What is the moment of inertia of a uniform circular disc of mass 500 g (0.5 kg) and radius 10 cm (0.1 m) about: 1. The diameter of the disc. 2. An axis tangent to the disc and parallel to its diameter. 3. An axis through the center of the disc and perpendicular to its plane?

What is the moment of inertia of a uniform circular disc of mass 500 g (0.5 kg) and radius 10 cm (0.1 m) about:

1. The diameter of the disc.

2. An axis tangent to the disc and parallel to its diameter.

3. An axis through the center of the disc and perpendicular to its plane?

A) Moment of inertia about diameter: \( 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about tangent axis: \( 6.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about center: \( 5 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)

B) Moment of inertia about diameter: \( 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about tangent axis: \( 0.00625 \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about center: \( 0.0025 \, \text{kg} \cdot \text{m}^2 \)

C) Moment of inertia about diameter: \( 6.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about tangent axis: \( 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about center: \( 5 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)

D) Moment of inertia about diameter: \( 0.005 \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about tangent axis: \( 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about center: \( 6.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)

Answer: B) Moment of inertia about diameter: \( 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about tangent axis: \( 0.00625 \, \text{kg} \cdot \text{m}^2 \)
Moment of inertia about center: \( 0.0025 \, \text{kg} \cdot \text{m}^2 \)

The moment of inertia of a uniform circular disc about its diameter is calculated as:

\[ I_{\text{diameter}} = \frac{1}{4} M R^2 \]

Substituting \( M = 0.5 \, \text{kg} \) and \( R = 0.1 \, \text{m} \):

\[ I_{\text{diameter}} = \frac{1}{4} \times 0.5 \times (0.1)^2 = 1.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \]

The moment of inertia about an axis tangent to the disc and parallel to its diameter is found using the parallel axis theorem:

\[ I_{\text{tangent}} = I_{\text{diameter}} + M R^2 \]

Substituting the values:

\[ I_{\text{tangent}} = 1.25 \times 10^{-3} + 0.5 \times (0.1)^2 = 6.25 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \]

The moment of inertia about the axis through the center of the disc and perpendicular to its plane is:

\[ I_{\text{center}} = \frac{1}{2} M R^2 \]

Substituting the values:

\[ I_{\text{center}} = \frac{1}{2} \times 0.5 \times (0.1)^2 = 2.5 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \]

MCQ On Motion In Plane | Newton's Laws Of Motion | CBSE, IIT JEE, NEET, CET | Physics PYQ

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What is the moment of inertia of a uniform circular disc of mass 500 g (0.5 kg) and radius 10 cm (0.1 m) about:

1. The diameter of the disc.

2. An axis tangent to the disc and parallel to its diameter.

3. An axis through the center of the disc and perpendicular to its plane?