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Milchstrabe
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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.
.Solving the Physics of a Seesaw Catapult
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The discussion revolves around the physics of a seesaw catapult, specifically how to calculate the necessary parameters for launching a hacky sack to a specific distance and height. The original poster seeks to determine the weight and height from which a mass must be dropped to achieve the required launch velocity, factoring in the seesaw's arm lengths and the effects of torque. Key points include the importance of calculating the force needed to bring the seesaw to equilibrium and the relationship between potential energy, kinetic energy, and torque in the system. Participants emphasize the need to consider the moment of inertia of the seesaw arms and the impact of gravity on the system's dynamics. Overall, the conversation highlights the complexity of the calculations involved in the project and the necessity of understanding fundamental physics concepts to successfully execute the experiment.
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Milchstrabe
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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 < -----
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 < -----
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whozum
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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.
\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
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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
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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.
v = \sqrt{2gh} and solving for h.
whozum
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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"
https://www.physicsforums.com/showthread.php?t=32417&highlight=angular+impulse"
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Milchstrabe
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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
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
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whozum
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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.
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
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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
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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 :)
Milchstrabe
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Ha i just realized the distance from the stool is 2.5 m the height is the same. So I'll probably need to do a 60-70 degree launch. Also as part of the project we need expert advice, and if it's okay with you, we may use you as one of our experts since you are majoring in physics. You wouldn't have to do anything, you've already done enough i'd just have to find a way to make this work since i'd need a signature. I'll talk to my teacher and see what I can do, I'll contact you when i find anything out.
whozum
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I feel so important. I'm very very far from an expert, an expert could atleast quantify the effect of angular impulse on this system. Did you ever solve for the value of v above?
Milchstrabe
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I'm working on it right this second, I had to change all my calculations to account for the REAL weight of the hacky (approx 50 grams) the new distance from the stool (2.5m) so I have an angle of about 77 now and I'll be back here soon to post my results for the V of the falling object.
Milchstrabe
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I'm getting VERY low velocities, maybe it's right, I'll see what you have to say.
The mass I may use will be 1 kg.
The Torque required to launch the hack (.05kg) at a rate of 7.85674 m/s is 11.0219 Nm
The pine wood I'm using is 381 grams for a 50 cm long piece.
.30504kg for longer board
.07626kg for shorter board
\tau_{grav} = \tau_{left} + \tau_{right}
\tau_{grav} = (.30504kg)(9.8)\frac{.4m}{2} - (.07626kg)(9.8)\frac{.1m}{2}
\tau_{netgrav} = .560511 Nm
Torque of Hacky Sack
\tau_{hackysack} = (.05)(9.8)(.4m)
\tau_{hackysack} = .196 Nm
Torque Required to launch hacky at 7.85674 m/s
KE = \frac {1}{2}m(v^2)
Ke = .5 (.05)(7.85674^2)
Ke = 1.54307
W = \tau\theta
1.54307 = \tau (.14\pi) (.14 pi is 25 degrees in radians, 25 degrees is the new sweep angle cause of the new launch angle)
\tau_{launch} = 11.0219
Total Clock Wise Torque (negative torque):
\tau_{clockwise} = -11.8157Nm
Counter Clockwise Torque (gravity on the board is included as a net torque above, so we don't need to worry about that)
\tau_{counter} = \tau_{obj} + \tau_{imp}
\tau_{counter} = (1kg)(9.8)(.1m) + \frac {(1kg)(V)}{.01}
11.8157985(clockwise torque) - .98(torque of obj) = \frac {V}{.01}
V = .108356 m/s
Does that seem small for a 1kg weight, considering everything?
The mass I may use will be 1 kg.
The Torque required to launch the hack (.05kg) at a rate of 7.85674 m/s is 11.0219 Nm
The pine wood I'm using is 381 grams for a 50 cm long piece.
.30504kg for longer board
.07626kg for shorter board
\tau_{grav} = \tau_{left} + \tau_{right}
\tau_{grav} = (.30504kg)(9.8)\frac{.4m}{2} - (.07626kg)(9.8)\frac{.1m}{2}
\tau_{netgrav} = .560511 Nm
Torque of Hacky Sack
\tau_{hackysack} = (.05)(9.8)(.4m)
\tau_{hackysack} = .196 Nm
Torque Required to launch hacky at 7.85674 m/s
KE = \frac {1}{2}m(v^2)
Ke = .5 (.05)(7.85674^2)
Ke = 1.54307
W = \tau\theta
1.54307 = \tau (.14\pi) (.14 pi is 25 degrees in radians, 25 degrees is the new sweep angle cause of the new launch angle)
\tau_{launch} = 11.0219
Total Clock Wise Torque (negative torque):
\tau_{clockwise} = -11.8157Nm
Counter Clockwise Torque (gravity on the board is included as a net torque above, so we don't need to worry about that)
\tau_{counter} = \tau_{obj} + \tau_{imp}
\tau_{counter} = (1kg)(9.8)(.1m) + \frac {(1kg)(V)}{.01}
11.8157985(clockwise torque) - .98(torque of obj) = \frac {V}{.01}
V = .108356 m/s
Does that seem small for a 1kg weight, considering everything?
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whozum
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I'm not sure if its too small, just try it and see. Notice though your dropping a really heavy object in comparison to your hackysack. It makes perfect sense that it won't need to be going very fast to launch an object 20x smaller with a 4x mechanical advantage. Actually come to think of it, target momentum is (mv)_{hs} = (0.05)(7.85) = 0.3925kgm/s [/tex], and impact momentum is (mv)_{kg} = (1)(.108) [/tex], mechanical advantage is 4x, so <br />
<br />
4(mv)_{kg} ~~ (mv)_{hs} which is almost true!<br />
<br />
\tau_{grav} = (.30504kg)(9.8)\frac{.4m}{2} - (.07626kg)(9.8)\frac{.1m}{2}<br />
<br />
Just one thing:<br />
Define a direction of positive torque. I&#039;m guessing from convention that counterclockwise (the long board rising) is positive, but here you say that the torque from gravity is positive, but its acting in a clockwise direction.<br />
<br />
Aside from that ,everything looks fantastic, I didnt check your math but I don&#039;t see a reason to.
Milchstrabe
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I say the torque of gravity is positive, but then when i add it to the other clockwise torques, i make it negative you see?
Anyways, height i'd have to drop the mass:
\Delta x = \frac {1}{2}a(t^2)
V = at + Vo
t = \frac {.108356m/s}{9.8m/s}
t = .011056 sec
\Delta x = \frac {1}{2}(9.8 m/s)(.011056^2)
\Delta x = .000599 m THhat is .05 cm high that i would need to drop the 1 kg... eh? something might be wrong.
IT doesn't make sense because if I re-calculate it with a mass falling of only 50 grams (same mass as the hacky sack) then it only needs a 2.3502 velocity to launch a 50 gram hacky at 7.85 m/s, lift an arm 4 times heavier than it, 20 degrees, i don't know. I think our clockwise torques are just fine based on basic definitions of torque etc... Unless our launch torque is wrong. I think something is more likely wrong with our counter clockwise torque forumulas...
Anyways, height i'd have to drop the mass:
\Delta x = \frac {1}{2}a(t^2)
V = at + Vo
t = \frac {.108356m/s}{9.8m/s}
t = .011056 sec
\Delta x = \frac {1}{2}(9.8 m/s)(.011056^2)
\Delta x = .000599 m THhat is .05 cm high that i would need to drop the 1 kg... eh? something might be wrong.
IT doesn't make sense because if I re-calculate it with a mass falling of only 50 grams (same mass as the hacky sack) then it only needs a 2.3502 velocity to launch a 50 gram hacky at 7.85 m/s, lift an arm 4 times heavier than it, 20 degrees, i don't know. I think our clockwise torques are just fine based on basic definitions of torque etc... Unless our launch torque is wrong. I think something is more likely wrong with our counter clockwise torque forumulas...
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whozum
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\tau_{counter} = (1kg)(9.8)(.1m) + \frac {(1kg)(V)}{.01}
Your impulse torque doesn't have units of torque. You forgot to add the .1m
11.8157985(clockwise torque) - .98(torque of obj) = \frac {V}{.01}
I get V = 1083 when I do this, but factoring in the .1 from above, Its still really high, 108.3. Are you sure you did your torque-work correct?
Two things I also noticed. The torque isn't constant while the angle is swept, since gravity isn't always perpendicular to the radius vector, do you want to take this into account? Also, when the KG hits the seesaw, its impulse torque won't be perpendicular to the seesaw, so you are losing some torque right there. Just a few thoughts.
Your impulse torque doesn't have units of torque. You forgot to add the .1m
11.8157985(clockwise torque) - .98(torque of obj) = \frac {V}{.01}
I get V = 1083 when I do this, but factoring in the .1 from above, Its still really high, 108.3. Are you sure you did your torque-work correct?
Two things I also noticed. The torque isn't constant while the angle is swept, since gravity isn't always perpendicular to the radius vector, do you want to take this into account? Also, when the KG hits the seesaw, its impulse torque won't be perpendicular to the seesaw, so you are losing some torque right there. Just a few thoughts.
Milchstrabe
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Are you sure about those numbers? watch the decimal places maybe? Because (11.8 - .98 )(.01) = V doesn't get you that big of a number. I won't take those things into account.
whozum
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Oh right my bad, I can't multiply.
Milchstrabe
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lol okay so if we both agree that hte number is too small, then what the heck is going on ::bangs head on desk::
whozum
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\tau_{clock} - \tau_{obj} = \tau_{imp}
10.8 = \frac{mv}{t} x
\frac{10.8t}{xm} = v
v = 1.08m/s
Try that. Sounds more reasonable.
I get 6cm.
10.8 = \frac{mv}{t} x
\frac{10.8t}{xm} = v
v = 1.08m/s
Try that. Sounds more reasonable.
I get 6cm.
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Milchstrabe
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That sounds more reasonable, 6 cm drop height.. approx: 2.5 inches. Maybe...
whozum
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The biggest source of error is assuming the torque angles are perpendicular. You're probably going to need a larger impact velocity to get it to work perfect.
Milchstrabe
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I'll cut the wood tomorrow and set it up and run some experiments to see what happens, and if it's not a whole lot off (meaning i don't have to adjust the drop height more then 3 cm or so, i'll do an error analysis, if not, i may factor it it in. However if i have to adjust by 10 cm + then we probably have a problem.
whozum
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Sounds good, have fun with it. Dont break the wood :)
Milchstrabe
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I tried setting something similar to this last night with the wood at the exact length and the ratio's were pretty close. The height of the fulcrum was a little less than it should be but I was just using my hand impacting the wood as the force. I found it EXTREMELY difficult to get much height out of the hacky sack, not even a meter. I was hitting it pretty hard too. Is there anyone you can verify that we did this right with?
Milchstrabe
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Yeah, I know it won't work with a kg of mass from that small of a height, using my hand, with correct rations and everything, and a lip to keep the hacky sack from releasing early, I still can't get it to go 2.5 m horizontally. The best I can get hitting it very hard is about 1 m.
Should I start a new thread in the advance physics forum and see what we can do?
Should I start a new thread in the advance physics forum and see what we can do?
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whozum
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Wouldnt hurt, but try to make some error analysis to pinpoint the problem..
whozum
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I wouldnt' be surprised if I missed something, but if we had a way to quantify the discrepancy it would be really helpful.
Milchstrabe
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I posted on the college forum just to see if we can see what others have to say... if anything lol I have AP test next week, all kinds of things to do for this physics project, this is the most important part at least. THe rest the calculations i'll have to do are all basic, so I'm good. Refresh me tho, I just hada brain fart, how do you solve for the acceleration of a ball rolling down a ramp knowing the angle of course.
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