A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis through its center?

Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2

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johanrusli johanrusli · Answer 2 · 2020-01-03T09:15:49+01:00

The cd's moment of inertia for rotation about a perpendicular axis through its center is about 5.3 × 10⁻⁵ kg.m²

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Further explanation

Let's recall Angular Momentum and Moment of Inertia of Cylinder formula as follows:

[tex]\boxed {L = I \omega}[/tex]

where:

L = angular momentum ( kg.m²/s )

I = moment of inertia ( kg.m² )

ω = angular frequency ( rad/s )

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

where:

I = moment of inertia ( kg.m² )

M = mass of object ( kg )

R = radius of object ( m )

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Given:

mass of cd = M = 25 g = 0.025 kg

diameter of cd = d = 13 cm = 0.13 m

radius of cd = R = d/2 = 0.13/2 = 0.065 m

Asked:

moment of inertia = I = ?

Solution:

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

[tex]I = \frac{1}{2} \times 0.025 \times 0.065^2[/tex]

[tex]\boxed{I = 5.3 \times 10^{-5} \texttt{ kg.m}^2}[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion