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How to calculate impulse required to move an object vertical

  • #1
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Homework Statement


[/B]
Suppose I have a stationary object of mass 'm' and I want to apply a momentary force in the vertical direction so that it just reaches the height 'h'.

So how do I calculate the impulse required in this case?

Also how do I exactly define the delta(t) for the momentary force. shall I just take 1 sec.


Homework Equations



Impulse = F*delta(t)

The Attempt at a Solution



Since I need to accelerate the object, I think the work done = m(a+g)h
but I am also not sure how to exactly define this delta(t), shall I just use 1 second?
Also how to get the value of 'a'?
 

Answers and Replies

  • #2
FactChecker
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Suppose you relate impulse to change in momentum. Then relate momentum to kinetic energy. Then relate kinetic energy to potential energy. Then relate potential energy to the required maximum height. That should give some equations you can use to solve the problem.
 
  • #3
rude man
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Also how do I exactly define the delta(t) for the momentary force. shall I just take 1 sec.
factchecker has shown you the way to get the answer.

As to the nature of the impulse, it's the force exerted over the time interval. Theoretically, in an applied impulse the force approaches infinity while the time interval approaches zero, with the product remaining constant and equal to the impulse.

You have to picture an arbitrarily large force Fapplied exerted over arbitrarily short time Δt on the mass, with impulse = Fapplied ⋅ Δt. Δt has to be arbitrarily short unless you take into account the fact that, during the force exertion, gravity is also acting on the mass, so the net force exerted over the finite time interval is reduced by gravity, as will be the imparted momentum, velocity, k.e., p.e. and height.

For example, if Δt = 1 sec. then the net applied impulse is (Fapplied - mg) x 1 sec assuming constant Fapplied.
 
  • #4
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Suppose you relate impulse to change in momentum. Then relate momentum to kinetic energy. Then relate kinetic energy to potential energy. Then relate potential energy to the required maximum height. That should give some equations you can use to solve the problem.
Hi, thanks a lot for your reply and really sorry for my late response. Ya, I used mgh=0.5mv^2 to get the velocity, and used it in Impulse = mv-0 to get the impulse. Thanks.
 
  • #5
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factchecker has shown you the way to get the answer.

As to the nature of the impulse, it's the force exerted over the time interval. Theoretically, in an applied impulse the force approaches infinity while the time interval approaches zero, with the product remaining constant and equal to the impulse.

You have to picture an arbitrarily large force Fapplied exerted over arbitrarily short time Δt on the mass, with impulse = Fapplied ⋅ Δt. Δt has to be arbitrarily short unless you take into account the fact that, during the force exertion, gravity is also acting on the mass, so the net force exerted over the finite time interval is reduced by gravity, as will be the imparted momentum, velocity, k.e., p.e. and height.

For example, if Δt = 1 sec. then the net applied impulse is (Fapplied - mg) x 1 sec assuming constant Fapplied.
Ya, I wanted to know what delta(t) to use. Is there any standard that delta(t) should be such that the particle doesn't move by more than say 1mm, or 0.1*diameter, or any such thing. Is there any way to know what is the max. delta(t) allowed.

When I use Impulse = Change in momentum = mv - 0, this delta(t) is not even required, but still I wanted to know if there is ant max. limit for delta(t).

Thanks a lot for your response, and sorry for my late reply.
 
  • #6
rude man
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Ya, I wanted to know what delta(t) to use. Is there any standard that delta(t) should be such that the particle doesn't move by more than say 1mm, or 0.1*diameter, or any such thing. Is there any way to know what is the max. delta(t) allowed.

When I use Impulse = Change in momentum = mv - 0, this delta(t) is not even required, but still I wanted to know if there is ant max. limit for delta(t).

Thanks a lot for your response, and sorry for my late reply.
Well, you 're right in that you don't need the delta function in this problem, and you're also right to wonder about finite approximations of that function.

There is no absolute minimum number like 1 mm or 0.1 diameter.

I've indicated that in your problem here the error in not using infinitesmally small time intervals is that your impulse of force times time is reduced by gravity acting during that time, so you wind up with a lower net momentum.

Rigorously speaking you have 2 problems, not 1. Say constant force F is applied for a time T :
1. solve my'' + mg = F for y(T) and y'(T) at the end of the finite time interval T; then
2. solve my'' + mg = 0 with those y(T) and y'(T) values as your initial conditions for the rest of the trajectory.
Notice how gravity subtracts from F in the 1st problem.

Another example: say you have a mass-spring system; you know the period of oscillations is 2π√(m/k) (hope I got that right), anyway, the rule of thumb here would be to keep the time of the impulse to not more than a few % of that period. In general, keep the time interval to << the smallest time constant in your system. Otherwise you again have 2 problems.

You presumably haven't had the Laplace or Fourier transform yet but the delta function is an easy way to determine linear system response. In your problem and all other similar, 2 problems are reduced to 1, and any initial conditions are automatically included in the transforms. Much more fun!
 
  • #7
FactChecker
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Ya, I wanted to know what delta(t) to use. Is there any standard that delta(t) should be such that the particle doesn't move by more than say 1mm, or 0.1*diameter, or any such thing. Is there any way to know what is the max. delta(t) allowed.

When I use Impulse = Change in momentum = mv - 0, this delta(t) is not even required, but still I wanted to know if there is ant max. limit for delta(t).
I don't think so. I think that converting impulse to change of momentum immediately is the only practical thing to do. Impulse is more of a theoretical concept where the delta(t) goes to 0. If you want to use a finite delta(t), you are using a finite force and acceleration rather than an impulse.
 
  • #8
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If you want to use a finite delta(t), you are using a finite force and acceleration rather than an impulse.
Hi, thanks so much. Ya, this makes so much sense. If I use finite delta(t) I will also need to use finite force, and the impulse concept will be disregarded a bit, I guess. Thanks a lot for pointing this out.

Edit: However, when there is an impulse, there is an acceleration too. Isn't it. Or impulse disregards this momentary acceleration?
 
  • #9
7
0
Well, you 're right in that you don't need the delta function in this problem, and you're also right to wonder about finite approximations of that function.

There is no absolute minimum number like 1 mm or 0.1 diameter.

I've indicated that in your problem here the error in not using infinitesmally small time intervals is that your impulse of force times time is reduced by gravity acting during that time, so you wind up with a lower net momentum.

Rigorously speaking you have 2 problems, not 1. Say constant force F is applied for a time T :
1. solve my'' + mg = F for y(T) and y'(T) at the end of the finite time interval T; then
2. solve my'' + mg = 0 with those y(T) and y'(T) values as your initial conditions for the rest of the trajectory.
Notice how gravity subtracts from F in the 1st problem.

Another example: say you have a mass-spring system; you know the period of oscillations is 2π√(m/k) (hope I got that right), anyway, the rule of thumb here would be to keep the time of the impulse to not more than a few % of that period. In general, keep the time interval to << the smallest time constant in your system. Otherwise you again have 2 problems.

You presumably haven't had the Laplace or Fourier transform yet but the delta function is an easy way to determine linear system response. In your problem and all other similar, 2 problems are reduced to 1, and any initial conditions are automatically included in the transforms. Much more fun!
Hi, thanks a lot for taking the time to respond in such details.

From the equations, I get it that first I solve for the portion the force is acting and find out the velocity reached, and then use this velocity as initial velocity for the remaining trajectory. Thanks a lot.

Ya I have no idea how to use Laplace or Fourier transform in such questions, may be will have a look as I progress.

Thanks a lot :-)
 

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