Falling objects, springs

1. Sep 11, 2004

aliendoom

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.

Last edited: Sep 11, 2004
2. Sep 11, 2004

Integral

Staff Emeritus
The peak velocity will occur when the mass first reaches the spring after its fall.

3. Sep 11, 2004

aliendoom

you need to rethink that.

Last edited: Sep 11, 2004
4. Sep 11, 2004

Tide

The maximum speed will occur when the gravitational force is balanced by the spring force: $x = -\frac{mg}{k}$
Before that point is reached the object is still accelerating and beyond that point the acceleration is upward.

5. Sep 11, 2004

aliendoom

thanks. anyone else?

6. Sep 11, 2004

Integral

Staff Emeritus
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. Sep 11, 2004

Tide

No, because the spring force is zero at that point and gravity is still acting.

8. Sep 11, 2004

aliendoom

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.

Last edited: Sep 11, 2004
9. Sep 11, 2004

Tide

Use energy conservation!

10. Sep 11, 2004

Integral

Staff Emeritus
Ok, ok...
Looks to me like once we hit the spring we must satisfy the DE

$$\ddot {x}m = -mg - kx$$

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

I'm workin on it.

Edit: inserted a lost factor.

Last edited: Sep 16, 2004
11. Sep 11, 2004

aliendoom

how? the spring continues to stretch after kx = mg.

12. Sep 11, 2004

Tide

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. Sep 11, 2004

aliendoom

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?

Last edited: Sep 11, 2004
14. Sep 11, 2004

Tide

That's part of it. Energy conservation tells you that
$$mv^2 + kx^2 + 2mgx = mv_0^2$$
is a constant. I've taken the zero point of the potential to be at the point of contact.

Last edited by a moderator: Sep 16, 2004
15. Sep 11, 2004

Integral

Staff Emeritus
From the solution of the DE I get
Vi = velocity at the top of the spring.
$$x(t) = \frac {mg} k Cos ( \sqrt { \frac k m}t) + V_i \sqrt {\frac m k} Sin (\sqrt {\frac k m } t) -\frac {mg} k$$
The velocity equation is:

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

$$t_m =\sqrt {\frac m k} Tan^{-1} (\frac g {V_i} \sqrt { \frac m k})$$

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

Edit the edit: All units are now happy!

Last edited: Sep 16, 2004
16. Sep 11, 2004

Tide

17. Sep 11, 2004

Integral

Staff Emeritus
Thats for sure!

Got 'em fixed now. A lost constant or 2 can really mess things up!

18. Sep 11, 2004

aliendoom

ok but shouldn't it be:

$$KE_{mass} + PE_{spring} = \Delta PE_{mass}$$

when $$v_{0}=0$$.

so,

$$1/2mv^2 + 1/2kx^2 = mg(h+x)$$

or

$$mv^2 + kx^2 - 2mg(h+x) = 0$$

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

in other words in your equation the distance in the $$kx^2$$ term shouldn't be the same as the distance in the $$2mgx$$ term, and it seems to me the sign of the $$2mg(h+x)$$ 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:

also, these equations seem to express the same thing I said in my previous post:

(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.)

Last edited by a moderator: Sep 16, 2004
19. Sep 11, 2004

Tide

The full equation would be
$$mv^2 + kx^2 + 2mgx = mv_0^2 + kx_0^2 + 2 mgx_0$$
where the subscripts refer to some initial value. I took $x_0$ to be zero which corresponds to the point of initial contact. My $v_0$ 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.

Last edited by a moderator: Sep 16, 2004
20. Sep 12, 2004

ehild

$$mv^2 + kx^2 -2mgx = mv_0^2$$
V is maximum when
$$x=\frac{mg}{k}\mbox{. } v_{max}=\sqrt{v_0^2+\frac{mg^2}{k}}.$$

ehild

Last edited by a moderator: Sep 16, 2004