What is the moment of inertia of a 5.00 kg 34.0 cm radius hoop about its normal axis?

1 Answer
Jul 25, 2015

I found: $I = M {R}^{2} = 0.578 k g {m}^{2}$

Explanation:

Have a look:

so $I = 5 \times {\left(\frac{34}{100}\right)}^{2} = 0.578 k g {m}^{2}$