Homework Help: Moment of Inertia plank Problem

1. The problem statement, all variables and given/known data

Two People are holding the ends of a uniform plank of lenth, l and mass, m. Show that if one person suddenly lets go, the load suppored by the other person suddenly drops from mg/2 to mg/4.

2. Relevant equations

Fnet=ma
L=Iw'
I=ml^2+I'


3. The attempt at a solution

I have done the first part of the problem. I am have trouble with the second part.

Before Plank Falls:

Fnet = ma = F1 + F2 -Fg = 0 (F1=F2, Fg=mg)

2F1=mg
F1=(mg)/2


After Plank Falls:

Fnet = ma = F1 - Fg = 0
F1=mg

 

After the person has dropped the plank, the force becomes a torque.

Fnet = N - mg = ma
ma = (ml^2)/12 * w' - mg [(ml^2)/2 - for a thin uniform plank]

I am now stuck on what to do with w'.

 

hey tiny-tim i have taken the moment about the centre:

[tex]\sum\tau[/tex]=0=ml^2/12*w'-mg*l/2
w'=(gl/2)*(12/l^2)
=(g6)/l

I am now confused what to do next, you said to take moments about the centre. Is it just :

[tex]\sum\tau[/tex]=0=ml^2/12*w' ?

Thanks

 

ok tiny-tim i think i got it, tell me if im right.

Fnet = F1 - Fg

Moment about End
[tex]\sum[/tex]=N - (l/2)Fg=0
0=Iw' - (l/2)(mg)
0=(ml^2/12)w' - (l/2)(mg)

w'= 6g/l

Moment about center of mass
[tex]\sum[/tex]= N*(l/2)
=Iw' * (l/2)
=(ml^2/12)(6g/l)*(l/2)
=(1/4)mg