A circular plate of side l has mass per unit area mew find its moment of inertia about an Axis passing through the centre and perpendicular to its plane

To find the moment of inertia of a circular plate with a uniform mass per unit area, we can use some fundamental principles of physics and calculus. The moment of inertia is a measure of an object's resistance to rotational motion about a given axis. For a circular plate of radius \( r \) and mass per unit area \( \mu \), we can derive the moment of inertia about an axis passing through its center and perpendicular to its plane.

Understanding the Components

First, let's clarify the variables involved:

• Radius (r): The distance from the center of the plate to its edge.
• Mass per unit area (μ): This is the mass of the plate divided by its area, typically expressed in kg/m².
• Moment of Inertia (I): This is what we are trying to calculate, representing how mass is distributed relative to the axis of rotation.

Calculating the Moment of Inertia

The formula for the moment of inertia \( I \) of a thin circular plate about an axis through its center and perpendicular to its plane can be derived using integration. The moment of inertia is given by the integral:

I = ∫ r² dm

Here, \( dm \) is the mass element, and \( r \) is the distance from the axis of rotation. For a circular plate, we can express \( dm \) in terms of the area element \( dA \) and the mass per unit area \( \mu \):

Since \( dm = \mu \cdot dA \), we can rewrite the integral as:

I = ∫ r² (μ dA)

Setting Up the Integral

To evaluate this integral, we can use polar coordinates, where the area element \( dA \) is given by:

dA = r' dr' dθ

Here, \( r' \) is the radial distance from the center, and \( θ \) is the angle. The limits for \( r' \) will be from 0 to \( R \) (the radius of the plate), and for \( θ \) from 0 to \( 2π \).

Evaluating the Integral

Substituting \( dA \) into our moment of inertia equation gives:

I = ∫₀²π ∫₀ᴿ (r')² (μ r' dr' dθ)

Now we can separate the integrals:

I = μ ∫₀²π dθ ∫₀ᴿ (r')³ dr'

The integral over \( θ \) evaluates to \( 2π \), and the integral over \( r' \) is:

∫₀ᴿ (r')³ dr' = \left[ \frac{(r')^4}{4} \right]₀ᴿ = \frac{R^4}{4}

Putting it all together, we have:

I = μ (2π) \left( \frac{R^4}{4} \right) = \frac{μπR^4}{2}

Relating Mass and Area

Next, we need to express the mass \( M \) of the plate in terms of the mass per unit area and the area of the plate:

M = μ \cdot A = μ \cdot πR²

Now, substituting \( μ \) back into our moment of inertia equation:

I = \frac{M}{πR²} \cdot \frac{πR^4}{2} = \frac{1}{2} MR²

Final Result

Thus, the moment of inertia of a circular plate about an axis through its center and perpendicular to its plane is:

I = \frac{1}{2} MR²

This result shows that the moment of inertia depends on both the mass and the square of the radius of the plate. This relationship is crucial in understanding how objects rotate and is widely applicable in various fields of physics and engineering.