.Solving the Physics of a Seesaw Catapult

In summary, the physics of a seesaw seem quite simple, however I've been stumped on a problem. The weight falling needs to have enough force to bring the seesaw to equilibrium, and it must be weighed down so it doesn't move. Additionally, the angular velocity of the seesaw must be calculated.
  • #1
Milchstrabe
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At first glance the physics of a seesaw seem quite simple, however I've been stumped on a problem. For part of my Physics project, I am incorporating a seesaw catapult where one weight will be dropped on one side launching the hacky sack on the other side. Simple right? Of course... but the hacky sack needs to land .76 m to the left, on a stool .76 m high.

I figured my launch angle to be approx... 80 degrees ( I can always change this) which would yeild me a 5.45 m/s vertical velocity and a .8 (approx) m/s horizontal velocity. The Tanget velocity would therefore need to be 5.55 m/s (approx). Now how can I work backwards to solve for how heavy and at what height the object would have to be dropped from in order to launch it at that velocity? Let's say the arm with the hacky sack is 4 times larger then the other arm, so we have 4 and 1. I'm guessing I'll have to use Torque, and force of course.

I gave it some thought tho, and in order for the hacky to even move the mass falling must have enough force to bring the seesaw to equilibrium. So once I calculate that force required, then anything more than that will begin to bring the acceleration and velocity of the hacky up from zero. Also I must find a way to factor in the weight of the arms, it will make a difference. Any help guys?! Sorry if it was confusing
 
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  • #2
No one can help? My calculus teacher didn't really know either, he knew it was a ratio of radii but that's obvious. My physics teacher hasn't helped me since it's part of my final project, I've done some extensive research online into catapults and such, but hav been unable to find what I need. So I turned here hoping for a response. Any help would be appreciated!
 
  • #3
Should I move this to a more advanced board? I asked my physics teacher and he said I need to talk to my calc teacher, whom also wasn't sure, so I have one more teacher who might know...
 
  • #4
You need to be more specific,

"but the hacky sack needs to land .76 m to the left, on a stool .76 m high."

Left of where?

"I figured my launch angle to be approx... 80 degrees"
relative to what? 80 degrees towards the other end of the seesaw, from the ground?

What exactly are you trying to find? The force downwards? The lengths of the arms? The weight of the projectile?
 
  • #5
"I gave it some thought tho, and in order for the hacky to even move the mass falling must have enough force to bring the seesaw to equilibrium. So once I calculate that force required, then anything more than that will begin to bring the acceleration and velocity of the hacky up from zero. Also I must find a way to factor in the weight of the arms, it will make a difference. Any help guys?! Sorry if it was confusing"

Isnt necessarily true, the projectile can fly before the seesaw reaches equilibrium (i assume by equilibrium you mean parallel to the ground) depending on the friction between the two objects.
 
  • #6
Sorry maybe I should have rephrased that, the force on the left needs to over power the force(s) on the right before it can move.

I'll make a picture real quick hold on.
 
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  • #7
What initial position is the seesaw in? What are the relative masses of each arm? Uniformity?
 
  • #8
Here is a very simple , but messy, diagram:

http://www.mplionhearts.com/My_pictures/catapult-diagram.jpg


I've also been looking at page 7 and 8 of this document, I don't understand their velocity equation:

Vo = l (subscript 2) * Theta
 
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  • #9
Sorry this was the document i was looking at:

http://www.algobeautytreb.com/trebmath35.pdf pages 7 and 8...
 
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  • #10
V = I_0 Theta(dot).

The dot signifies the time derivative of theta, which is omega (angularvelocity).

If your going to actually do this demonstration, itwill be much harder to do the calculations as youwillhave to take several more factors into account, such as the inertia of the beams, air resistance perhaps, and more specifically the places where the object falls and where the hackysack is.

Why don't you go ahead and spell out what you've got, we'll work from there.
 
  • #11
Okay the weight will be landing on one extreme of the beam and the hacky will be on the other extreme of the beam. The Length of the beam on the side with the hacky will be maybe 4 times larger than the side with the weight falling. I'm going to ignore air resistance as it won't have enough of an impact to make a huge difference. If you can tell me what calculations I need to do in order to solve for the Angular Velocity then I should be okay. Let's not work with numbers right now, and instead work with variables thatI can plug in later.

My biggest problem is that I don't understand how the object falling relates to what goes on once it impacts. Aside from the obvious of course. Yet the force of the object falling will always be the same... however different heights affect the beam differently. So what involves velocity and mass, Momentum, however I don't see how I can use momentum to calculate anything I need either. I'm very lost with this. I'd really like to be able to figure out how I Can do this.
 
  • #12
Consider the energy of the falling object, [itex] PE = mgh [/itex]. As h increases, the potential energy increases, correct? This PE is then converted to KE as it falls, and with given h, the KE will impart a certain amount of energy into the beam. You'll need to know the mass of the entire beam system, aside from the shaft that holds it up, to find the clockwise torque due to the right side being 4x longer. Don't forget to include the projectile.

Once you can find that, you'll find the torque required to put the beam in motion.
 
  • #13
So once the mass impacts the beam most of the energy has been converted into kinetic energy with a small amount of potential energy still left. So this kineteic energy is then transferred over to the hacky sack, but you're right, I would need to find the torque first, let's give some theoretical numbers first. Let's say the beam is equally distrubted in mass, with a total mass of 1 kg. The mass falling is 3kg, and the hacky sack is .05 kg. Let's say the height of the mass falling is .5m. Left arm 5 cm right arm 20 cm

GPE would therefore be: 14.7

Kinetic energy at impact would be (approx): 14.7 ( a little less)

net Torque: 1.461Nm
 
  • #14
You want the launch angle to be 80 degrees. Find the displacement from the original position the seesaw will be into its final position. You'll need to specify the seesaw height too.
 
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  • #15
Lets change the launch angle to 60 degrees becausethe seesaw is so small, the fulcrum would only be .8 cm high with an 80 degree launch. With a 60 degree launch the height of the fulcrum would be 4.33 cm. Also I'm thinking of starting with the seesaw level to make calculations a little easier. So the arc length would be .1047 m of travel distance for the hacky.
 
  • #16
If you start the seesaw level, then your going to need a force to balance the seesaw initially, since the long side is significantly heavier than the short. Consider this.

Otherwise, now that you found the height you should be able to find the different torques that the seesaw will experience when a certain amount of energy (from a falling object) is imparted on it. If you're having difficulty try energy analysis. Some assumptions and simplifications will greatly help you here.

Find the new flight time with the new launch angle, from there you can use energy analysis again to find the amount of energy needed to bring the ball to the peak of its trajectory, and that's the energy you'll want to drop on the seesaw.
 
  • #17
How would the Kinetic Energy tranfer from one end of the beam to the other if you knew the n et torque and everything and the kinetic energy at impact. Are they both the same? Is the kinetic energy of the hacky the same ? Or is it affected depending on the torque?
 
  • #18
[tex] KE = W = \tau \theta [/tex]

The work done on it will equal the torque applied to it times the angle it sweeps, or

The work done to it will equal the kinetic energy of the ball as it impacts the system.
 
  • #19
The torque on the left would be the same as the torque of the right if they were the same length. Find the moment of inertia of each arm, if you know the torque of one end, you can find the torque of the other end.

[tex] T = I\alpha = \frac{1}{12}mr^2\alpha [/tex]

[tex] r_{right} = 4r_{left} [/itex]
 
  • #20
so would the kinetic energy be the same for both masses, just one would be in the opposite direction?
 
  • #21
No, the energy is also being used to rotate the seesaw.
 
  • #22
Oh I think I got it, now that i saw that you said work can be calculated by multiplying the torque by the angle that it sweeps. I'll make some simple calculations and find out. Question tho, since the arms have mass, to make this simple, how can I convert the arms masses into their weight equivilant at the end of each arm? That way we could pretend the arm's are nearly massless and add to the weights at each end.
 
  • #23
The moment of inertia of a rod is [itex] \frac{1}{12} mR^2 [/itex]

The moment of inertia of a swing with a massless arm is [itex] mR^2 [/itex]

Dont forget to include the hackey sack in your calculations. You can use superposition to calculate moments of inertia.
 
  • #24
What exactly does the moment of inertia tell me.

Torque = r (arm) x F right?

and if i find that the torque of the arm that is being impacted is 2.94, how would I find the torque of the arm that is 4 x longer with a .05 kg mass on the end of it?

KE = 1/2m(v^2)

so KE=5.789 on impact
So I'm still very CONFUSED on how to find the KE of the hacky once it reaches the top of the swing. Thanks for being patient.
 
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  • #25
Your getting too far ahead of yourself:

find the torque about the axis of rotation of each arm DUE TO GRAVITY using [itex] \tau = I \alpha [/itex]

You should get two torques in opposite directions, one for each arm. Find the angular acceleration [itex] \alpha [/itex] for each arm, then sum them to find the net angular acceleration. The dropped object must induce a torque great enough that the acceleration overcomes the angular acceleration from gravity. Do you understand this?

Once again I'll ask, are you actually going to perform the experiment?
 
  • #26
Yes I understand it. However I don't understand moment of inertia, we never really touched on inertia in or class. Yes I will be doing this which is why I prefer to have all the calculations. I could be a butt and just build it and adjust it until it works then use photogates to measure the velocity, but that's very cheap and no fun.
 
  • #27
Ok, my first suggestion is to go read up on moment of inertia. I am sure www.wikipedia.org has a great tutorial on it.

Second, if you are going to conduct this experiment, there are a lot of other factors you need toconsider about the actual seesaw. Youneed it to be abel to absorb all this energy and transfer it well. It would need to be rigid, strong, but relatively light. I can help you with the calculations, but they might not give you an experiment that's practical.

I could just tell you about M of I, but that would take away from the learning segment of your experiment, right?
 
  • #28
I will be making the beam as light and as rigid as possible, I'm not going to worry about other things, I'll just do an error analysis. I will still be building before I do my final calculations, but if I know hwo to do the calculations it will be much easier to figure out why it works once I am done. I don't expect it to be completely accurate, that's ok.

Okay I read up on moment of inertia and I think I understand it now. However you say earlier that:

[tex] T = I\alpha = \frac{1}{12}mr^2\alpha [/tex]

so since we are trying to find alpha, would you find torque using another method, then set it equal to that and solve for alpha?
 
  • #29
Yup.
The effect of gravityon the board can be simplified by finding thecenter of mass of the rod. Assuming its uniform, you know the center of mass is halfway along each rod. You can then use the equation for torque you mentioned above to find the torque, and then angular acceleration.
 
  • #30
Lol okay, so the center of mass on a 50cm rod will be 15 cm to the right (towards the longer arm) of the fulcrum if I have one arm at 10cm and one at 40. For some reason, I'm not functioning tonight though, and with all the stress I'm under I'm not surprised. How would I find the Force from gravity on each arm knowing the center of gravity and the mass of the rod?

EDIT sorry let me rephrase that, Torque is measured by distance from the fulcrum (radius) x Force (in this case gravity). Yet the mass of the arm could be divided up infinately, with an infinate number of distances, which gets us into calculus. But I know you're trying to show me how to make it so this is not necessary, I'm just not understanding it.
 
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  • #31
The best way to do this is to find the center of mass of each rod about the axis of rotation.

Forexample, say that the right rod, the long one is of length x = 16cm. The center ofmass is then at x=8cm, because it is uniform.

The torque can be expressed as

[tex] \tau = F R = F \frac{x}{2} [/tex] for any uniform rod length x, the cm will be at x/2

the force is simply F = mg

[tex] \tau = m g \frac{x}{2} [/tex]

The moment of inertia of a rod of length x is

[tex] I = \frac{1}{12} mx^2 [/tex]

and we know that [itex] \tau = I\alpha [/itex] so [itex] \alpha = \frac{\tau}{I} [/itex] and so

[tex] \alpha = \frac{\tau}{I} = \frac{mg\frac{x}{2}}{\frac{1}{12}mx^2} = 6 \frac{g}{L} [/itex] if I didnt make any mistakes.

Do you knowwhere to go from here?
when is this due?
 
  • #32
OOH the center of mass of each arm, I gotcha now, I didnt understand what you were saying before lol. I feel stupid. I have till may 11 but this is just part of the project, the last part of it. It is basically something that we start, and then let it run through and the goal is to cover everything we have learned this year and get the hacky sack on the center of the stool (you can't have anything go over the stool). This is the most complex part of the whole aparatus, I'm working backwards.

Momentum
Torque
Force and Motion
Projectile Motion
Constant Acceleration
Friction
Work, Energy, and Power
Circular Motion
Constant Velocity

Can all be covered just in the catapult part of the mechanism lol. We have 11 units to cover, that is 8 of them. However I believe we need to have a separate mechanism that specifically demonstrates each of these :( lol.
 
  • #33
Okay I want to move on now, but I got my Angular acceleration to be 147.924...

right arm length : 40cm (.4m)
right arm mass: 80g (.08kg)

Do I need to factor in the left arm too or something? Because 147.92 4does not seem right at all, expecially since i converted to meters and kg's.
 
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  • #34
Remember you have to do the same for the left arm and add the two, to find the torque of the entire system. Also remember since the torques are acting in opposite directions, one will be negative. How did you get angular velocity? All the above gives angular acceleration.

Also, whenever you give a number, units are just as important as the number.

What was your launch angle? How far is the stool? I suggest figuring out these two questions before advancing with the seesaw.
 
  • #35
Okay so I got a new torque for the entire thing, but With inertia, do I do the same thing, one will be negative and one will be positive, or are they both positive.
 

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