Question

1. use the formula $i = mr^2$. show your work

a 40 kg child sits 0.5 m from the center of a seesaw. b. the same child moves to 1.0 m from the center. given: $m = 40$ kg, $r = 0.5$ m
unknown: $i$
$i = 40×(0.5)^2$
$= 40×0.25 = 10$ $kg·m^2$
$i = 40×1.0^2 = 40$ $kg·m^2$

1. by what factor did the moment of inertia change when the distance doubled?

answer: ______

Explanation:

Step1: Calculate initial moment of inertia

$I_1 = m r_1^2 = 40 \times (0.5)^2 = 10 \ \text{kg·m}^2$

Step2: Calculate final moment of inertia

$I_2 = m r_2^2 = 40 \times (1.0)^2 = 40 \ \text{kg·m}^2$

Step3: Find the change factor

$\text{Factor} = \frac{I_2}{I_1} = \frac{40}{10} = 4$

Answer:

4