Moment of Inertia plank Problem

  • Thread starter vsharma88
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  • #1
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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
 

Answers and Replies

  • #2
tiny-tim
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this plank is not constant

Hi vsharma88! :smile:
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? :wink:

Hint: what is the angular acceleration? :smile:
 
  • #3
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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
tiny-tim
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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! :smile:

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

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). :smile:
 
  • #5
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Hi vsharma88! :smile:

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

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). :smile:
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
 
  • #6
tiny-tim
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hey tiny-tim i have taken the moment about the centre:

[tex]\sum\tau[/tex]=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. :wink:

Then take moments about the end, and compare the two equations to get N/g. :smile:
 
  • #7
9
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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
 
  • #8
tiny-tim
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Hi vsharma88! :smile:

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:​
 

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