How do I calculate the peak velocity of a 1kg object dropped 1m onto a spring?

[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.

 

[Integral]

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

 

[aliendoom]

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

you need to rethink that.

 

[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.

 

[aliendoom]

i know the peak velocity occurs when F = -kx + mg = 0.

The maximum speed will occur when the gravitational force is balanced by the spring force..

thanks. anyone else?

 

[Integral]

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?

 

[Tide]

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.

 

[aliendoom]

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.

 

[Tide]

Use energy conservation!

 

[Integral]

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.

 

[aliendoom]

Use energy conservation!

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

 

[Tide]

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?

 

[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?

 

[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.

 

[Integral]

From the solution of the DE I get
V_i = 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!

 

[Tide]

Your units aren't right! :-)

 

[Integral]

Your units aren't right! :-)

Thats for sure!

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

 

[aliendoom]

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.

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:

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.)

 

[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.

 

[ehild]

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.

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

ehild

 

[Tide]

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

ehild

Gravity pulls downward so the gravitational potential must have the positive sign. Your version has gravity pushing upward.

 

[Integral]

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

$$x(0)=0$$

$$\dot {x}(0)= - V_i$$

Solution:
let
$$\lambda^2 = \frac k m$$

$$x(t)=\frac g {\lambda^2} \cos(\lambda t) - \frac {V_i} \lambda \sin(\lambda t)- \frac g {\lambda^2}$$

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

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

$$\sin(\lambda t) = \frac g H$$

$$\cos(\lambda t) = \frac {V_i \lambda} H$$

so:

$$x_{max} = - \frac g {\lambda^2} = -\frac {mg} k$$

$$V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k$$

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!

 

[ehild]

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.

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

 

[ehild]

$$x_{max} = - \frac g {\lambda^2} = -\frac {mg} k$$

$$V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k$$

Actually, they are rather

$$x_{min} = - \frac g {\lambda^2} = -\frac {mg} k$$

$$V_{min}= - \sqrt { {V_i} ^2 + \frac {m g^2} k$$



ehild

 

[Integral]

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

$$x_{v_{max}}$$

 

[Integral]

I have not done any clean up algebra but the minimum velocity V=0 will occur at

$$x = \frac {2V_i g} {\lambda H} - \frac g {\lambda ^2}$$

Where H and $$\lambda$$ are as defined in my previous post.

 

[aliendoom]

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.

your equation just doesn't make sense. the term $$mv^2$$ on the left side of the equals sign is larger than the term $$mv_{0}^2$$ on the right side because v is the peak velocity. and since $$kx^2$$ is positive that means the term $$2mgx$$ must be a negative quantity in order to balance out the equation.

i guess the response is that the downward direction is the negative direction, so x is negative which would make the term $$2mgx$$ negative. that's way too confusing. in fact that 'problem' even fooled you because you said my description:

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.

was only part of it. my description yields the same results as your equation once your equation is 'corrected' with the negative sign, although it took me quite awhile to figure out why they didn't agree when I knew they must.

thanks for the help.

 

[Tide]

It makes perfect sense! :-)

I took the "up direction" to be positive x and the ball hits the pendulum at x = 0. You can see that without the spring it is precisely what you would write for an object moving in the gravity field.
$$mv^2 + 2mgx = mv_0^2$$
If the ball goes up it loses speed and if it goes down it acquires speed. This is pretty much the standard way of writing energy conservation.

Now I add the spring. It's logical to place the spring before compression with its top end at x = 0 and its potential is $kx^2$. Including it gives the equation I wrote:
$$mv^2 + kx^2 + 2mgx = mv_0^2$$
Clearly, as the ball continues its downward motion the potential due to the spring increases (tending to decrease the speed) while the gravitational potential continues to decrease (tending to increase the speed).

 

[ehild]

ehild,

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

Well, how can be a quantity negative if its minimum is zero?

(You mix velocity and magnitude of the velocity.)

ehild

 

[Integral]

The negative sign IS the direction. In my set up the coordinate system x increases up. So a negative velocity is going down. Note that in the initial conditions the initial velocity is negative. When speaking of velocities we must specify direction and magnitude, in this case negative means down. when looking for the min or max velocity it only makes sense to speak of the magnitude. Direction does not matter.

 

[ehild]

The negative sign IS the direction. In my set up the coordinate system x increases up. So a negative velocity is going down. Note that in the initial conditions the initial velocity is negative. When speaking of velocities we must specify direction and magnitude, in this case negative means down. when looking for the min or max velocity it only makes sense to speak of the magnitude. Direction does not matter.

The question was peak velocity. Cannot it be positive as well? What do you think will happen after the body reaches that maximum negative velocity? I admit that "maximum peak velocity" means velocity of maximum magnitude, but this body moves first downward then upward so its peak velocities are both positive and negative, aren't they?

ehild

 

[HallsofIvy]

What makes you think the object will move faster going upward than downward?

 

[ehild]

What makes you think the object will move faster going upward than downward?

Oh my... My English must be very bad. I can not make myself understand. No, I did not say that the peak speed is higher upward than downward.
Well, Integral said that the maximum velocity was negative.

$$V_{max}= - \sqrt { {V_i} ^2 + \frac {m g^2} k$$

I wanted to say that the body reaches the same maximum SPEED both downward and upward, and I argued about the minus sign in Integral's post. It should be

$$V_{max}= \pm \sqrt { {V_i} ^2 + \frac {m g^2} k$$

if we speak about velocity, and just

$$V_{max}= \sqrt { {V_i} ^2 + \frac {m g^2} k$$

if it is speed.

ehild

 

[Integral]

Ideally each time the mass/spring combination oscillates past the point I called x_max it will have the same velocity. My solution is specific to a TIME at the TIME I found the ball is moving down. Notice that my solution is oscillatory in nature, so with a bit more effort I could find a sequence of times at which the ball passes through the point at which it has a maximum velocity.

Now, if it is not an ideal spring and losses are occurring which are not captured in our simple model, then when the spring/ball returns it will NOT be traveling at the same speed. So I could argue that the ONLY time the ball/spring system will have that maximum velocity is on the way down the first time.

Notice that the energy based solution the final result is found after taking the square root, by all rights you should take the negative root as the correct solution.