Falling objects, springs

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  • #26
Integral
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I have not done any clean up algebra but the minimum velocity V=0 will occur at

[tex] x = \frac {2V_i g} {\lambda H} - \frac g {\lambda ^2}[/tex]

Where H and [tex]\lambda[/tex] are as defined in my previous post.
 
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  • #27
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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.
your equation just doesn't make sense. the term [tex]mv^2[/tex] on the left side of the equals sign is larger than the term [tex]mv_{0}^2[/tex] on the right side because v is the peak velocity. and since [tex]kx^2[/tex] is positive that means the term [tex]2mgx[/tex] 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 [tex]2mgx[/tex] 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. :smile:
 
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  • #28
Tide
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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.
[tex]mv^2 + 2mgx = mv_0^2[/tex]
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 [itex]kx^2[/itex]. Including it gives the equation I wrote:
[tex]mv^2 + kx^2 + 2mgx = mv_0^2[/tex]
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).
 
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  • #29
ehild
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Integral said:
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
 
  • #30
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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.
 
  • #31
ehild
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Integral said:
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
 
  • #32
HallsofIvy
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What makes you think the object will move faster going upward than downward?
 
  • #33
ehild
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HallsofIvy said:
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.

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

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

if we speak about velocity, and just

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

if it is speed.

ehild
 
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  • #34
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Ideally each time the mass/spring combination oscillates past the point I called xmax 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.
 

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