# Moment of Inertia plank Problem

1. Oct 26, 2008

### vsharma88

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

2. Oct 27, 2008

### tiny-tim

this plank is not constant

Hi vsharma88!
Nooo … the acceleration is not 0, is it?

Hint: what is the angular acceleration?

3. Oct 27, 2008

### vsharma88

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'.

4. Oct 28, 2008

### tiny-tim

Hi vsharma88!

(btw, you can't say "the force becomes a torque" … a force is a force, a torque is a torque … torque = force times distance )

Take moments about the end to find the angular acceleration.

Then take moments about the centre to find the normal force (using that value of angular acceleration).

5. Oct 29, 2008

### vsharma88

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

$$\sum\tau$$=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 :

$$\sum\tau$$=0=ml^2/12*w' ?

Thanks

6. Oct 29, 2008

### tiny-tim

Nooo … the weight has no moment about the centre … you should have the normal force, N, in there, instead.

Then take moments about the end, and compare the two equations to get N/g.

7. Oct 29, 2008

### vsharma88

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

Fnet = F1 - Fg

$$\sum$$=N - (l/2)Fg=0
0=Iw' - (l/2)(mg)
0=(ml^2/12)w' - (l/2)(mg)

w'= 6g/l

$$\sum$$= N*(l/2)
=Iw' * (l/2)
=(ml^2/12)(6g/l)*(l/2)
=(1/4)mg

8. Oct 29, 2008

### tiny-tim

Hi vsharma88!

Sorry, but this is wrong:
and this is wrong:
I'm going to bed now … :zzz:​