Question

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

Short answer

The short answer to the question is: The moment of inertia of the disc about an axis passing through its center and normal to the disc is \(I₂ = (1/2)MR²\). The correct option is \(\mathrm{B}\\(1 /2) \mathrm{MR}^{2}\).

Step-by-step solution

Write the given information.

We know that the moment of inertia, I₁, of the disc about any of its diameter is (1/4)MR², and we need to find the moment of inertia, I₂, about an axis passing through the center of the disc and normal to the disc.

Write Parallel Axis Theorem formula

According to the Parallel Axis Theorem: I₂ = I₁ + Md²

Substitute given values into the formula.

We know I₁ = (1/4)MR² and d = R/2. Let's substitute these values into the formula: I₂ = (1/4)MR² + M(R/2)²

Simplify the equation.

Now, let's simplify the equation: I₂ = (1/4)MR² + M(R²/4) I₂ = MR²(1/4 + 1/4)

Solve for I₂.

Finally, let's solve for I₂: I₂ = MR²(1/2) The moment of inertia of the disc about an axis passing through its center and normal to it is: \(\boxed{I₂ = (1/2)MR²}\) Therefore, the correct answer is \(\mathrm{B}\\(1 /2) \mathrm{MR}^{2}\).