How is the moment of inertia of a uniform disk rotating about its diameter derived using integration?
Conditions
- The disk is uniform.
- The rotation axis is one of its diameters (e.g., the y-axis).
- Polar coordinates are used with the angle measured from the x-axis.
Reasoning, step by step
- Establish a coordinate system where the rotation axis aligns with the y-axis.
- Define the surface density and the infinitesimal area element .
- Express the differential mass element as .
- Determine the squared distance from the area element to the rotation axis as .
- Set up the double integral for the moment of inertia over the disk region .
- Separate the integral into angular and radial components.
- Evaluate the integrals.
- Combine the result with the total mass to simplify to .
Example
The script states: 'To perform the integration, the problem utilizes surface density ρ and polar coordinates. An infinitesimal area element dσ is defined, converting to a mass element dm = ρdσ. The squared distance from this element to the rotation axis is expressed as r²sin²θ. By integrating over the entire disk region D, the double integral separates into angular and radial components. Evaluating these yields a result that, when combined with the total mass M, simplifies to the final formula MR².'
Common misconceptions
- Using instead of for the distance to the y-axis, though symmetry means both diameters give the same result.
- Forgetting to include the Jacobian when converting the area element to polar coordinates ().
- Confusing the moment of inertia about the diameter with the moment of inertia about the central vertical axis ().
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