.Solving the Physics of a Seesaw Catapult

[whozum]

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

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

the force is simply F = mg

\tau = m g \frac{x}{2}

The moment of inertia of a rod of length x is

I = \frac{1}{12} mx^2

and we know that \tau = I\alpha so \alpha = \frac{\tau}{I} and so

\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.<br /> <br /> Do you knowwhere to go from here?<br /> when is this due?

 

[Milchstrabe]

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.

 

[Milchstrabe]

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.

 

[whozum]

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.

 

[Milchstrabe]

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.

 

[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 I = \frac{1}{3}mR^2 [/tex].<br /> <br /> 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.

<br /> \frac {\tau}{I} = <br /> <br /> \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?