Q.A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate the moment of inertia of the dumb-bell when rotated about an axis passing through its centre and perpendicular to the length.

Rod's own MI = $\frac{1}{12}(60)(20)^2=2000$ g cm^2; each sphere's own MI ($\frac{2}{5}(25)(10)^2=1000$ g cm^2) plus parallel-axis shift over distance 20 cm (10 cm half-rod + 10 cm sphere radius, since the sphere is mounted AT the rod's end) gives …

The dumb-bell's rotation axis passes through its centre (the rod's midpoint), perpendicular to the rod's length.

Rod's own moment of inertia about this axis (Table 3, perpendicular to length, through centre): $I_{rod}=\frac{1}{12}m_{rod}L^2=\frac{1}{12}(60)(20)^2=\frac{1}{12}(60)(400)=2000\text{ g cm}^2$

Each sphere is mounted AT an end of the rod, meaning its own centre sits a further sphere-radius (10 cm) beyond the rod's end (10 cm from the rod's midpoint), so each sphere's centre is at a distance $d=10+10=20$ cm from the rotation axis. Each sphere's OWN moment of inertia, about an axis through its own centre (parallel to the rotation axis): $I_{sphere,own}=\frac{2}{5}m_{sphere}r_{sphere}^2=\frac{2}{5}(25)(10)^2=\frac{2}{5}(25)(100)=1000\text{ g cm}^2$ Using the parallel …

Find the rod's own moment of inertia directly (standard formula), then find each sphere's own moment of inertia about its own centre and shift it to the rotation axis using the parallel-axes theorem, taking the s …

Placing each sphere's centre exactly AT the rod's end (distance 10 cm from the axis) instead of a further sphere-radius beyond it (distance 20 cm), which under-counts the parallel-axis s …