A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system? (b) What is its rotational kinetic energy?

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Md Ariful Islam · Answer 1 · Apr 06, 2024 05:52:07

Given:

Mass of the disk, $ m_1 = 2.0 \, \text{kg} $
Radius of the disk, $ r_1 = 50 \, \text{cm} = 0.50 \, \text{m} $
Mass of the annular cylinder, $ m_2 = 1.0 \, \text{kg} $
Inner radius of the annular cylinder, $ r_{\text{inner}} = 20 \, \text{cm} = 0.20 \, \text{m} $
Outer radius of the annular cylinder, $ r_{\text{outer}} = 30 \, \text{cm} = 0.30 \, \text{m} $
Angular velocity, $ \omega = 10 \, \text{rev/s} $

To find the moment of inertia of the system:

The moment of inertia of the disk ($ I_1 $) is given by:

\[ I_1 = \frac{1}{2} m_1 r_1^2 \]

The moment of inertia of the annular cylinder ($ I_2 $) is given by:

\[ I_2 = \frac{1}{2} m_2 (r_{\text{outer}}^2 + r_{\text{inner}}^2) \]

Now, let's calculate $ I_1 $ and $ I_2 $ and then sum them up to find the total moment of inertia $ I $ of the system.

(b) To find the rotational kinetic energy ($ K $):

Rotational kinetic energy is given by:

\[ K = \frac{1}{2} I \omega^2 \]

Substituting the value of $ I $ calculated in part (a), we can find $ K $.