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krek1111 [17]
3 years ago
5

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 throu

gh its center?

Physics
2 answers:
faltersainse [42]3 years ago
8 0
<span>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</span>
Leya [2.2K]3 years ago
4 0

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

\texttt{ }

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Momentum</em><em> and </em><em>Moment of Inertia of Cylinder</em><em> formula as follows:</em>

\boxed {L = I \omega}

<em>where:</em>

<em>L = angular momentum ( kg.m²/s )</em>

<em>I = moment of inertia ( kg.m² )</em>

<em>ω = angular frequency ( rad/s )</em>

\texttt{ }

\boxed { I = \frac{1}{2} M R^2 }

<em>where:</em>

<em>I = moment of inertia ( kg.m² )</em>

<em>M = mass of object ( kg )</em>

<em>R = radius of object ( m )</em>

\texttt{ }

<u>Given:</u>

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

<u>Asked:</u>

moment of inertia = I = ?

<u>Solution:</u>

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

I = \frac{1}{2} \times 0.025 \times 0.065^2

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

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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