A thin, uniform rod has length L and mass M.
a. Find by integration its moment of inertia about an axis perpendicular to the rod at one end.
b. Find by integration its moment of inertia about an axis perpencicular to the rod at its center of mass.
When any object is made to rotate about some axis, the net torque "T", required to give the object an angular acceleration 'A', is related to 'I', the moment of inertia of the object about that axis, by the expression: 'T = I A'. If the object is very small, with mass 'm', and distance 'r' from the axis, then its moment of inertia, 'I', about that axis is 'r^2 m'. This is a scalar quantity, so if a system consists of more than one small mass, then we must add the individual contributions, as expressed by: 'Itotal = I1 + I2 + I3 ... = r1^2 m1 + r2^2 m2 + r3^2 m3 ...' and so on, or, more simply, 'I = Sum (ri^2 mi)'. This works for small masses, ('point masses') ...
The expert examines the moment of inertia of a thin uniform rod by integration.