Falling objects, springs

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  • #1
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hi,

how can I figure out the peak velocity of 1kg dropped 1m onto a spring? i know the peak velocity occurs when F = -kx + mg = 0.
 
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  • #2
Integral
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The peak velocity will occur when the mass first reaches the spring after its fall.
 
  • #3
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The peak velocity will occur when the mass first reaches the spring after its fall.
you need to rethink that.
 
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  • #4
Tide
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The maximum speed will occur when the gravitational force is balanced by the spring force: [itex]x = -\frac{mg}{k}[/itex]
Before that point is reached the object is still accelerating and beyond that point the acceleration is upward.
 
  • #5
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aliendoom said:
i know the peak velocity occurs when F = -kx + mg = 0.
tide said:
The maximum speed will occur when the gravitational force is balanced by the spring force..
thanks. anyone else?
 
  • #6
Integral
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The instant the mass hits the spring it begins to decelerate. Perhaps we are not speaking of the same speed. I am referring to the mass, and you?
 
  • #7
Tide
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Integral said:
The instant the mass hits the spring it begins to decelerate. Perhaps we are not speaking of the same speed. I am referring to the mass, and you?
No, because the spring force is zero at that point and gravity is still acting.
 
  • #8
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tide said:
You can certainly take it from there to find the speed.
in my initial post i tried to indicate how far i had gotten. i can get x. i can also get the velocity before the spring starts stretching. what i need is the peak velocity. that's the part i'm having trouble with.
 
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  • #9
Tide
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Use energy conservation!
 
  • #10
Integral
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Ok, ok...
Looks to me like once we hit the spring we must satisfy the DE

[tex] \ddot {x}m = -mg - kx [/tex]


Find the velocity expression from this and you are done. :)

I'm workin on it.

Edit: inserted a lost factor.
 
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  • #11
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tide said:
Use energy conservation!
how? the spring continues to stretch after kx = mg.
 
  • #12
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aliendoom said:
how? the spring continues to stretch after kx = mg.
No, the spring is being compressed. Remember, you're DROPPING the object onto the spring.

Do you know what the potential energy is for a spring and how to calculate gravitational potential energy?
 
  • #13
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a flicker of a light bulb. 1/2kx^2 is the potential energy of the spring, so subtracting that from the change in potential energy of the falling mass should give me the kinetic energy. am i right?
 
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  • #14
Tide
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That's part of it. Energy conservation tells you that
[tex]mv^2 + kx^2 + 2mgx = mv_0^2[/tex]
is a constant. I've taken the zero point of the potential to be at the point of contact.
 
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  • #15
Integral
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From the solution of the DE I get
Vi = velocity at the top of the spring.
[tex] x(t) = \frac {mg} k Cos ( \sqrt { \frac k m}t) + V_i \sqrt {\frac m k} Sin (\sqrt {\frac k m } t) -\frac {mg} k [/tex]
The velocity equation is:

[tex] \dot {x}(t) =\sqrt {\frac m k}g Sin(\sqrt{\frac k m}t) + V_i Cos{(\sqrt {\frac k m}t)}[/tex]
The time at which the max velocity occurs will be

[tex] t_m =\sqrt {\frac m k} Tan^{-1} (\frac g {V_i} \sqrt { \frac m k})[/tex]


Simply plug this time into the velocity equation to find the Max velocity.

Edit the edit: All units are now happy!
 
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  • #16
Tide
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Your units aren't right! :-)
 
  • #17
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Tide said:
Your units aren't right! :-)
Thats for sure!

Got 'em fixed now. A lost constant or 2 can really mess things up!
 
  • #18
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Tide said:
That's part of it. Energy conservation tells you that
[tex]mv^2 + kx^2 + 2mgx = mv_0^2[/tex]
is a constant. I've taken the zero point of the potential to be at the point of contact.
ok but shouldn't it be:

[tex]KE_{mass} + PE_{spring} = \Delta PE_{mass}
[/tex]

when [tex]v_{0}=0[/tex].

so,

[tex]1/2mv^2 + 1/2kx^2 = mg(h+x)[/tex]

or

[tex]mv^2 + kx^2 - 2mg(h+x) = 0[/tex]

where h is the drop distance and x is the spring compression.

in other words in your equation the distance in the [tex]kx^2[/tex] term shouldn't be the same as the distance in the [tex]2mgx[/tex] term, and it seems to me the sign of the [tex]2mg(h+x)[/tex] term should be negative. as far as i know, energy isn't a vector quantity, so i'm not sure how this would result in different signs:

I've taken the zero point of the potential to be at the point of contactp
also, these equations seem to express the same thing I said in my previous post:

1/2kx^2 is the potential energy of the spring, so subtracting that from the change in potential energy of the falling mass should give me the kinetic energy.
(how do you get two lines of equations between a set of tags? i tried \\ at the end of a line and the equations just ran together.)
 
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  • #19
Tide
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The full equation would be
[tex]mv^2 + kx^2 + 2mgx = mv_0^2 + kx_0^2 + 2 mgx_0[/tex]
where the subscripts refer to some initial value. I took [itex]x_0[/itex] to be zero which corresponds to the point of initial contact. My [itex]v_0[/itex] is the speed of the object just as it hits the spring which you can calculate if you know the height above the spring when it was released.
 
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  • #20
ehild
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Tide said:
That's part of it. Energy conservation tells you that
[tex]mv^2 + kx^2 + 2mgx = mv_0^2[/tex]
is a constant. I've taken the zero point of the potential to be at the point of contact.
[tex]mv^2 + kx^2 -2mgx = mv_0^2[/tex]
V is maximum when
[tex]x=\frac{mg}{k}\mbox{. }
v_{max}=\sqrt{v_0^2+\frac{mg^2}{k}}.[/tex]

ehild
 
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  • #21
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ehild said:
[tex]mv^2 + kx^2 -2mgx = mv_0^2[/tex]
V is maximum when
[tex]x=\frac{mg}{k}\mbox{. }
v_{max}=\sqrt{v_0^2+\frac{mg^2}{k}}.[/tex]

ehild
Gravity pulls downward so the gravitational potential must have the positive sign. Your version has gravity pushing upward.
 
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  • #22
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Tide said:
Gravity pulls downward so the gravitational potential must have the positive sign. Your version has gravity pushing upward.
Isn't that just a matter of coordinate system?

I couldn't leave this one alone. Here are the results of the DE solution.
Problem statement:
[tex] \ddot {x} = - kx - mg [/tex]

[tex] x(0)=0 [/tex]

[tex] \dot {x}(0)= - V_i [/tex]

Solution:
let
[tex] \lambda^2 = \frac k m [/tex]

[tex]x(t)=\frac g {\lambda^2} \cos(\lambda t) - \frac {V_i} \lambda \sin(\lambda t)- \frac g {\lambda^2}[/tex]

[tex] \dot {x}(t) = - V_i \cos(\lambda t) - \frac g \lambda \sin(\lambda t) [/tex]
The maximum velocity occurs when:
[tex] \tan(\lambda t) = \frac g {V_i \lambda} [/tex]

From the value of the Tan we get:
[tex] H= \sqrt { ({V_i \lambda})^2 + g^2}[/tex]

[tex] \sin(\lambda t) = \frac g H [/tex]

[tex] \cos(\lambda t) = \frac {V_i \lambda} H [/tex]

so:

[tex] x_{max} = - \frac g {\lambda^2} = -\frac {mg} k[/tex]

[tex] V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k [/tex]

which is in agreement with the energy solution up to the choice of coordinate system.

The main difference is that the DE solution takes over a page to write out while the energy solution can be done in a few lines!
 
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  • #23
ehild
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Tide said:
Gravity pulls downward so the gravitational potential must have the positive sign. Your version has gravity pushing upward.
You are right if you take x positive upward. But aliendoom took it on the opposite way.

aliendoom said:
hi,

how can I figure out the peak velocity of 1kg dropped 1m onto a spring? i know the peak velocity occurs when F = -kx + mg = 0.
ehild
 
  • #24
ehild
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Integral said:
[tex] x_{max} = - \frac g {\lambda^2} = -\frac {mg} k[/tex]

[tex] V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k [/tex]
Actually, they are rather

[tex] x_{min} = - \frac g {\lambda^2} = -\frac {mg} k[/tex]

[tex] V_{min}= - \sqrt { {V_i} ^2 + \frac {m g^2} k [/tex]

:smile:

ehild
 
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  • #25
Integral
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ehild,

Why do you say that? I think the min velocity will be 0.

Give me a bit and I can tell you where that occurs.

Edit: ok I will admit that it should be

[tex]x_{v_{max}}[/tex]
 
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