Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :- Show (dL')/(dt)=(sumr_i') xx (dp')/(dt)` Further, show that (dL')/(dt)=`tau'_(ext)` where `tau'_(ext)` is the sum of all external torques acting on the system about the centre of mass.

Separation of mmotion of a systme of particles into motion of the centre of mass and motion about the centre of mass Show vecp = vec p_i + m_i vecV where vecp_i is the momentum of the ith particle

Separation of mmotion of a systme of particles into motion of the centre of mass and motion about the centre of mass Show K = K + 1/2 MV^2 , where k is the total kinetic energy of the system of particles, K is the total kinetic energy of the system when the particle velcoities are taken with respect to the centre of mass and MV^2//2 is the kinetic energy of the translation of the system as a whole (i,e of the centre of mass motion of the system).

Separation of mmotion of a systme of particles into motion of the centre of mass and motion about the centre of mass Show vec L = L = vec L + vecM R xx vecV where vec L = Sigma vecr_i xx vecp_i is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember vecr_i = vecr_i - vecR , rest f the notation is the standard notation used in the chapter. Note vecL and vecMR xx vecV can be said to be angular moemnta respectively about and of the centre of mass of the system of particles.

The centre of mass of a system of particles moves as if all the _________ of the system was concentrated at the _________ of mass and all the external forces acting on the system were applied directly at this point.

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? Wil you require external forces to sustain the motion?

Two particles , each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2 R . The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod

Different points in earth are at slightly different distancesfrom the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the C.M. (centre of mass ) causing rotation around an axis thorugh the C.M. For the earth-sun system (approximating the earth as a uniform density sphere)

Two particles, each of mass mand speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.