What is the rotational inertia of a solid iron disk of mass 41.0 kg with a thickness of 5.00 cm and radius of 30.0 cm, about an axis perpendicular to the disk and passing through its center?
Question
What is the rotational inertia of a solid iron disk of mass 41.0 kg with a thickness of 5.00 cm and radius of 30.0 cm, about an axis perpendicular to the disk and passing through its center?
Solution
The rotational inertia (I) of a solid disk about an axis perpendicular to the disk and passing through its center is given by the formula:
I = 0.5 * m * r^2
where: m is the mass of the disk, r is the radius of the disk.
Given: m = 41.0 kg, r = 30.0 cm = 0.3 m (since 1 m = 100 cm).
Substituting these values into the formula, we get:
I = 0.5 * 41.0 kg * (0.3 m)^2 I = 0.5 * 41.0 kg * 0.09 m^2 I = 1.845 kg*m^2
So, the rotational inertia of the disk is 1.845 kg*m^2.
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