# Rotational KE question

sheepcountme

## Homework Statement

A thin square slab with sides of length 0.550 m and mass 1.80 kg is suspended so it can freely rotate about a horizontal axis of rotation perpendicular to the surface at one corner. The square is held so its top and bottom edges are horizontal, and then is released. What is the angular velocity of the square when the point opposite the pivot is as low as possible?

## Homework Equations

KE = ½Iω^2
IP = ICM + Mh^2 (parallel axis theorem)

## The Attempt at a Solution

I need help setting it up. I feel like I should be determining the angle to find out how far the sheet moved..am I on the right track?

## Answers and Replies

Homework Helper
The change in potential energy = change in the rotational kinetic energy.

So you need to find the change in potential energy the center of gravity undergoes from the horizontal position to the lowest position.

sheepcountme
Okay, so setting up the problem...

½Iω^2=½mv^2

So then switching things around I get

ω^2=(½(1.8)v^2)/(½I)

So then to find I I use Ip=Icm +mh^2 and h will be the diagonal of the square.
And for v?? I don't know..

Homework Helper
Okay, so setting up the problem...

½Iω^2=½mv^2

So then switching things around I get

ω^2=(½(1.8)v^2)/(½I)

So then to find I I use Ip=Icm +mh^2 and h will be the diagonal of the square.
And for v?? I don't know..

Not change in translational ke, change in pe.

At the horizontal position what is the height of the center of mass (COM) of the square? At the lowest point what the height of the COM, both with respect to a line passing through the stationary corner.

Also what is Ic for the slab?

sheepcountme
So.. ½Iω^2=mgh

and then for I, I= 1/12 x m (a^2+b^2) + mh^2

So altogether: ½(1/12 x m (a^2+b^2) + mh^2)(ω^2)=mgh

For h I used the difference in measurement between the diagonal and side since this seems to me like the distance the square would have fallen.

I ended up with an angular velocity of 3.153 after plugging everything in but this was incorrect.