Question

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:

• A. \(\dfrac{4mr^2}{5}\)
• B. \(\dfrac{7mr^2}{5}\)
• C. \(\dfrac{14mr^2}{5}\)
• D. \(\dfrac{5mr^2}{14}\)

Hint:

Tangent is a distance \(r\) from each centre. Find \(\tfrac{2}{5}mr^2 + mr^2\) per sphere, then double it.

Answer 1

The Correct Option is C

Step 1: The moment of inertia of a solid sphere about an axis through its centre is\[I_{cm} = \frac{2}{5}mr^2.\]
Step 2: The tangent passes through the point where the two spheres touch. This tangent line is at a perpendicular distance \(r\) from the centre of each sphere (it just grazes each sphere's surface).

Step 3: Using the parallel axis theorem for one sphere about this tangent,\[I_{1} = I_{cm} + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2.\]
Step 4: Both spheres are the same distance \(r\) from the tangent, so the second sphere gives the same value \(I_2 = \dfrac{7}{5}mr^2\). Total moment of inertia is\[I = I_1 + I_2 = 2\times\frac{7}{5}mr^2 = \frac{14}{5}mr^2.\]This is option (C).\[\boxed{I = \frac{14mr^2}{5}}\]