RaySix

Post

Moment of Inertia of Solid Disc

at December 4th 2023, 8:25:17 pm.

Question: A 2.0 kg solid disc with a radius of 0.5 m is rotating about an axis perpendicular to the disc and passing through its center. Calculate the moment of inertia of the disc.

Answer: The moment of inertia, I, for a solid disc rotating about an axis perpendicular to the disc and passing through its center is given by the formula:

I = \frac{1}{2} m r^2

Where: I = moment of inertia m = mass of the disc r = radius of the disc

Given: m = 2.0 kg r = 0.5 m

Substitute the given values into the formula:

I = \frac{1}{2} \cdot 2.0\, \text{kg} \cdot (0.5\, \text{m})^2

I = \frac{1}{2} \cdot 2.0\, \text{kg} \cdot 0.25\, \text{m}^2

I = 0.5\, \text{kg} \cdot 0.25\, \text{m}^2

I = 0.125\, \text{kg m}^2

Therefore, the moment of inertia of the disc is 0.125 kg m^2.