Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is $$\frac{x}{4}Ma^2$$. Then, the value of $$x$$ will be _____.

Solution & Explanation

Let each identical disc have a mass $$M$$ and a diameter $$a$$. The radius ($$r$$) of each disc is given by:

$$r = \frac{a}{2}$$

The axis $$OO'$$ lies in the plane of the discs. It passes vertically straight through the centers of the top and bottom discs, and acts as a tangent line to the left and right discs.

Let us find the individual contributions of each disc to the total moment of inertia about $$OO'$$:

• For the top and bottom discs (2 discs): The axis $$OO'$$ passes directly through their diameters. The moment of inertia of a uniform disc about its diameter is:

  $$I_{\text{diameter}} = \frac{1}{4}Mr^2 = \frac{1}{4}M\left(\frac{a}{2}\right)^2 = \frac{1}{16}Ma^2$$

  Combined contribution for both discs: $$2 \times \frac{1}{16}Ma^2 = \frac{1}{8}Ma^2$$

• For the left and right discs (2 discs): The axis $$OO'$$ is tangent to these discs in their plane. According to the Parallel Axis Theorem ($$I = I_{\text{diameter}} + Md^2$$), where the distance to the axis is $$d = r = \frac{a}{2}$$:

  $$I_{\text{tangent}} = \frac{1}{4}Mr^2 + M r^2 = \frac{5}{4}Mr^2 = \frac{5}{4}M\left(\frac{a}{2}\right)^2 = \frac{5}{16}Ma^2$$

  Combined contribution for both discs: $$2 \times \frac{5}{16}Ma^2 = \frac{5}{8}Ma^2$$

Summing up the contributions of all four discs gives the total moment of inertia ($$I_{\text{total}}$$) of the system:

$$I_{\text{total}} = \frac{1}{8}Ma^2 + \frac{5}{8}Ma^2 = \frac{6}{8}Ma^2 = \frac{3}{4}Ma^2$$

Comparing our result to the given expression template from the problem statement, $$\frac{x}{4} Ma^2$$:

$$\frac{x}{4}Ma^2 = \frac{3}{4}Ma^2 \implies x = 3$$

Correct Answer Value: 3