Small blocks, each with mass ?m?, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end.

Solution 32E Step 1: Equation for the moment of inertia, I = MR 2 Where, M - Mass of the body R - Distance of the mass from the axis Step 2: a) in the first case, we can represent the system as in the diagram 2 Then, the moment of inertia of mass “m” at the end, I end= m(L/2) Iend= mL /4 One more mass is held at another end. Therefore, the total moment of inertia, I’ = 2 × mL /4 2 I’ = mL /2 Since, the axis is passing through the center, the moment of inertia due to central mass will be zero. 2 The moment of inertia of the system, I’ = mL /2