Problem 648 - JEE Physics

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 \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

Step-by-Step Solution

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 1: 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.

Step 2: Write Parallel Axis Theorem formula

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

Step 3: 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

Step 4: Simplify the equation.

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

Step 5: 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 \(\mathrm{B}\\(1 /2) \mathrm{MR}^{2}\).

Key Concepts

Parallel Axis Theorem

The Parallel Axis Theorem is a fundamental concept in understanding how the moment of inertia can be adjusted when considering an axis that is not through the center of mass. It allows for the calculation of a complex moment of inertia from a simpler known one.

Here's how it works:

• The theorem states that if you know the moment of inertia (\(I_c\)) of an object about an axis through its center of mass, you can find the moment of inertia (\(I\)) about any other parallel axis by adding the product of the mass (\(M\)) of the object and the square of the distance (\(d\)) between the two axes.
• Mathematically, it's expressed as: \[I = I_c + Md^2\]

In the context of a circular disc, this theorem is handy because you often know the inertia about the diameter or center, and need to calculate it about a different axis.

With the given exercise, the theorem helps us correctly calculate the moment of inertia when the axis is normal to the disc and goes through the center, as seen when applying the provided formula.

uniform circular disc

A uniform circular disc is a simple yet profound object in physics, particularly when discussing rotation and inertia. These discs are characterized by having uniform mass distribution across their surface.

Let's dive into some key characteristics:

• They have a constant mass density, meaning anywhere on the disc, the mass remains the same relative to its size.
• The geometry is perfectly circular, which simplifies calculations like those for the moment of inertia.

In terms of physics, this uniformity makes it easier to apply formulas and derive results.

When dealing with real-world applications like wheels or pulleys, uniform circular discs are a great model to study rotational dynamics efficiently.

axis of rotation

The axis of rotation is crucial when calculating the moment of inertia. It defines the line around which the object, like our circular disc, spins or rotates.

Why is it important?

• The position and orientation of this axis influence the rotational behavior and inertia of the object.
• Changing the axis can significantly alter the moment of inertia, as seen with the Parallel Axis Theorem.

In the exercise, we look at two axes:

• One through the diameter – giving us a baseline inertia of \((1/4)MR^2\).
• Another is perpendicular to the disc at its center – resulting in the simpler calculation.

Thus, understanding the axis of rotation aids in determining how an object will behave under rotational forces, making it essential for engineers and scientists working with rotating systems.