Two spheres each of mass M and radius R/2 are | vedclass

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(A) The moment of inertia of the system about the axis $YY'$ passing through the center of the right sphere is the sum of the moments of inertia of bo

Vedclass · Accepted Answer

$\frac{21}{5} M R^2$

Vedclass · Answer

(A) The moment of inertia of the system about the axis $YY'$ passing through the center of the right sphere is the sum of the moments of inertia of both spheres about this axis.For the right sphere,the axis passes through its center,so its moment of inertia is $I_1 = \frac{2}{5} M (R/2)^2 = \frac{2}{5} M (R^2/4) = \frac{1}{10} M R^2$.For the left sphere,the axis is at a distance $d = 2R$ from its center. Using the parallel axis theorem,its moment of inertia is $I_2 = I_{cm} + Md^2 = \frac{2}{5} M (R/2)^2 + M(2R)^2 = \frac{1}{10} M R^2 + 4 M R^2 = \frac{41}{10} M R^2$.The total moment of inertia is $I = I_1 + I_2 = \frac{1}{10} M R^2 + \frac{41}{10} M R^2 = \frac{42}{10} M R^2 = \frac{21}{5} M R^2$.

Two spheres each of mass $M$ and radius $R/2$ are connected by a massless rod of length $2R$ as shown in the figure. The moment of inertia of the system about an axis passing through the center of one of the spheres and perpendicular to the rod is:

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