Question

A solid sphere and a disc of same mass \(M\) and radius \(R\) are kept such that their curved surfaces are in contact and their centers lie along the same horizontal line. The moment of inertia of the two body system about an axis passing through their point of contact and perpendicular to the plane of the disc is:

• A. \( \frac{53 MR^2}{20} \)
• B. \( \frac{39 MR^2}{10} \)
• C. \( \frac{29 MR^2}{10} \)
• D. \( \frac{9 MR^2}{10} \) \bigskip

Hint:

the axis is through the point of contact, we need to use the parallel axis theorem to shift the axis from the center to the point of contact.

Answer 1

The Correct Option is C

We are given that a solid sphere and a disc have the same mass \(M\) and radius \(R\), and their centers lie along the same horizontal line. Their curved surfaces are in contact, and we are tasked with finding the moment of inertia of the two-body system about an axis passing through their point of contact and perpendicular to the plane of the disc. Step 1: Moment of inertia of a solid sphere The moment of inertia of a solid sphere about an axis through its center is given by: \[ I_{\text{sphere, center}} = \frac{2}{5} M R^2 \] However, since the axis is through the point of contact, we need to use the parallel axis theorem to shift the axis from the center to the point of contact. The distance between the center of the sphere and the point of contact is \(R\), so the moment of inertia about the new axis is: \[ I_{\text{sphere}} = I_{\text{sphere, center}} + M R^2 = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 \] Step 2: Moment of inertia of a disc The moment of inertia of a disc about an axis through its center and perpendicular to the plane of the disc is: \[ I_{\text{disc, center}} = \frac{1}{2} M R^2 \] Using the parallel axis theorem again, the distance between the center of the disc and the point of contact is \(R\), so the moment of inertia of the disc about the new axis is: \[ I_{\text{disc}} = I_{\text{disc, center}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2 \] Step 3: Total moment of inertia The total moment of inertia of the system is the sum of the moments of inertia of the sphere and the disc: \[ I_{\text{total}} = I_{\text{sphere}} + I_{\text{disc}} = \frac{7}{5} M R^2 + \frac{3}{2} M R^2 \] To combine these terms, we find a common denominator: \[ I_{\text{total}} = \frac{14}{10} M R^2 + \frac{15}{10} M R^2 = \frac{29}{10} M R^2 \] Thus, the moment of inertia of the system about the given axis is \( \frac{29 MR^2}{10} \).