Asked by Anonymous
What is the rotational inertia of a solid iron disk of mass 45 kg, with a thickness of 3.44 cm and radius of 19.0 cm, about an axis through its center and perpendicular to it?
Answers
Answered by
Damon
r = .19 meter
I = (1/2) m r^2 is moment of inertia of disk
I = (1/2)(45)(.19)^2
That is the rotational moment of inertia, I suspect what you want. If you really want the rotational inertia you must multiply that by omega, the angular velocity in radians per second.
I = (1/2) m r^2 is moment of inertia of disk
I = (1/2)(45)(.19)^2
That is the rotational moment of inertia, I suspect what you want. If you really want the rotational inertia you must multiply that by omega, the angular velocity in radians per second.
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