Moment of Inertia of Solid Sphere about Diameter: Derivation

TITLE: Moment of Inertia of Solid Sphere about Diameter: Derivation DESCRIPTION: Learn the step-by-step derivation of the moment of inertia of a solid sphere about its diameter using integration and the parallel-axis theorem.

Concept Overview

This question tests the understanding of rotational motion, specifically the calculation of the moment of inertia for a continuous mass distribution. We will derive the moment of inertia of a solid sphere about its diameter using the method of integration, breaking the sphere into simpler components whose moments of inertia are known or easily calculable. This process also implicitly utilizes the concept of summing up infinitesimal contributions to find the total property.

Step 1: Visualize the solid sphere and choose a coordinate system. Let's consider a solid sphere of uniform mass density $\rho$ and radius $R$. We'll place its center at the origin $(0,0,0)$ and align the diameter of rotation along the z-axis. The equation of the sphere is $x^2 + y^2 + z^2 = R^2$.

Step 2: Divide the sphere into infinitesimal elements for integration. A convenient way to integrate is to consider thin disks perpendicular to the axis of rotation (the z-axis). Imagine slicing the sphere into infinitesimally thin disks of thickness $dz$ at a distance $z$ from the center.

Step 3: Determine the properties of an infinitesimal disk. Each disk is essentially a cylinder with radius $r$ and thickness $dz$. The radius $r$ of the disk at a height $z$ can be found from the sphere's equation: $r^2 + z^2 = R^2$, so $r^2 = R^2 - z^2$. The volume of this infinitesimal disk is $dV = \pi r^2 dz = \pi (R^2 - z^2) dz$.

Step 4: Calculate the mass of the infinitesimal disk. The mass $dm$ of this disk is its volume multiplied by the mass density $\rho$: $dm = \rho dV = \rho \pi (R^2 - z^2) dz$

Step 5: Determine the moment of inertia of the infinitesimal disk about its diameter. The moment of inertia of a solid disk of mass $m$ and radius $r$ about its central axis (perpendicular to its plane) is given by $dI_{disk} = \frac{1}{2} m r^2$. In our case, the mass is $dm$ and the radius is $r$. So, the moment of inertia of our infinitesimal disk about the z-axis (which is a diameter for this disk) is: $dI_z = \frac{1}{2} (dm) r^2$ Substituting the expressions for $dm$ and $r^2$: $dI_z = \frac{1}{2} [\rho \pi (R^2 - z^2) dz] (R^2 - z^2)$ $dI_z = \frac{1}{2} \rho \pi (R^2 - z^2)^2 dz$

Step 6: Integrate to find the total moment of inertia of the sphere. To find the total moment of inertia $I$ of the sphere about the diameter, we need to sum up the moments of inertia of all such disks from $z = -R$ to $z = R$: $I = \int_{-R}^{R} dI_z = \int_{-R}^{R} \frac{1}{2} \rho \pi (R^2 - z^2)^2 dz$ $I = \frac{1}{2} \rho \pi \int_{-R}^{R} (R^4 - 2R^2 z^2 + z^4) dz$ Since the integrand is an even function, we can integrate from $0$ to $R$ and multiply by 2: $I = \frac{1}{2} \rho \pi \times 2 \int_{0}^{R} (R^4 - 2R^2 z^2 + z^4) dz$ $I = \rho \pi \left[ R^4 z - \frac{2R^2 z^3}{3} + \frac{z^5}{5} \right]_{0}^{R}$ $I = \rho \pi \left( R^4(R) - \frac{2R^2 (R)^3}{3} + \frac{(R)^5}{5} - 0 \right)$ $I = \rho \pi \left( R^5 - \frac{2R^5}{3} + \frac{R^5}{5} \right)$ $I = \rho \pi R^5 \left( 1 - \frac{2}{3} + \frac{1}{5} \right)$ $I = \rho \pi R^5 \left( \frac{15 - 10 + 3}{15} \right)$ $I = \rho \pi R^5 \left( \frac{8}{15} \right)$

Step 7: Express the moment of inertia in terms of the total mass of the sphere. The total mass $M$ of the sphere is its volume ($V = \frac{4}{3} \pi R^3$) multiplied by its density $\rho$: $M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right)$ From this, we can express $\rho$ as: $\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3M}{4 \pi R^3}$ Now substitute this expression for $\rho$ back into the equation for $I$: $I = \left( \frac{3M}{4 \pi R^3} \right) \pi R^5 \left( \frac{8}{15} \right)$ $I = \frac{3M \pi R^5 \times 8}{4 \pi R^3 \times 15}$ $I = \frac{24 M R^2}{60}$ $I = \frac{2}{5} M R^2$

Key Takeaways:

• The moment of inertia of a continuous body is found by integrating the contributions of infinitesimal mass elements ($dI = r^2 dm$).
• A solid sphere can be effectively divided into thin disks perpendicular to the axis of rotation for integration.
• The moment of inertia of a solid disk about its central axis is $\frac{1}{2}mr^2$.
• The final result for the moment of inertia of a solid sphere about its diameter is $\frac{2}{5}MR^2$.

Answer: $\boxed{\frac{2}{5}MR^2}$