- #1
Bucky
- 82
- 0
A uniform cube of side 2a is placed in unstable equilibrium with one edge in contact with a rough horizontal table and allowed to fall. Show that, if w is the angular velocity when a face of the cube comes into contact with the table, then
w^2 = 3/4 (root2 - 1) g/a ,
provided the table is sufficiently rough to prevent sliding.
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?