A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the y axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the y axis and look toward the origin at the disk). The disk makes one complete rotation every 0.8 s. What is the rotational angular momentum of the disk

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MechEngineer MechEngineer · Answer 1 · 2020-04-16T03:48:23+02:00

Answer:

[tex]29.69 kgm^2/s[/tex]

Explanation:

So suppose the axis of rotation is perpendicular to the surface of the disk, then the moment of inertia can be calculated as the following:

[tex]I = mr^2/2 = 21 * 0.6^2/2 = 3.78 kgm^2[/tex]

We can convert the rotation speed in term of 0.8 seconds per revolution to the angular velocity knowing that each revolution is 2π

[tex]\omega = 2\pi / 0.8 = 7.85 rad/s[/tex]

Then the rotational angular momentum of the disk is:

[tex]\omega I = 7.85 * 3.78 = 29.69 kgm^2/s[/tex]

In case the axis of rotation is parallel with the surface, the moment of inertia would have a formula of:

[tex]I = m(3r^2 + h^2)/12 = 21*(3*0.6^2 + 0.5^2)/12 = 2.3275 kg m^2[/tex]