Question

2.Derive moment of inertia of a solid sphere

Answer 1

Answer on Question #63869, Physics / Mechanics | Relativity

Derive moment of inertia of a solid sphere.

Answer:

Direct calculation of the moment of inertia of the body about the axis reduced to the evaluation of the integral

$$J = \int r ^ {2} d m$$

Where $r$ is the distance of elemental mass $dm$ to the axis of rotation.

Calculate the moment of inertia of the sphere (in spherical coordinate’s $r, \theta, \varphi$)

$$dm = \frac {m}{V} \, d V = \frac {m}{V} r ^ {2} \sin \theta \cdot d r \cdot d \theta \cdot d \varphi$$

Here $m$ is the mass of bullet, $V$ is its volume.

Since

$$\rho = r \sin \theta$$

Then

$$dJ = \rho ^ {2} \cdot d m = \frac {m}{V} r ^ {4} \sin^ {3} \theta \cdot d r \cdot d \theta \cdot d \varphi$$

So

$$J = \frac {m}{V} \int_ {0} ^ {R} r ^ {4} \, dr \int_ {0} ^ {2 \pi} d \varphi \int_ {0} ^ {\pi} \sin^ {3} \theta \cdot d \theta = \frac {m}{V} \cdot \frac {R ^ {5}}{5} \cdot 2 \pi \cdot \frac {4}{3}$$

Since

$$V = \frac {4}{3} \pi R ^ {3}$$

Finally

$$J = \frac {2}{5} m R ^ {2}$$

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