Question

4. Three identical solid spheres, each with a radius 10.0 cm and a mass 0.25 kg, are fastened to a 4.0 kg rod with a length of 1.0m as shown. The rotational inertia about one end of the rod of this array is:

**I'm taking a general physics course. Can you be clear with the equations you are using & the steps please, thank you!!

Answer 1

Okay, I am assuming the all the three sphere are at same distance from each other with one sphere at one of rod and one sphere at other end and one sphere at the middle of rod.

______________

first, we find moment of inertia of rod about the end

we can use parallel axis method

I = I_cm + md²

I = ML² /12 + M * (L/2)²

I = ML² / 3

I = 4 * 1² / 3

I = 4/3 Kg.m²

Now,

moment of inertia of sphere which is placed at the end of rod about which axis is passing.

I = 2/5 * m * r² = 2/5 * 0.25 * 0.1² = 0.001 kg.m²

moment of inertia of middle sphere is

I = 2/5 * m * r² + m * L² / 4

I = 0.001 + 0.25 * 0.25

I = 0.0635 Kg.m²

moment of inertia of sphere at last end of rod,

I = 2/5 * m * r² + m * L²

I = 0.251 Kg.m²

Now,

just add them all

I = (4/3) + 0.001 + 0.0635 + 0.251

I = 1.648 Kg.m²