.Solving the Physics of a Seesaw Catapult

[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?

 

[whozum]

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

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

\tau_{net} = \tau_{grav} - \tau_{obj} - \tau_{impulse} [/itex]<br /> <br /> The impulse torque will be the torque from the object&#039;s impact, due to the transfer of its kinetic energy. Were assuming it sticks.<br /> <br /> The object torque will be the extra torque applied by the dropped object once it attaches.<br /> <br /> 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.

\tau_{impulse} = F_{impulse}x_{rod}[/itex]<br /> <br /> Now you want \tau_{net} to be such that when it goes through a displacement \theta [/itex] it will have imparted energy E into the hackysack so that E = 0 at the peak of its trajectory. &lt;br /&gt; &lt;br /&gt; Do you know how far you want it to go?&lt;br /&gt; Can you find how high you want it to go?&lt;br /&gt; Can you find the angle that the hackysack will sweep while being accelerated by the seesaw? &lt;br /&gt; Can you find the potential energy the hackysack has in regards to the seesaw and the potentially applied torque?&lt;br /&gt; 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.

 

[whozum]

Again, define your angles, if it is starting flat, then its starting angle is 0 degrees.

 

[Milchstrabe]

Yes, and it's release angle of the beam is at 10 and the hacky will be traveling at an 80 degree angle at the release point

 

[whozum]

The math is the same either way. If you start parallel, you're going to have less angle to sweep through, and will not produce as much energy as if you started on the floor.

 

[Milchstrabe]

Okay well we'll start on the floor then, so make the sweep angle 20.

 

[whozum]

You're going to need to recalculate your fulcrum height, height at launch, and swept angle then. Let's say the height is 1cm, then your starting angle will be -arcsin(1/4) to the horizontal.

Also note that the height of the fulcrum determines the max launch angle, so we need to find that first.

How high is the target? How far is the target? Your launch angle depends on this. 80 degree launch is almost all horizontal, so if the target is considerably higher than the seesaw, you'll want a much smaller launch angle.

 

[Milchstrabe]

no no no, you're interperting me wrong. The launch angle is 80 degrees, as if you wer eto draw a right angle up from the ground, so 80 degrees is extremely vertical. I've done all my calculations for the projectiles flight. I'm still just trying to figure out the impulse torque, so i can get the net torque, so i can get the work done.

 

[whozum]

Ok show me your launch calculations and stuff. Also show me your seesaw with the before and after angles.

 

[Milchstrabe]

Actually let me change my launch calculations and i'll get back to you. I have to go someplace but if you can let me know a clear way to calculate the impulse torque, that will greatly help.

The see saw starts at a 10 degree angle, with the right side down, and ends at a 10 degreen angle with the right side up. so the total distance swept is 20 degrees

 

[whozum]

Ok, there's two ways you can do this. You can set a target launch distance, and therefore would need to find the impulse torque, or you can set the impulse torque, and find the correct launch distance.

I thin kthe first way, which is what I just asked you to do, is easier. If you know the energy needed to raise it a certain height, then you can find the required net torque. We already know the other two torques, and we would know the net torque needed to attain a certain energy, so solving we would find the impulse torque.

The impulse torque we want will depend on the height we want. Don't worry about it for now, we'll deal with the net torque and energy analysis first.

 

[Milchstrabe]

Time:
(vertical distance)
X = 1/2(g)(t^2)

Max height I want: 5 feet ( 1.52 m)

1.52 = .5(9.8)(t^2)

t = .55675 (to max height)

Time to Fall 2.5 feet to final height:

.76 = .5(9.8)(t^2)

t = .39382

Total flight time:

.39382 + .55675

.95 sec

Vertical Velocity:

V = at + Vv

0 = 9.8(.55675) + Vv

Vv = 5.45 m/s

Horizontal Velocity

Vh(t) = X

X = .76m

V(.95) = .76

Vh = .7995 m/s

Tanget Velocity:

a^2 + b ^2 = c^2

Vh^2 + Vv^2 = Vt^2

.7995^2 + 5.45^2= Vt^2

Vt = 5.5144 m/s

Kinetic Energy Required for a 5 gram hacky sack:

KE = 1/2mv^2

KE = .5(.005)5.514^2

KE = .076

Energy required to raise beam? hmm not sure yet... one min

 

[whozum]

Time:
(vertical distance)
X = 1/2(g)(t^2)

Max height I want: 5 feet ( 1.52 m)

1.52 = .5(9.8)(t^2)

t = .55675 (to max height)

The object has a nonzero initial velocity, you can't make this simplification.

Time to Fall 2.5 feet to final height:

.76 = .5(9.8)(t^2)

t = .39382

Cool.)

Total flight time:

.39382 + .55675

.95 sec

Vertical Velocity:

V = at + Vv

0 = 9.8(.55675) + Vv

Vv = 5.45 m/s

Horizontal Velocity

Vh(t) = X

X = .76m

V(.95) = .76

Vh = .7995 m/s

Tanget Velocity:

a^2 + b ^2 = c^2

Vh^2 + Vv^2 = Vt^2

.7995^2 + 5.45^2= Vt^2

Vt = 5.5144 m/s

Kinetic Energy Required for a 5 gram hacky sack:

KE = 1/2mv^2

KE = .5(.005)5.514^2

KE = .076

Energy required to raise beam? hmm not sure yet... one min

The initial calculation isn't correct, which messes up the rest.

Give the numbers for:
Max height
Height of landing zone
Distance of landing zone from seesaw

 

[Milchstrabe]

Oh you're right about the initial velocity, i don't know why i overlooked that, crap.

.76m away from catapult

Thereis no limit for max height, but to make the mass we're dropping less i'll just make it 5-6 feet.

.76m high

 

[Milchstrabe]

Actually I considered the initial velocity. First I calculated how high the object would need to go at the angle of 80 knowing the stool is .76 m high and .76 m away. I derived a nasty little equation to do this, then simplified it greatly:

arctan ( 2x + square Root(2x)square Root(2x-.76)) = Theta

with theta = 80
x(vertical height) = 1.52 m yup

V^2 = Vo^2 + 2(a)x

plugging in x a and knowing the final velocity for the maximum height would need to be 0, we get Vo = 5.45 m/s as vertical velocity. So it's the same.

 

[whozum]

Just to make sure were on the same page, is this right?

http://www.public.asu.edu/~hyousif/csaw.JPG

 

[whozum]

Actually I considered the initial velocity. First I calculated how high the object would need to go at the angle of 80 knowing the stool is .76 m high and .76 m away. I derived a nasty little equation to do this, then simplified it greatly:

arctan ( 2x + square Root(2x)square Root(2x-.76)) = Theta

with theta = 80
x(vertical height) = 1.52 m yup

V^2 = Vo^2 + 2(a)x

plugging in x a and knowing the final velocity for the maximum height would need to be 0, we get Vo = 5.45 m/s as vertical velocity. So it's the same.

We aren't starting at x = 0, but that equation can still work. Just let x be the vertical distance between the release point and the top of the trajectory.

 

[Milchstrabe]

Yeah I understand, I'm ignoring the height of the release arm when it releases just for the time being, I totally understand that i'll need to factor it in tho. And yes your diagram is right. What are you majoring in?

 

[Milchstrabe]

SO what's next, do i need to find the energy required to move the arm that far?

 

[whozum]

I'm a physics sophomore-junior at ASU. You?

The energy to move the seesaw is a function of the torque, which we don't know yet. We want to find the required v_0 to let the hackysack reach its target, ocne we find that, we'll have the energy required by the seesaw. If we have the energy, we can find the torque we need.

 

[Milchstrabe]

Well we have that, as long as we ignore the height of the see saw, it will make a difference, but i just want to try some calculations so i know how to do it. Above I hae the kinetic energy and Velocity required.

 

[whozum]

Ok then you know that KE = \Delta PE [/tex], and you know PE = \tau\theta[/tex]&lt;br /&gt; &lt;br /&gt; You know theta, so find the required \tau_{net}. Once you know that, then you&amp;#039;ll know the required \tau_{impulse} [/tex]

 

[Milchstrabe]

Now I have a question, is \tau_{obj} the torque of the object falling, after it hits? Because if it is, then I have to make up a mass for that object.

 

[Milchstrabe]

I'll be going to ASU next year and I'll be majoring in Chemical Engineering :D


KE = \frac {1}{2}mv^2

KE = \frac {1}{2}(.005)5.514^2

KE = .076

20 degree angle sweep = \frac {\pi}{9}

KE = \Delta PE

W = \tau\theta

.076 = \tau(\frac {\pi}{9})

\tau_{net} = .217724 < -----

 

[whozum]

Set the positive torque direction to be counter clockwise:

\tau_{net} = \tau_{grav} + \tau_{obj} + \tau_{imp}

\tau_{net} = m_{l}g\left(\frac{x_l}{2}\right) - m_{r}g\left(\frac{x_r}{2}\right)} + m_{obj}gx_{l} + \frac{m_{obj}v_{obj}}{t_{impact}}x_l

Put all your knowns on one side

\frac{\tau_{net} - \tau_{grav}}{x_l}} = m_{obj}g + \frac{m_{obj}v_{obj}}{t_{impact}}

Pick a mass, it doesn't really matter what mass you pick, since we can make up for it by letting \frac{v}{t} be larger. Then:

\frac{\tau_{net} - \tau_{grav}}{x_l}} - m_{obj}g= \frac{mv_{obj}}{t_{imp}}

The rigth side is an approximation to the instantaneous force the falling obj will exert on the seesaw, you can probably find a much better approximation by doing some rotational collision research but my estimate would be take t_{imp} to be 0.01s and find v.

This process assumes that the falling object will lose all of its velocity when it hits the seesaw which isn't true. I would try to find a better description of this interaction before proceeding, maybe your physics teacher can help you. We're trying to approximate the force of impact so we can find the impulse torque.

 

[Milchstrabe]

My physics teacher won't help because it is for the final project, I see where we run into trouble tho. I'm not sure what to do because this is going to get increasingly more and more complicated. hmm.. time of an impact, I'm not sure how that is defined, but i'll go with .01 s. I hope assuming the velocity drops to zero when it impacts won't change things a whole lot. Is there a way I can assume it stays the same? or just drops off a little

 

[whozum]

If you want to assume that all the velocity is lost, then if oyu set impact time to 0.01 you can just plug in the numbers for it and get a required velocity. You can then approximate the height by usign

v = \sqrt{2gh} and solving for h.

 

[whozum]

Searching this forum for 'angular impulse' I found a ton of people with similar questions to yours, I would recommend doing that and reading through to gain af ull understanding.
https://www.physicsforums.com/showthread.php?t=32417&highlight=angular+impulse"

 

[Milchstrabe]

Okay so I need to consider the net torque of gravity, and since it's clockwise we'll make it negative. Then we have to consider the torque of the hack sack sitting on the arm, and that is also clockwise, so we'll make it negative. Torque of impact is counter clockwise, so we'll make that positive, and torque of the object once it hits is also counter clockwise so we'll make it positive. Since the arm needs to rotate counter clockwise in order to launch the hacky sack, we'll make the torque required to launch it positive. So if we add the negative torques to the positive counter clockwise torques and we know how much positive torque we need, then we can solve for velocity, kinda like you said right? yeah... lol

so wouldn't it be:

\tau_{net} = \tau_{imp} + \tau_{object} - \tau_{hacky} - \tau_{netgrav}

Actually I guess it's the same i just made the negatives already included, and also included the torque of the hacky sack. nm

 

[whozum]

You pretty much summed up the last 30 posts in this thread :)

Keep in mind the assumptions were making here:
The object sticks to the seesaw on collision.
Impact time is relatively small, and linear momentum effects are negligable.
The seesaw itself doesn't absorb any energy from any kind of impact
Air resistance is negligable.
The target is large nough to catch the hackysack
The hackysack is completely rigid.

 

[Milchstrabe]

Yeah i realize that, the one I'm worried about is impact time, I mean i knew velocity had a role, but I didn't know there was impact torque, seems kinda weird. Because you're assuming that the velocity only has an effect for .01 sec. Hmm... And is the equation for impact torque right, I assume it is, but the more time it takes to impact, the smaller the torque becomes.

 

[whozum]

This is true. The force required to dissipate the energy will be much greater if you reduce the amount of time allowed to dissipate the energy. If you look around for things similar to this you'll find af ewo n this forum. there was a good one on domino's in another thread. You could always start a new thread and ask about the angular impulse, I am sure a lot of people are intimidated by 100post threads :)