Question

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.

Answer 1

Answer:

$I_t=4.95\ kg.m^2$

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

$I=\dfrac{ML^2}{12}$

The moment of inertia of point mass about center of rod

I'= mr² +  mr²

So total moment of inertia

$I_t=\dfrac{ML^2}{12}+2mr^2$

By putting the values

$I_t=\dfrac{ML^2}{12}+2mr^2$

$I_t=\dfrac{2.4\times 3^2}{12}+2\times 0.7\times 1.5^2$

$I_t=4.95\ kg.m^2$