Question

A 15-cm-diameter CD has a mass of 25 g.A. What is the CD's moment of inertia for rotation about a perpendicular axis through its center?B. What is the CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk?

Answer 1

Solution :

The moment of inertia for a circular disc about its centre is given by

$I=\frac{mr^2}{2}$

Given , radius of CD, $r=\frac{d}{2}$

$r=\frac{1.5 \times 10^{-2}}{2}$

  $=7.5 \times 10^{-2} \ m$

A).

The moment of inertia of CD about its centre is

$I_{centre}=\frac{mr^2}{2}$

          $=\frac{25 \times 10^{-3}\times (7.5 \times 10^{-2})^2}{2}$

          $= 7\times 10^{-5} \ kg \ m^2$

B).

The moment of inertia about parallel axis of CD with respect to its central axis is given by the parallel axis theorem,

$I_{p}=I_c + md^2$

where, d is the distance between the parallel axis and central axis

Therefore, moment of inertia of CD about its edge is

$I_p=I_c +md^2$

   $=\frac{mr^2}{2}+mr^2$

   $=\frac{3}{2}mr^2$

   $=\frac{3(25 \times 10^{-3})\times (7.5 \times 10^{-2})^2}{2}$

   $=2.1 \times 10^{-4} \ kg \ m^2$

Answer 2

(a) The  CD's moment of inertia for rotation about a perpendicular axis through its center is $7.031 \times 10^{-5} \ kgm^2$

(b) The CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk is $2.11 \times 10^{-4} \ kgm^2$

The given parameters;

• diameter of the CD, d = 15 cm
• radius of the CD, r = 7.5 cm
• mass of the CD, m = 25 g = 0.025 kg

(A) The  CD's moment of inertia for rotation about a perpendicular axis through its center is calculated as follows;

$I_c = \frac{1}{2} mr^2\\\\I_c = \frac{1}{2} \times 0.025 \times (0.075)^2\\\\I_c = 7.03 1 \times 10^{-5} \ kgm^2$

(B) The CD's moment of inertia for rotation about a perpendicular axis through the edge of the disk is calculated as;

$I_p = I_c + mr^2\\\\I_p = (7.031 \times 10^{-5}) \ + \ 0.025 \times (0.075)^2\\\\I_p = 2.11 \times 10^{-4} \ kgm^2$

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