Question

Moment of inertia of a solid cone about its vertical axis is:

• A. \(MR^2/10\)
• B. \(3MR^2/10\)
• C. \(5MR^2/10\)
• D. \(7MR^2/10\)

Hint:

It's helpful to memorize the moments of inertia for common shapes: - Thin Rod (center): \(ML^2/12\) - Hoop (central axis): \(MR^2\) - Solid Cylinder/Disk (central axis): \(MR^2/2\) - Solid Sphere (center): \(2MR^2/5\) - Solid Cone (central axis): \(3MR^2/10\)

Answer 1

The Correct Option is B

The moment of inertia \(I\) for a solid cone with mass \(M\) and base radius \(R\) about its central axis of symmetry is \( I = \frac{3}{10}MR^2 \). This result from classical mechanics is obtained by integrating the moments of inertia of constituent infinitesimal circular disks, where a disk of mass \(dm\) and radius \(r\) has a moment of inertia of \(\frac{1}{2}r^2 dm\).