.Solving the Physics of a Seesaw Catapult

[whozum]

The inertia is a scalar. The angular acceleration is a vector, and has a direction.

 

[Milchstrabe]

Scalar meaning... sorry, by the way, I'm in arizona too.

 

[whozum]

Dont worry about scalars and vectors. The inertia doesn't have a direction, it just has a magnitude. The acceleration however has a magnitude and a direction (clockwise, counterclockwise)

Where in?

 

[Milchstrabe]

Okay same numbers, factoring the entire beams moment of inertia and the different torques I get a Angular Acceleration of 32.9 M/s^2 does this sound more reasonable...?

It still doesn't seem right.

Net Torque of arms = .0686

Moment of inertia = .00208

 

[whozum]

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

$$I =\frac{1}{12}mR^2 = \frac{1}{12} (.08)(.4)^2 = 1.066x10^{-3}$$

$$\tau = m g \frac{x}{2} = (.08)(9.8)(.2) = 0.1568Nm$$

$$\alpha = \frac{\tau}{I} = \frac{0.1568}{1.066x10^{-3}} = 147.09**\frac{rad}{sec^2}$$

Cool?

 

[whozum]

Left side:
arm length = 10cm (.1m)
arm mass = 20g (0.02kg)


$$I = \frac{1}{12}mR^2 = \frac{1}{12}(.02)(.1)^2 = 1.67x10^5$$

$$\tau = mg\frac{x}{2} = (.02)(9.8)(.05) = 0.0098Nm$$

$$\alpha = \frac{\tau}{I} = \frac{.0098}{1.67x10^5} = 586.83\frac{rad}{sec^2}$$

$$\alpha_{net} = \sum \alpha = \alpha_1 + \alpha_2 = -147.09 + 586.83 = 439.74\frac{rad}{sec^2}$$

 

[Milchstrabe]

Right but I thought I had to factor in both arms?

 

[whozum]

I'm sorry man, the moment of inertia for a rod about its end is [itex] I = \frac{1}{3}mR^2 [/tex].

The angular acceleration for each end will be a quarter of what itwas, and the net angular acceleration will be 109.935 rad/sec^2.

 

[whozum]

If you want to workonthis some more tonight, the next step is to figure out the height of the see saw. If you want a launch angle of 80 degrees like you said, then use some geometry to find the angle between the seesaw and the horizontal. You want them to be touching at this point. Use somemore geometry to find the height of the seesaw.

Then using the torque due to gravity, find a mass that when dropped from a certain height will exert enough torque to overcome gravity, and swing the seesaw around with enough speed to launch the projectile the desired distance. whether the object will be dropped straight down or perpendicularto the resting position of the seesaw will make a big difference, so keep in mind the difference between theoretical and actual results.

 

[Milchstrabe]

What I find confusing about this is the fact that the angular acceleration is in radians? Expecially since we're only talking about gravity here with no other masses on the beam yet. With a rate of change in velocity of 109 rad/s^2 That is very fast. If it were degree's it would make more sense, but I don't understand howthat can be possible in radians.

 

[whozum]

It is pretty high but think about it. You're letting an object fall about an axis. This is only the acceleration at the point where the seesaw is parallel to the ground, you would have to use some calculus to find the acceleration as a function of time, but initially, there is a lot of force pulling the seesaw down. 109rad/s^2 is about 20 revolutions per sec, if the acceleration was constant.

 

[Milchstrabe]

But we haven't factored anything aside from the mass of the beams have we?

 

[whozum]

No, the mass and radius are the only things that play a role in this, remember the seesaw is only rotating ~30 degrees til it hits the floor.

 

[Milchstrabe]

I know that, but we haven't included the mass falling, or the mass of the hacky sack yet, just the mass of the arms and length etc. So that's why I think the number is too high? Maybe not?

 

[whozum]

Thats the seesaw's natural rotation from gravity. You want the falling object to counter this rotation, aka to apply a torque greater than that applied by gravity. I posted about this last night, check back.

 

[Milchstrabe]

Is there a problem or is it just me?

We want the angular acceleration to be 0.

Right now with no extra masses included we have it at at 110.339 rad/sec (yours is just slightly different). However in order to get this to equal zero we need to change both the inertia and the torque of the shorter side. The length of the beam will remain the same, at .1 m, but the mass will be different now (in addition to the mass of the beam)

$$I = \frac{1}{3}mR^2\$$

$$\tau = mgR$$

$$\frac{\tau}{I} = \alpha$$

And in order for the two angular accelerations to balance out, they both need to be equal, right now the one on the right (longer beam) is 36.8075 rad/sec.

$$\frac {\tau}{I} = \frac {m(9.8)\frac {.1}{2}}{\frac{1}{3}m.1^2} = 36.8075$$

 

[whozum]

Your seesaw is going to be built out of one rod, with a fulcrum placed somewhere towards the left side, so how are you going to change the mass of your left side? If you want the torque to be zero, both sides will have to be of same length. I don't understand what your trying to do, you seem to be changing the problem.

Why do you want it to equal zero?

 

[Milchstrabe]

We calculated the angular acceleration with just the beam mass length and the fulcrum at its left posistion. And the angular acceleration was 109 rad/sec, if it is that much, then it has to be a clockwise acceleration (since the longer beam and more mass is on the right side right now), we need to counter that acceleration until it is zero and then make it counter clockwise right? I'm totally lost.

 

[whozum]

If one end of your seesaw is going to be longer than the other, then there's no way you can balance the seesaw, and get the angular acceleration to be zero. Why are you trying to do this?
Answer me this:
What is the torque on the beam due to gravity?

 

[Milchstrabe]

Net torque is : .147 Nm

 

[whozum]

Clockwise or anticlockwise? If thast the torque, what's the minimum torque required to overcome this and get it to accelerate the other direction?

 

[Milchstrabe]

It's clockwise and you'd need .147 Nm of torque in the opposite direction to counter it.

 

[whozum]

Ok good, the next step is to figure out the height of the see saw. If you want a launch angle of 80 degrees like you said, then use some geometry to find the angle between the seesaw and the horizontal. You want them to be touching at this point. Use somemore geometry to find the height of the seesaw.

 

[Milchstrabe]

The height of the fulcrum would be 1.786 cm and the height of where the hacky is being released will be 8.682 cm.

 

[whozum]

Alright it gets a bit complicated now. You want to find the velocity of the hackysack that will let it fly into the chair. This velocity depends on how fast the seesaw is swinging when it releases it, which depends on how hard the object hits it.

You know that the net torque will be the sum of the torque from gravity and the torque of the object, and you know the torque applied of the object falling will depend on how high it falls.

First off, is this object going to stick to the seesaw or fall off?

 

[whozum]

$$\tau_{net} = \tau_{grav} + \tau_{opp}$$

$$\tau_{opp} = \tau_{obj} + \tau_{impulse}$$
so

[tex] \tau_{net} = \tau_{grav} - \tau_{obj} - \tau_{impulse} [/itex]

The impulse torque will be the torque from the object's impact, due to the transfer of its kinetic energy. Were assuming it sticks.

The object torque will be the extra torque applied by the dropped object once it attaches.

Can you find equations for these in terms of m,g, x, and h?

 

[Milchstrabe]

Is torque from gravity the net torque that we found earlier? The effect of gravity on both arms torques gave us a Tnet of .147? Or is Torque grav just the right arm? I'm Guessing that torque gravity is the torque given from the gravity pulling on both arms.

And how can I get Impulse into a torque? If impulse is defined by:

$$p = F t$$

and Angular Impulse is defined by:

$$L = \tau t$$

So far I have

$$\tau_{net} = \tau_{grav} - \tau_{obj} - \tau_{impluse}$$

$$\tau_{net} = |m_1g(\frac {x_1}{2}) - m_2g(\frac {x_2}{2})| - m_{obj}g(x_{obj}) - \tau_{impulse}$$

 

[whozum]

Alright, I'm not sure about this part but, by definitions:

$$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt}$$

My estimate would be that $dt$ would be the time it would take the object to travel its own length at the given $dv$.

Then given that we can find the force applied (apprxoimately) and by knowing the radius, can find the torque applied.

[tex]\tau_{impulse} = F_{impulse}x_{rod}[/itex]

Now you want $\tau_{net}$ to be such that when it goes through a displacement [tex] \theta [/itex] it will have imparted energy $E$ into the hackysack so that $E = 0$ at the peak of its trajectory.

Do you know how far you want it to go?
Can you find how high you want it to go?
Can you find the angle that the hackysack will sweep while being accelerated by the seesaw?
Can you find the potential energy the hackysack has in regards to the seesaw and the potentially applied torque?
Can you translate that energy into kinetic energy of the hackysack?

 

[Milchstrabe]

Okay you're loosing me, I'm not sure about what the first formula tells me, and the I can get all the answers to the questions on the bottom except the last 2.

My estimate would be that dt would be the time it would take the object to travel its own length at the given dv

That is confusing Its own lenght, however the measurement from the top of the object, to the bottom of the object?

 

[whozum]

The first line is the calculus explaining Why Ft = mv. If you don't understand it, no worries.

For the last two questions,

$$W = \tau\theta$$, right?

As the seesaw turns through angle $\theta$, it is imparting energy into the hackysack. Once it has turned through $\theta$, the hackysack releases from the seesaw and no longer is receiving energy from it. If the hackysack gained all its energy from the seesaw, then what is it's energy in relation to the energy imparted to it by the seesaw?

 

[Milchstrabe]

Oh gotcha, so $$\tau$$ in that equation is the net torque correct? And once we know that, we already have the angle that it sweeps (10 degrees) and that would give us the energy. And when the hacky sack is released it has mostly kinetic energy. But finding the impulse is what I'm still confused about.

 

[whozum]

When the hackysack is released, all of its kinetic energy is from the seesaw. Thus

$$KE = \tau\theta$$

Are you sure its ten degrees? that seems really low. Can you show the geometry?

 

[Milchstrabe]

The angle that it sweeps? If its release angle is 80 degrees, then the angle the beam sweeps must be 10. It's starting parallel

 

[whozum]

How are you going to keep it steady? I was undert he impression we were starting with the hackysack side touching the floor.

 

[Milchstrabe]

Well it makes calculations simpler if I start it parallel, so i'll have some mass underneath the right arm to support it.