Calculating spring constant

[Unity+]

I keep getting results that don't make sense.

 

So, the idea is I have to find the spring constant only using a meter stick, cross-bow(rubber band), and tape. What I did was I shot the cross-bow straight horizontally and when shot 5 times I got the average distance it took for the ping pong ball(.0025 kg) to hit the ground.

 

Here are the known variables with the experiment:

 

d = 4.471m

h = .83m

g = 9.81(given)

x(spring stretched) = .17m

 

Now, my logic was to use the height and distance the ping pong ball went to find the time it took for the ball to hit the ground.

 

[math].83m = \frac{1}{2}(9.81)t^{2}[/math]

[math].83m = (4.905)t^{2}[/math]

[math].1692 = t^{2}[/math]

[math]t = .4113s[/math]

 

Then, I used this time and the distance in the x direction to find the horizontal speed.

 

[math]v = \frac{4.471}{.4113} = 10.87m/s[/math]

 

Then, I setup the equation of kinetic, elastic potential, and potential gravitational energy because the ping pong ball is being shot with a rubberband(crossbow) from a height of .83m.

 

[math]\frac{1}{2}k(.17)^{2} + (.0025kg)(9.81)(.83m) = \frac{1}{2}(.0025)(10.87m/s)^{2}+(.0025kg)(9.81)(.83m)[/math]

[math]\frac{1}{2}k(.17)^{2}= \frac{1}{2}(.0025)(118.1569)[/math]

[math]\frac{1}{2}k(.17)^{2}= \frac{1}{2}(.0025)(118.1569)[/math]

[math].01445k= \frac{1}{2}(.0025)(118.1569)[/math]

[math].01445k= .1477[/math]

[math]k= 10.22...[/math]

 

When I did it in this post, it seems I got a reasonable answer now. However, I have a feeling that on the first step of the energy equations I do not put the potential gravitational energy party on the left side of the equation.

 

EDIT: Found an error.

Edited April 7, 2014 by Unity+

[imatfaal]

1. Really silly way to find k - shoot straight upwards and time. Or even better just hang masses off band.

2. You can really measure the distance a ping-pong ball flies to a millimetre precision??

3. If this is a real experiment you lose tonnes of marks for not including uncertainties.

4. Why are you including grav pot energy in the first place? It acts orthogonally to initial flight (ie spring force direction) and thus cannot contribute. You can just equate kinetic energy and spring potential energy to get the same answer

5. There are multiple ways to improve experiment that may get extra marks if mentioned in write-up. Air resistance is gonna play a huge part and screw up results.

[studiot]

Was this another of those experiments designed by that well known Physicist, JC (John Clees) from the Ministry of Silly Questions?

 

It rather reminds me of a GCE practical from long ago when we had to measure surface tension by floating a needle and then dragging it up the meniscus with a magnet.

 

 

Comments?

 

I can't easily think of a worse projectile. Pingpong balls are designed for maximum interaction with the air. You need something small and dense to minimise air interaction.

With the proviso that air resistance can be ignored (not at all true here) your kinematics equations are correct.

 

However there are several questions relating to your energy analysis.

 

Well there was not diagram so I can't tell was acting double or single. Any catapault/bow arrangement applies the tension double, but at an angle.

So is the constant calculated for a double band or single?

 

Further is your x distance in the strain energy term in your equation correct or should it be 0.17 cos of some angle?

That is: Do the forces in the elastic act at an angle to the displacement?

Again a diagram would help.

 

 

Finally concerning imatfaal,s point 4.

Energy is a scalar so does not act orthogonally to anything.

Your energy balance is lacking a term as is your kinematic analysis.

This does not matter in the kinematic analysis since the term is zero at launch, but does affect the energy balance.

 

The intial vertical velocity is zero.

Upon striking the ground the ball has acquired a vertical velocity, whose kinetic energy exactly equals the loss of potential energy due to the vertical drop.

Edited April 7, 2014 by studiot

[imatfaal]

Was this another of those experiments designed by that well known Physicist, JC (John Clees) from the Ministry of Silly Questions?

 

It rather reminds me of a GCE practical from long ago when we had to measure surface tension by floating a needle and then dragging it up the meniscus with a magnet.

 

 

Comments?

...

 

Quite Agree Quite Agree

 

 

...

 

Finally concerning imatfaal,s point 4.

Energy is a scalar so does not act orthogonally to anything.

Your energy balance is lacking a term as is your kinematic analysis.

This does not matter in the kinematic analysis since the term is zero at launch, but does affect the energy balance.

 

The intial vertical velocity is zero.

Upon striking the ground the ball has acquired a vertical velocity, whose kinetic energy exactly equals the loss of potential energy due to the vertical drop.

 

I should have said gravity (rather than gpe) was orthogonal to the initial horizontal force you are quite right - but as gravity is indeed orthogonal you can ignore it entirely - it cannot affect the matter. The Energy balance is not missing a term - it has extra terms. All you need is 1/2mv^2 = 1/2kx^2 - ie in the horizontal x axis only - gravity acting orthogonally can do no work in the x direction; so no need to confuse matters by including GPE terms.

[studiot]

 

Comments?

 

Reading my previous post I realise you might I thought I was inviting comments on JC, so I apologise for the sloppy posting. I meant that was the start of my comments on the physics of the question.

 

 

The Energy balance is not missing a term - it has extra terms. All you need is 1/2mv^2 = 1/2kx^2 - ie in the horizontal x axis only - gravity acting orthogonally can do no work in the x direction; so no need to confuse matters by including GPE terms.

 

 

Conservation of energy (from which the balance equation is derived) applies to the total energy. You cannot just take a bit of the system's energy and equate it to another bit. Just taking the "horizontal energy" implies we can resolve energy into components, which is not true. Any of the energies can be converted to any of the others, given the right circumstances. If we do ignore some energy we should always show why; for instance there are no magnetic forces acting, or if we can say that a particular energy is constant, we can ignore it. Neither the gravitational PE nor the kinetic energy due vertical motion are contant.

 

 

Similarly the calculation of the time of flight stems from the constant acceleration equation s=ut+0.5ft²

Had the zero term been included here the nonzero term in the energy balance might not have been missed.

 

This may be considered a philosophical point or excessively pedantic, but many questions are fluffed simply due to missing part of the system.

Edited April 7, 2014 by studiot

[imatfaal]

 

...

 

Conservation of energy (from which the balance equation is derived) applies to the total energy. You cannot just take a bit of the system's energy and equate it to another bit. Just taking the "horizontal energy" implies we can resolve energy into components, which is not true. Any of the energies can be converted to any of the others, given the right circumstances. If we do ignore some energy we should always show why; for instance there are no magnetic forces acting, or if we can say that a particular energy is constant, we can ignore it. Neither the gravitational PE nor the kinetic energy due vertical motion are contant.

 

Whilst we are neglecting air resistance (which I agree with a pingpong ball is ridiculous) - there is no force acting on the ball in the horizontal direction after launch. The spring potential energy of 1/2kx^2 is converted to kinetic energy of 1/2mv^2 - that is the only energy balance you need as other forces are acting orthogonally. I am not claiming we can resolve energies into components - but you seem to be implying that we cannot do it with forces; and in fact we must do it for forces. "Neither the gravitational PE nor the kinetic energy due vertical motion are contant." and the vertical motion (once the time has been calculated which was a completely separate section) has absolutely nothing to do with answer. Take a time of one second to fall - work out energy balances with gravity set at 10ms^-2. 15ms^-2 and 20ms^-2 - they will all be the same, they cannot be anything else. Gravity acts in the y direction the spring acted in the x - nothing will change the fact that a force acting at right angles cannot do work that affects the horizontal motion - and if there is no work being done then energy in must equal energy out.

This may be considered a philosophical point or excessively pedantic, but many questions are fluffed simply due to missing part of the system.

 

But just as many are fluffed by un-necessary complication.

[studiot]

 

But just as many are fluffed by un-necessary complication.

 

 

It's not unnecessary in this case since the original analysis was flawed.

 

Further this analysis is also suspect

 

 

The spring potential energy of 1/2kx^2 is converted to kinetic energy of 1/2mv^2 - that is the only energy balance you need as other forces are acting orthogonally.

 

What is x?

 

 

I have already said we need a diagram.

 

How many elastic forces are acting on the ball?

 

In my diagram the orientation could be vertical, in which case there are opposing vertical forces to consider

 

It could be horizontal or somewhere in between. It then becomes necessary to prove that there is no exchange of energy between horizontal and vertically acting agents. See my comment on the pendulum at the end.

 

I also am not sure what the dimension x represents, because if it is as I have shown then it is not the elastic extension of the band.

 

 

As to conservation of energy and resolution into components.

 

I do not think for one minute that I have anywhere implied you cannot or should not resolve vectors including forces.

My intention, since it is an important technique, has always been quite the reverse. I' ve tried to carefully avoid say horizontal or vertical energy, rather to indicate what may be correctly attributed (resolved) into such directions.

 

If you consider the humble pendulum and gaily state that you may gaily separate the energies into horizontal and vertical you will come to inappropriate conclusions.

The total (mechanical) energy is constant. In the centre of the swing it is all KE and the motion is purely horizontal.

At the ends of the swing it is all PE and the motion is purely vertical. Everywhere else it is some combination of both.

 

So there is an example mechanical system where gravitational PE and the energy due to horizontal motion are interchanged.

Edited April 7, 2014 by studiot

[imatfaal]

In the pendulum you have a force (tension which acts along the the line of the string) which has a component that acts in the horizontal direction. This can do work and the force can change the horizontal component of the velocity. You seem to be saying that it is gravity which provides the horizontal force - it does not; it is the tension in the string. Gravity alone cannot change the horizontal motion (until you get to very large scales and orbits etc. )

 

In this problem there is no such force. Considering GPE is superfluous and the whole of the vertical direction is a waste of time and potentially confusing - and doing extra sums can only lead to further complication.

[studiot]

 

Considering GPE is superfluous

 

How can you say that? Are you not considering it by saying it does not work?

 

All I have asked for is sufficient justification to exclude it.

 

I was taught to start static analyses with the words

 

"As in equilibrium."

 

At the tiem I thought "what an unnecessary fuss.

 

But the number of poster I have seen here here carrying out inappropriate equilibrium analyses on systems that are not in equilibrium justifies it in my hindsight.

 

You have also talked about gaining extra marks for a good discussion.

 

You can also earn extra marks (or at least prevent loss of marks) by justification statements so even if (as here) the strain energy calculation went awry the statement

 

Total energy of system = Strain energy + Potential Energy + Kinetic Energy = A constant

 

Would earn part (or even half) marks.

 

And in the real world outside proper annotation and documentation can help understand the flow some time alter.

[imatfaal]

 

How can you say that? Are you not considering it by saying it does not work?

 

...

 

It is superfluous - it is not wrong but it is un-necessary. Show me a way that gravity accelerates an object in the x direction without another force present (the normal, tension etc); it doesn't, it cannot, it acts in the y direction, and has no x component. Mechanics problems work by analysing that which can make a difference.

[studiot]

 

Show me a way that gravity accelerates an object in the x direction without another force present (the normal, tension etc); it doesn't, it cannot, it acts in the y direction, and has no x component.

 

 

I am trying to keep this discussion friendly. I think we are agreed on the merits and demerits of the problem in hand.

It's just in the presentation we can surely agree to differ.

 

 

Unity+ can then have the (luxury) of two different points of view to choose from.

He is a first rate student and I have a lot of confidence in him.

 

 

 

I didn't say otherwise, however since you ask I believe it's called fluid pressure.

 

A vertical force (acceleration) due to gravity becomes a horizontal force (acceleration) within the fluid.

 

It is also true to a less obvious extent that in solids a vertical force (due to gravity) becomes a horizontal shear force.

Edited April 7, 2014 by studiot

[imatfaal]

 

 

I am trying to keep this discussion friendly. I think we are agreed on the merits and demerits of the problem in hand.

It's just in the presentation we can surely agree to differ.

 

 

Unity+ can then have the (luxury) of two different points of view to choose from.

He is a first rate student and I have a lot of confidence in him.

 

 

Quite agree - I wouldn't dream of arguing in most threads like this. I am sure Unity can avoid confusion caused by two duffers trading blows

I didn't say otherwise, however since you ask I believe it's called fluid pressure.

 

A vertical force (acceleration) due to gravity becomes a horizontal force (acceleration) within the fluid.

 

It is also true to a less obvious extent that in solids a vertical force (due to gravity) becomes a horizontal shear force.

 

But again there are other forces present (buoyancy and the normal I presume). Thus you need to take account of them. In this case there are none other than gravity and that acts only downwards.

[Unity+]

I can't easily think of a worse projectile. Pingpong balls are designed for maximum interaction with the air.

I know, but it was the item of requirement. In the report I wrote, I stated that in order to get a more accurate result for k the air resistance would have to be considered by determining the force it had on the ping pong ball.

 

I think I got insight on the problem. Thanks for all the information.