(a) The amount of work required to change the rotational rate is 0.0112 J.
(b) The decrease in the rotational inertia when the outermost particle is removed is 64.29%.
<h3>
Moment of inertia of the rod</h3>
The moment of inertia of the rod from the axis of rotation is calculated as follows;
I = md² + m(2d)² + m(3d)²
where;
- m is mass = 10 g = 0.01 kg
- d = 3 equal division of the length
d = 6/3 = 2 cm = 0.02 m
I = md²(1 + 2² + 3²)
I = 14md²
I = 14(0.01)(0.02)²
I = 5.6 x 10⁻⁵ kg/m³
<h3>Work done to change the rotational rate</h3>
K.E = ¹/₂Iω²
K.E = ¹/₂(5.6 x 10⁻⁵)(60 - 40)²
K.E = ¹/₂(5.6 x 10⁻⁵)(20)²
K.E = 0.0112 J
<h3>Percentage decrease of rotational inertia when the outermost particle is removed</h3>
I₂ = md² + m(2d)²
I₂ = 5md²
ΔI = 14md² - 5md²
ΔI = 9md²
η = (ΔI/I) x 100%
η = (9md²/14md²) x 100%
η = 64.29 %
Learn more about rotational inertia here: brainly.com/question/14001220
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