Rotation of Rigid Bodies:
Let us consider a couple of questions involving rigid body rotation.
(1) A meter scale is held vertically with one end on the floor and is allowed to fall. Assuming that the end on the floor does not slip, what will be the linear velocity of the other end when it strikes the floor?
(a) 2.7m/s (b) 3.1m/s (c) 5.4m/s (d) 9.8m/s (e) 11.2m/s
When the meter scale is allowed to fall, its gravitational potential energy gets converted into rotational kinetic energy so that we have
mgl/2 = ½ I ω ² where ‘l’ is the length(1metre for a metre scale), ‘m’ is the mass, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the scale. l/2 appears in the potential energy expression since the centre of gravity of the scale is initially at a height l/2. You should note that the moment of inertia of the scale is about the end in contact with the floor and is equal to ml²/3.
From the above equation, ω =√(3g/l).
The linear velocity of the feree end of the scale = ωl = √(3gl). On substituting for l(=1) and g(=9.8) the linear velocity is 5.4m/s [Option (c)].
(2) An impulsive force F acting for a short time interval ∆t is applied at one end of a thin uniform bar of mass M and length L, in a direction perpendicular to the lengthof the bar. The angular velocity with which the bar will rotate is
(a) F∆t/4ML (b) F∆t/2ML (c)2F∆t/ML (d) 4F∆t/ML (e) 6F∆t/ML
The impulse received by the bar is F∆t which is equal to the linear momentum supplied. The bar will rotate about its centre of mass. The ‘lever arm’ for the angular momentum is L/2 so that we have, (L/2)F∆t = Iω = ML²ω /12. So, ω =6F∆t/ML

Circular motion of electrons- Angular Momentum and magnetic dipole moment:
If an electron (charge ‘e’, mass ‘m’) is in uniform circular motion with angular velocity ‘ω’ the orbital angular momentum of the electron about a perpendicular axis passing through the centre of the circle is mωr^2 (=mvr). Often you will encounter questions regarding the direction of the angular momentum vector. You should note that the orbital angular momentum of the electron is directed perpendicular to the plane of the orbit.
The magnetic dipole moment of an orbiting electron is eωr^2/2 (= evr/2). [Magnetic dipole moment = IA = (e/T)×πr^2 = (eω/2π)×πr^2 = eωr^2/2=evr/2. Here I is the equivalent current, T is the orbital period and r is the orbital radius].
Now consider the following M.C.Q. which appeared in Kerala Engineering Entrance Test paper of 2006:
Two electrons (each of charge = e and mass = m) are attached one at each end of a light rigid rod of length 2r. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the angular momentum about the axis of rotation to the magnetic dipole moment of the system is
(a) 2me/3 (b) e^2/2m (c) 2×specific charge of electron (d) 5m/2e (e) 2m/e
The ‘light rigid rod’ in the problem could have been dispensed with.
Angular momentum of two electrons = 2mωr^2. As shown above, the magnetic dipole moment of two orbiting electrons = 2 eωr^2/2. The ratio of angular momentum to the magnetic dipole moment = 2m/e. [Option(e)].
Note that this result is independent of the number of electrons.
You should also note that the ratio of the magnetic moment to the angular momentum of a charged particle in angular motion is called gyromagnetic ratio.