Question

Moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters is \(\frac{MR^2}{4}\). The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, \(\frac{x}{2}MR^2\). The value of \(x\) is _________.

Hint:

Remember the perpendicular and parallel axis theorems. They are essential for calculating moments of inertia about different axes.

Answer 1

Correct Answer: 3

Step 1: Apply the Perpendicular Axis Theorem

The moment of inertia of a disc about its diameter is given as \(I_d = \frac{MR^2}{4}\). According to the perpendicular axis theorem, the moment of inertia of a planar lamina about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two perpendicular axes in the plane that intersect the perpendicular axis at its point of intersection with the lamina. So, for a disc, the moment of inertia about an axis through its center and perpendicular to the plane is:

\[ I_c = I_d + I_d = 2 \left(\frac{MR^2}{4}\right) = \frac{MR^2}{2}. \]

Step 2: Apply the Parallel Axis Theorem

The moment of inertia about an axis normal to the disc and passing through a point on its edge can be found using the parallel axis theorem:

\[ I_e = I_c + MR^2 \]

where \(I_e\) is the moment of inertia about the edge, \(I_c\) is the moment of inertia about the center, and \(R\) is the distance between the two parallel axes (which is the radius of the disc in this case).

Step 3: Calculate \(I_e\)

Substitute \(I_c = \frac{MR^2}{2}\):

\[ I_e = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2. \]

Step 4: Find the Value of \(x\)

The moment of inertia about the edge is given as \(\frac{x}{2}MR^2\). We have found that \(I_e = \frac{3}{2}MR^2\). Therefore, \(x = 3\).

Conclusion: The value of \(x\) is 3.