Moment of Inertia plank Problem

[vsharma88]

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.

Homework Equations

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]

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?

 

[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'.

 

[tiny-tim]

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).

 

[vsharma88]

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]

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.

 

[vsharma88]

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

Fnet = F1 - Fg

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

 

[tiny-tim]

Hi vsharma88!

Sorry, but this is wrong:

Moment about End
…
0=(ml^2/12)w' - (l/2)(mg)

and this is wrong:

∑ = N*(l/2)
=Iw' * (l/2)

I'm going to bed now … :zzz:​