A uniform bar has two small balls glued to its ends. The bar is 3.00 m long and with mass 2.40 kg , while the balls each have mass 0.700 kg and can be treated as point masses. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of An abstract sculpture. Part A Find the moment of inertia of this combination about an axis perpendicular to the bar through its center. Express your answer in kilogram-meters squared.

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Netta00 Netta00 · Answer 1 · 2019-11-19T08:18:45+01:00

Answer:

[tex]I_t=4.95\ kg.m^2[/tex]

Explanation:

Given that

L = 3 m

M = 2.4 kg

m=0.7 kg

r= L/2 = 1.5 m

The moment of inertia of rod about it center

[tex]I=\dfrac{ML^2}{12}[/tex]

The moment of inertia of point mass about center of rod

I'= mr² + mr²

So total moment of inertia

[tex]I_t=\dfrac{ML^2}{12}+2mr^2[/tex]

By putting the values

[tex]I_t=\dfrac{ML^2}{12}+2mr^2[/tex]

[tex]I_t=\dfrac{2.4\times 3^2}{12}+2\times 0.7\times 1.5^2[/tex]

[tex]I_t=4.95\ kg.m^2[/tex]