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

`I_t=4.95\ kg.m^2`

Step-by-step 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=(ML^2)/(12)`

The moment of inertia of point mass about center of rod

I'= mr² + mr²

So total moment of inertia

`I_t=(ML^2)/(12)+2mr^2`

By putting the values

`I_t=(ML^2)/(12)+2mr^2`

`I_t=(2.4* 3^2)/(12)+2* 0.7* 1.5^2`

`I_t=4.95\ kg.m^2`