# Motion of Rigid Bodies

1. Mar 5, 2006

### Bucky

Well here's what I've done. I think I've missed the mark big time with this one...

MI = 2/3 Ma^2
= 2/3 m(2a)^2
= 14/3 ma^2

Initial Energy = 0 + mgh
KE + PE

Energy Just Before Impact = ((1/2 Iw^2)/-KE) + 0/PE

By Energy Conservation

1/2 Iw^2 = mgh (h=2a)
Iw^2 = 4mga
w^2 = (4mga)/I

w^2 = 4mga/(14/3)ma^2

w^2 = 12/14 (g/a)

can anyone point out where I started going wrong?

2. Mar 5, 2006

### Staff: Mentor

Not sure what you are doing here. What's the moment of inertia of a cube about one edge? (Depending on what you know, you may need to use the parallel axis theorem.)

What's the change in gravitational PE as the cube falls from its initial position to its final position? (Hint: What happens to the center of mass?)

3. Mar 5, 2006

### Bucky

ok it seems im wrong from the word go..i only have the equation of innertia for an axis about the centre of the cube parallel to an axis (which is the one i used).

so using the parallel axis thoerem (which oddly enough we've used for previous questions but i've never recalled being taught it) i get..

MI = 2/3 Ma^2 + Ma^2
= 5/3 Ma^2

that sound right?

as for the second part I'm not sure what you mean. the PE would fall to 0 as the KE rises to equal the PE?

4. Mar 5, 2006

### Staff: Mentor

No. The distance from the center to the edge is not a.

How does the height of the center of mass change from initial to final position?

5. Mar 5, 2006

### Bucky

ah right..a is the distance to a side ,not an edge...so its a triangle you need to make...

...so the distance from centre to an edge is

d^2 = a^2 + a^2
=2a^2
d = root (2a^2)
= root 2 a?

again with part 2 im not sure how you mean. the height of the centre to the table will change reletive to the angle the edge of the box makes with the table, and the length of the side. should i try to find the distance it travels?

6. Mar 5, 2006

### Staff: Mentor

Right.

To find the change in PE you need to know how far the center of mass of the box fell. Realize that if you measure with respect to the table surface, the PE of the box goes from $mgh_1$ to $mgh_2$ where h is the height of center of mass above the table. (It's not zero at any point.)

7. Mar 5, 2006

### Bucky

sigh, im totally lost on this question.

i've used the value for d to work out I...

I = 2/3ma^2 + M(root 2 a^2)^2
= 2/3ma^2 + 2ma^2
= 7/3ma^2

but i dont know how to find the height above the table of the centre of the cube, or how that is even relevant. Wouldn't it depend on the angle made by the edge of the box? since no angle is given, do you just use a variable to represent it or asume its at 45 degrees?

8. Mar 5, 2006

### Staff: Mentor

2 + 2/3 = ?

When the box falls, you need to find the change in its PE to calculate its KE at the point when its side hits the table.

The initial position of the box is one of unstable equilibrium. That means it's balanced right on its edge.

9. Mar 5, 2006

### Bucky

er yeah...should be 8/3 ma^2....

so we take the distance from the centre to the table to initially be root 2a^2 (since the distance from the centre to any edge is that)

and the distance from the centre to the table is 'a' when it has fallen.

so...

PE at start = mgh
=mg root 2a^2

PE at end = mgh
=mga

change in PE = change in KE

therefore, if initial KE is zero (yeah? since its starting from rest?)

change in PE = KE when cube hits table?

EDIT ok ..

change in PE =
PE1 - PE2
mg root2a^2 - mga
root2a^2 - a

therefore, KE = root 2a^2 - a

is this right? it sounds right going by the answer, so i just have to use this somewhere?

Last edited: Mar 5, 2006
10. Mar 5, 2006

### Bucky

never mind, got it. thanks for your help :)