# Moment of Inertia plank Problem

## Homework Statement

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.

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

## 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

tiny-tim
Homework Helper
this plank is not constant

Hi vsharma88!
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.After Plank Falls:

Fnet = ma = F1 - Fg = 0
Nooo … the acceleration is not 0, is it?

Hint: what is the angular acceleration?

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'.

tiny-tim
Homework Helper
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'.
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).

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).
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

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

$$\sum\tau$$=0=ml^2/12*w'-mg*l/2
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.

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

tiny-tim
Homework Helper
Hi vsharma88!

Sorry, but this is wrong: