Moment of Inertia of ceiling fan

[Colleen Walsh]

Homework Statement

A ceiling fan consists of a small cylindrical disk with 5 thin rods coming from the center. The disk has mass md = 2.9 kg and radius R = 0.22 m. The rods each have mass mr = 1.4 kg and length L = 0.83 m.
1) What is the moment of inertia of each rod about the axis of rotation?

Homework Equations

I=(sum of)mr^2

The Attempt at a Solution

I tried using the above equation multiple different ways, with just the mass and length of the rods, with the mas of the rod and the disk and the length. With the length plus the radius. The homework I have won't accept a wrong answer but I just don't really understand what I am doing wrong could someone please help?

 

[gneill]

Hello Colleen,
Welcome to Physics Forums!

Can you provide some detail for one of your attempts? What sort of calculation are you doing?

 

[Colleen Walsh]

I did I=mr^2=1.4(0.83)^2=0.96446kgm^2 that was my first attempt and it didn't work however upon looking further into my notes I found the equation I=(1/3)mr^2=(1/3)(1.4)(0.83)^2=0.3215 and that did work, but I don't quite understand it.

 

[haruspex]

I did I=mr^2=1.4(0.83)^2=0.96446kgm^2 that was my first attempt and it didn't work however upon looking further into my notes I found the equation I=(1/3)mr^2=(1/3)(1.4)(0.83)^2=0.3215 and that did work, but I don't quite understand it.

I = mr² is for a point mass m at radius r. A rod is not like that. If you treat it as a line of little point masses and consider their average value of r² you get the 1/3 factor. Have you learned how to integrate xⁿ?

 

[gneill]

I did I=mr^2=1.4(0.83)^2=0.96446kgm^2 that was my first attempt and it didn't work however upon looking further into my notes I found the equation I=(1/3)mr^2=(1/3)(1.4)(0.83)^2=0.3215 and that did work, but I don't quite understand it.

Okay, the moment of inertia depends upon how the mass is distributed with respect to the axis of rotation. The equation I=(1/3)mr^2 that you found refers to the moment of inertia of a rod about an end of the rod, rather than the rod's center.

Here the axis of rotation is located beyond the end of the rod (the rod is fixed to the perimeter of the disk, and the axis of rotation is at the center of the disk).

So you'll need to investigate the Parallel Axis Theorem to find out how to adjust the moment of inertia about the rod's center to an axis that is located elsewhere, or, you can go back to basics and set up the integral from the definition of moment of inertia to derive the correct expression. Either way will work.

 

[Colleen Walsh]

Okay thank you both, I think I understand it now.