\)
4. \((4,4,4)\)

Two rotating bodies \(A\) and \(B\) of masses \(m\) and \(2m\) with moments of inertia \(I_A\) and \(I_B\) \((I_B>I_A)\) have equal kinetic energy of rotation. If ${}_{}$\(L_A\) $$and \(L_B\) be their angular momenta respectively, then:
1. \(L_{A} = \frac{L_{B}}{2}\)
2. \(L_{A} = 2 L_{B}\)
3. \(L_{B} > L_{A}\)
4. \(L_{A} > L_{B}\)${}_{}$

A uniform circular disc of radius \(50~\text{cm}\) at rest is free to turn about an axis that is perpendicular to its plane and passes through its centre. It is subjected to a torque that produces a constant angular acceleration of \(2.0~\text{rad/s}^2.\) Its net acceleration in \(\text{m/s}^2\) at the end of \(2.0~\text s\) is approximately:

A disc of radius \(2~\text{m}\) and mass \(100~\text{kg}\) rolls on a horizontal floor. Its centre of mass has a speed of \(20~\text{cm/s}\). How much work is needed to stop it?

The rotational analouge of equation, \(F=\dfrac{mdv}{dt} \) is:

A string is wrapped along the rim of a wheel of the moment of inertia \(0.10~\text{kg-m}^2\) and radius \(10~\text{cm}.\) If the string is now pulled by a force of \(10~\text N,\) then the wheel starts to rotate about its axis from rest. The angular velocity of the wheel after \(2~\text s\) will be: