Two spheres each of mass M and radius frac{R} | vedclass

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(A) Let the two spheres be $P$ and $Q$. The axis of rotation passes through the centre of sphere $Q$ and is perpendicular to the rod.The moment of ine

Vedclass · Accepted Answer

$\frac{21}{5}MR^2$

Vedclass · Answer

(A) Let the two spheres be $P$ and $Q$. The axis of rotation passes through the centre of sphere $Q$ and is perpendicular to the rod.The moment of inertia of the system is $I_{	ext{net}} = I_Q + I_P + I_{	ext{rod}}$.$1$. For sphere $Q$: The axis passes through its centre. The moment of inertia of a solid sphere about its central axis is $I_Q = \frac{2}{5} M r^2$,where $r = \frac{R}{2}$.$I_Q = \frac{2}{5} M \left(\frac{R}{2}ight)^2 = \frac{2}{5} M \left(\frac{R^2}{4}ight) = \frac{1}{10} MR^2$.$2$. For sphere $P$: The axis is at a distance $d = 2R$ from its centre. Using the parallel axis theorem,$I_P = I_{	ext{cm}} + Md^2$.$I_P = \frac{2}{5} M \left(\frac{R}{2}ight)^2 + M(2R)^2 = \frac{1}{10} MR^2 + 4MR^2 = \frac{41}{10} MR^2$.$3$. For the rod: Since the rod is massless,$I_{	ext{rod}} = 0$.Total moment of inertia $I_{	ext{net}} = I_Q + I_P = \frac{1}{10} MR^2 + \frac{41}{10} MR^2 = \frac{42}{10} MR^2 = \frac{21}{5} MR^2$.

Two spheres each of mass $M$ and radius $\frac{R}{2}$ are connected at their centres with a massless rod of length $2R$. What will be the moment of inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod?

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