.Solving the Physics of a Seesaw Catapult

[whozum]

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

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

[tex]\tau_{net} = \tau_{grav} - \tau_{obj} - \tau_{impulse} [/itex]<br /> <br /> The impulse torque will be the torque from the object'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?[/tex]

 

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

 

[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 [itex]KE = \Delta PE [/tex], and you know [itex]PE = \tau\theta[/tex]<br /> <br /> You know theta, so find the required $$\tau_{net}$$. Once you know that, then you'll know the required [itex]\tau_{impulse} [/tex][/itex][/itex][/itex]