Center of mass, forces and momentum

  • #1
1) A force F = A/t^2 acts on an object. At time t = t_o its momentum is 0. What is its momentum after a very long time?
First I used Newton's second law:
dP/dt = A/t^2
Then
t^2 dP = A dt
dP = A/t^2 dt
And I took the integral of both sides, the left from 0 to P, the right from t_o to t...
P = -A/t + A/t_o
So as t-> infinity, P -> A/t_o. That's what I got. Is this correct?

2) Picture: http://www.brokendream.net/xh4/com.JPG
A wire is bent like a triangle with side lengths d at 45 degrees from the horizontal.
What is the y co-ordinate of the center of mass?
What I thought to do is to treat the center of mass as the centroid, taking the elements to be the length of each wire.
So I did:
y_com = [d*(d sin 45)/2 + d*(d sin 45)/2]/(2d) = (d sin 45)/2 = 0.35355d.
Is this correct?

3) Picture: http://www.brokendream.net/xh4/kidswing.JPG
With what force must the child pull down on the string if the combined mass of the child and the swing is W? Neglect friction from the pulley.
I'm guessing, from F = ma, 2T - W = 0 and thus T = W/2 is the force?

Thank you.
 
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Answers and Replies

  • #2
Hootenanny
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Pseudo Statistic said:
1) A force F = A/t^2 acts on an object. At time t = t_o its momentum is 0. What is its momentum after a very long time?
First I used Newton's second law:
dP/dt = A/t^2
Then
t^2 dP = A dt
dP = A/t^2 dt
And I took the integral of both sides, the left from 0 to P, the right from t_o to t...
P = -A/t + A/t_o
So as t-> infinity, P -> A/t_o. That's what I got. Is this correct?

I'm not sure about your method here, why have you attempted to seperate variables when there is only one? (I'm assuming A is a constant) In which case it would be sufficent to do;

[tex] P = \int \frac{dP}{dt} \; dt = \int \frac{A}{t^2} \; dt[/tex]

And evaluate it. Also, I think your question may require a qualative answer, such as; "When t is very large, the momentum is very small".

~H
 
  • #3
Hootenanny
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Pseudo Statistic said:
3) Picture: http://www.brokendream.net/xh4/kidswing.JPG
With what force must the child pull down on the string if the combined mass of the child and the swing is W? Neglect friction from the pulley.
I'm guessing, from F = ma, 2T - W = 0 and thus T = W/2 is the force?

Thank you.

Spot on, vector sum of force must equal zero.

~H
 
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  • #4
Hootenanny said:
I'm not sure about your method here, why have you attempted to seperate variables when there is only one? (I'm assuming A is a constant) In which case it would be sufficent to do;

[tex] P = \int \frac{dP}{dt} \; dt = \int \frac{A}{t^2} \; dt[/tex]

And evaluate it. Also, I think your question may require a qualative answer, such as; "When t is very large, the momentum is very small".

~H
Dunno, guess I'm just used to doing this from the way my physics teacher does it. :P
Anyway, it is a quantitative answer. (It was on a test)
 
  • #5
Hootenanny
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Ahh, sorry I didn't see this line;

dP = A/t^2 dt

I thought you where trying to integrate this
t^2 dP = A dt
for a moment :confused:. Anyway, I worked through it and my answer agrees with yours.

~H
 
  • #6
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Yes Pseudostatic, your method and answer are absolutely correct.

[tex] P = \int \frac{dP}{dt} \; dt = \int \frac{A}{t^2} \; dt[/tex]And evaluate it.
Yes Hoot, that's exactly what the OP has done. The additional term in his final expression is the value of the constant of integration which can be evaluated by letting t= t_0 and thereby P= 0
For the final answer as the OP did let t -> inf. and u know the rest.

Arun

Edit: Ah I guess it was just a small misunderstanding there .
 
  • #7
Anyone know about my center of mass question? :)
 
  • #8
mukundpa
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This is not a triangular sheet and hence the center of mass will not be at centroid. What is the ratio of the mass of three sides?
 
  • #9
Doc Al
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Pseudo Statistic said:
2) Picture: http://www.brokendream.net/xh4/com.JPG
A wire is bent like a triangle with side lengths d at 45 degrees from the horizontal.
What is the y co-ordinate of the center of mass?
What I thought to do is to treat the center of mass as the centroid, taking the elements to be the length of each wire.
So I did:
y_com = [d*(d sin 45)/2 + d*(d sin 45)/2]/(2d) = (d sin 45)/2 = 0.35355d.
Is this correct?
If your wire "triangle" just has two sides (the top two), then this is correct.

(What you did has nothing to do with the centroid of a triangle though. :smile: )
 
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  • #10
Well, yeah, it's just those two sides. :D
Great.... thanks. Now I know I got atleast 3 multiple-choice questions right on my physics exam. :)
 

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