The moment of inertia of a pair of spheres, each having mass $m$ and radius $r$, kept in contact, about the tangent passing through the point of contact is:

Question

A thin circular disc of mass $ M $ and radius $ R $ is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity $ \omega $. If another disc of the same dimensions but of mass $ M/2 $ is placed gently on the first disc co-axially, then the new angular velocity of the system is:

Answer 1

Think of the two touching spheres as sitting side by side. The point of contact is the single point where their surfaces meet, and the tangent line drawn there is perpendicular to the line joining the two centres.

For any solid sphere the perpendicular distance from its centre to a tangent of its own surface is exactly the radius $r$. So this tangent is a distance $r$ from each centre.

Moment of inertia of a solid sphere about a diameter is $\dfrac{2}{5}mr^2$. Shifting the axis out to a tangent a distance $r$ away, by the parallel axis theorem, gives for one sphere\[I_{\text{tangent}} = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2.\]There are two identical spheres, each the same distance $r$ from the same tangent line, so we simply double it:\[I = 2\times\frac{7}{5}mr^2 = \frac{14}{5}mr^2.\]\[\boxed{I = \frac{14mr^2}{5}}\]

The moment of inertia of a pair of spheres, each having mass \(m\) and radius \(r\), kept in contact, about the tangent passing through the point of contact is:

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The Correct Option is C

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