# I Rotational Inertia

1. Apr 26, 2017

### kacey

A 24 kg solid door is 220 cm tall, 95 cm wide. What is the door's moment of inertia for rotation about a vertical axis inside the door, 17 cm from one edge?

I've looked at examples of how to do this problem. They connect it to center mass but I get confused when determining what the center mass is. They also talk about parallel axis theorem, but my professor has not introduced this approach.

The formula that I have been using is I= (1/3)mr^2

2. Apr 26, 2017

### kacey

If you're like me and look at other examples for help:
I was using the wrong eq. use I=(1/12)mL^2 + m (L/2 -d)^2
L is the length of the door and d is the distance from the edge

3. Apr 28, 2017

### CWatters

That looks right.

From here..
http://hedberg.ccnysites.cuny.edu/img203207/moments-wide.png

The moment of inertia about the center of mass (Icm) of the door is..
Icm = (1/12)mL2

The centre of mass will be the middle of the door (L/2 from the edge) for a door of uniform thickness.

Then the parallel axis theorem says..
I = Icm + mx2

where x is the distance between the centre of mass and the required axis. In your case x = (L/2)-d so the total moment of inertial is

I = (1/12)mL2 + m((L/2)-d)2.......................................(1)

You should also be able to get the same answer without using parallel axis theorem...

You could also approach this problem by pretending you had TWO doors, a narrow one (width d) and a wide one (width=L-d) connected together at a common hinge and add together the moments of inertia of each.

The moment of inertia of a door about it's edge is..
Ih = (1/3)mL2

In this case you have two doors of different width L..

The narrow one..
In = (1/3)md^2
The wide one..
Iw = (1/3)m(L-d)2

So the total moment of inertial would be..
I = In + Iw

I = (1/3)md^2 + (1/3)m(L-d)2......................................(2)

I'll let you prove (1) and (2) are the same.