Question

1. determine the rotational inertia of a hoop with a mass of 6 kg and a radius of 0.75 m.

1.35 kg·m²
3.375 kg·m²
1.69 kg·m²
0.68 kg·m²

Explanation:

Step1: Recall the formula for rotational inertia of a hoop

The rotational inertia \( I \) of a hoop (a thin-walled circular ring) rotating about an axis perpendicular to its plane and passing through its center is given by the formula \( I = MR^2 \), where \( M \) is the mass of the hoop and \( R \) is its radius.

Step2: Identify the given values

We are given that the mass \( M = 6 \, \text{kg} \) and the radius \( R = 0.75 \, \text{m} \).

Step3: Substitute the values into the formula

Substitute \( M = 6 \) and \( R = 0.75 \) into the formula \( I = MR^2 \):
\[
I = 6 \times (0.75)^2
\]

Step4: Calculate \( (0.75)^2 \)

First, calculate \( (0.75)^2 \):
\[
(0.75)^2 = 0.75 \times 0.75 = 0.5625
\]

Step5: Calculate the rotational inertia

Now, multiply this result by the mass \( 6 \):
\[
I = 6 \times 0.5625 = 3.375 \, \text{kg·m}^2
\]

Answer:

\( 3.375 \, \text{kg·m}^2 \) (corresponding to the option "3.375 kg·m²")