Calculate the moment of inertia of a circular disc of radius 10 cm, thickness 5 mm, and uniform density 8 g/cc about a transverse axis through the center of the disc.

Calculate the moment of inertia of a circular disc of radius 10 cm, thickness 5 mm, and uniform density 8 g/cc about a transverse axis through the center of the disc.

A) \( 4.5 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \)
B) \( 6.8 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \)
C) \( 8.2 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \)
D) \( 9.6 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \)

Answer: B) \( 6.8 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \)

The moment of inertia \( I \) about a transverse axis through the center of the disc can be calculated using the formula:

\[ I = \frac{1}{2} M R^2 \]

Where:

• Mass of the disc, \( M \): The mass is the product of the volume and density:
\[ \text{Volume} = \pi R^2 t \] \[ \text{Volume} = \pi \times (0.1)^2 \times 0.005 = 0.00157 \, \text{m}^3 \] \[ \text{Density} = 8 \, \text{g/cc} = 8000 \, \text{kg/m}^3 \] \[ M = V \times \rho = 0.00157 \times 8000 = 12.56 \, \text{kg} \]
• Moment of inertia calculation:
\[ I = \frac{1}{2} M R^2 \] \[ I = \frac{1}{2} \times 12.56 \times (0.1)^2 \] \[ I = 6.8 \times 10^{-2} \, \text{kg} \cdot \text{m}^2 \]

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Calculate the moment of inertia of a circular disc of radius 10 cm, thickness 5 mm, and uniform density 8 g/cc about a transverse axis through the center of the disc.