Question

question 50
a turntable in the shape of a flat uniform disk has a radius 0.500 m and a mass of 0.120 kg. the turntable is initially rotating at 4.00 rad/s about a vertical axis through its center. suddenly, a 2.50 × 10⁻² kg spider crawls onto the turntable at a point near the outer edge. treat the spider as a particle. what is the moment of inertia of the turntable?
○ 0.0900 kg·m²
○ 0.0300 kg·m²
○ 0.0600 kg·m²
○ 0.0150 kg·m²

Explanation:

Step1: Recall Moment of Inertia for Disk

The formula for the moment of inertia \( I \) of a uniform disk about its central axis is \( I = \frac{1}{2}MR^2 \), where \( M \) is the mass and \( R \) is the radius.

Step2: Substitute Values

Given \( M = 0.120 \, \text{kg} \) and \( R = 0.500 \, \text{m} \). Plug into the formula:
\( I = \frac{1}{2} \times 0.120 \, \text{kg} \times (0.500 \, \text{m})^2 \)
First, calculate \( (0.500)^2 = 0.250 \, \text{m}^2 \).
Then, \( \frac{1}{2} \times 0.120 \times 0.250 = 0.015 \times 0.250? \) Wait, no: \( \frac{1}{2} \times 0.120 = 0.060 \), then \( 0.060 \times 0.250 = 0.015 \)? Wait, no, wait: \( (0.500)^2 = 0.25 \), so \( 0.120 \times 0.25 = 0.03 \), then \( \frac{1}{2} \times 0.03 = 0.015 \)? Wait, no, I messed up. Wait, \( \frac{1}{2}MR^2 = 0.5 \times 0.120 \times (0.5)^2 \). Let's compute again: \( 0.5 \times 0.120 = 0.06 \); \( (0.5)^2 = 0.25 \); \( 0.06 \times 0.25 = 0.015 \, \text{kg·m}^2 \). Wait, but the options include 0.0150 kg·m². Wait, but let me check again. Wait, maybe I miscalculated. Wait, \( M = 0.120 \, \text{kg} \), \( R = 0.500 \, \text{m} \). So \( I = 0.5 \times 0.120 \times (0.500)^2 \). \( (0.500)^2 = 0.250 \). Then \( 0.5 \times 0.120 = 0.060 \). Then \( 0.060 \times 0.250 = 0.015 \, \text{kg·m}^2 \). Wait, but the options: 0.0150 kg·m² is an option. Wait, but let me check the calculation again. Wait, maybe I made a mistake. Wait, \( 0.120 \times 0.250 = 0.030 \), then \( 0.5 \times 0.030 = 0.015 \). Yes, that's correct. So the moment of inertia of the turntable (disk) is \( 0.0150 \, \text{kg·m}^2 \).

Answer:

0.0150 kg·m² (the last option: \( \boldsymbol{0.0150 \, \text{kg·m}^2} \))