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

This is a standard result from classical mechanics. The moment of inertia \(I\) of a solid cone of mass \(M\) and base radius \(R\) about its central axis of symmetry (the vertical axis passing through the apex and the center of the base) is given by the formula: \[ I = \frac{3}{10}MR^2 \] This can be derived by integrating the moment of inertia of infinitesimal circular disks that make up the cone. The moment of inertia of a disk of mass \(dm\) and radius \(r\) is \(\frac{1}{2}r^2 dm\). Integrating this from the apex to the base yields the final result.