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Math Help - Year-Long Physics Project

  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    I forgot I didn't explain my project very well, here is my experimental design:

    Defining Question: By how much will a ball's speed change when its size, weight, and how hot it is outside changes?

    Hypotheses: The size of the ball will be the greatest factor in deciding the average acceleration of air resistance. The second important factor is the density of the ball. Finally, the least affecting factor will be the temperature of the air.

    Independent Variables:
    The Density of the ball: 5-10 levels per ball size, control will be unaltered density (I won't fill the control with weight)

    The Outside Temperature: 4-6 levels, control to be decided

    The Size of the Ball: 4-5 levels, Control will be the middle sized balls.

    Dependent Variable:
    The average acceleration of air resistance

    Number of Trials:
    4 trials per density

    Constants:
    The Starting Velocity of the ball.
    The Angle the ball is released.
    The Ground Level
    The Starting Location

    Procedure:
    1. Set up the trebuchet in a level area
    2. Place the first ball in the trebuchet
    3. Launch, makeing sure to measure the starting velocity of the ball
    4. Time how long it takes for the ball to hit the ground
    5. Measure the distance between the landing point and the point of release
    6. Discover the average acceleration for that time (using equation a=2(d/t)-2v, where d is the change in distance in meters, v is the starting velocity in meters per second, t is the time in seconds, and a is the average acceleration of the ball)
    7. Repeat until you have 4 tests of that ball
    8. Use different density balls while repeating steps 3-7 until all densities have been used.
    9. Change the ball size and repeat 2-8
    10. Repeat step 9 until all ball sizes have been used
    11. Repeat steps 1-10 when the temperature changes
    12. Repeat step 11 until sufficient data is collected
    Sounds reasonable. You'll be VERY busy taking data, but as it's a "year long project"...

    I object to the equation in step 6:

    Quote Originally Posted by Quick View Post
    6. Discover the average acceleration for that time (using equation a=2(d/t)-2v, where d is the change in distance in meters, v is the starting velocity in meters per second, t is the time in seconds, and a is the average acceleration of the ball)
    Where did you get this from? Among other problems I can think of, the units on the RHS don't match those on the LHS....

    d = vt + (1/2)at^2

    a = \frac{2d - 2vt}{t^2} = (2d)/t^2 - 2v/t

    The other problem you have is that this is NOT the equation for average acceleration. The v in the equation needs to be the horizontal component of the initial velocity. Then this equation is for the horizontal component of the average acceleration. I showed you how to find the vertical component of the average velocity in a previous post.

    -Dan
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  2. #17
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    Where did you get this from? Among other problems I can think of, the units on the RHS don't match those on the LHS....

    d = vt + (1/2)at^2

    a = \frac{2d - 2vt}{t^2} = (2d)/t^2 - 2v/t

    The other problem you have is that this is NOT the equation for average acceleration. The v in the equation needs to be the horizontal component of the initial velocity. Then this equation is for the horizontal component of the average acceleration. I showed you how to find the vertical component of the average velocity in a previous post.

    -Dan
    Yes I know it's supposed to be horizontal, I decided that the vertical air resistance would be nice to have, but I wouldn't be able to tell the max height of the curve (because equations are all theoretical without air resistance).

    And I manipulated the equation to the one above, unfortunately I forgot that the time was squared
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  3. #18
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    2) Acceleration divided by velocity gives units of 1/s, not density!
    Yeah, I know. The only problem is that I need a way to compare them accurately even if the starting velocity is different by a 100th or so
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  4. #19
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Quick View Post
    Yeah, I know. The only problem is that I need a way to compare them accurately even if the starting velocity is different by a 100th or so
    I have an idea, tell me if this is wrong (I could imagine it is), the problem I'm worried about is that acceleration increases with the square of velocity...

    note: subscript 1 means "of the first object" and v means starting velocity

    You have 1 object with v_1=10 and a_1=5

    You have the second object with v_2=20 and a_2=16

    Since I can't compare the two accelerations until they have the same starting velocity, I'll divide both v_2 and a_2 by k such that \frac{v_2}{k}=v_1 and then compare the accelerations.

    If that doesn't work then is there a way to compare them?

    PS: the starting velocities will be much, much closer in real life.
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  5. #20
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    I have an idea, tell me if this is wrong (I could imagine it is), the problem I'm worried about is that acceleration increases with the square of velocity...

    note: subscript 1 means "of the first object" and v means starting velocity

    You have 1 object with v_1=10 and a_1=5

    You have the second object with v_2=20 and a_2=16

    Since I can't compare the two accelerations until they have the same starting velocity, I'll divide both v_2 and a_2 by k such that \frac{v_2}{k}=v_1 and then compare the accelerations.

    If that doesn't work then is there a way to compare them?

    PS: the starting velocities will be much, much closer in real life.
    Hmmmm.... I see what you are trying to get at now.

    The problem is not so much that the initial velocities are different, but that we don't know the velocity curve. This means we don't know how to compare the averages. (It will not, unfortunately, be as simple as a linear transformation like you suggest.) What I am not able to say is if you procedure is "good enough" for a decent approximation.

    If we assume a linear resistance term, I can calculate it using Calculus:
    a = - \frac{1}{m}cv where v = v_0e^{-ct/m}
    (This is first the equation of motion and second the solution to the equation.)

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-cv_0}{m}e^{-ct/m} \,dt, where T is the time of flight.

    \bar{a} = \frac{-cv_0}{mT} \cdot \left ( - \frac{m}{c} \right ) \left ( e^{-cT/m} - 1 \right )

    \bar{a} = -\frac{v_0}{T} \left ( 1 - e^{-cT/m} \right )

    For the quadratic resistance case:
    a = -\frac{1}{m}cv^2 for v = \frac{mv_0}{v_0ct + m}

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-c}{m} \left ( \frac{mv_0}{v_0ct + m} \right )^2 \, dt

    \bar{a} = - \frac{cmv_0^2}{T} \int_0^T \frac{1}{(v_0ct + m)^2} \, dt

    \bar{a} = - \frac{cv_0^2}{cv_0T+m}

    As you can see in both of these equations, simply scaling v0 is not going to be enough because of the T dependence. You'll just have to get the numbers and see how they turn out.

    If you wish to use these equations, the velocity solutions are from "Analytical Mechanics" 4th ed. by Fowles, pg. 49. You can say I assisted you in finding the equations for the average acceleration.

    -Dan
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  6. #21
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    Hmmmm.... I see what you are trying to get at now.

    The problem is not so much that the initial velocities are different, but that we don't know the velocity curve. This means we don't know how to compare the averages. (It will not, unfortunately, be as simple as a linear transformation like you suggest.) What I am not able to say is if you procedure is "good enough" for a decent approximation.

    If we assume a linear resistance term, I can calculate it using Calculus:
    a = - \frac{1}{m}cv where v = v_0e^{-ct/m}
    (This is first the equation of motion and second the solution to the equation.)

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-cv_0}{m}e^{-ct/m} \,dt, where T is the time of flight.

    \bar{a} = \frac{-cv_0}{mT} \cdot \left ( - \frac{m}{c} \right ) \left ( e^{-cT/m} - 1 \right )

    \bar{a} = -\frac{v_0}{T} \left ( 1 - e^{-cT/m} \right )

    For the quadratic resistance case:
    a = -\frac{1}{m}cv^2 for v = \frac{mv_0}{v_0ct + m}

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-c}{m} \left ( \frac{mv_0}{v_0ct + m} \right )^2 \, dt

    \bar{a} = - \frac{cmv_0^2}{T} \int_0^T \frac{1}{(v_0ct + m)^2} \, dt

    \bar{a} = - \frac{cv_0^2}{cv_0T+m}

    As you can see in both of these equations, simply scaling v0 is not going to be enough because of the T dependence. You'll just have to get the numbers and see how they turn out.

    If you wish to use these equations, the velocity solutions are from "Analytical Mechanics" 4th ed. by Fowles, pg. 49. You can say I assisted you in finding the equations for the average acceleration.

    -Dan
    What does \bar{a} mean, average? So you're saying there is no way to compare them? I can handle this by getting rid of 1-3 decimal places to make them have the same starting velocity but it is frustrating.

    Thanx for your help, I'll have more questions when I start to collect the data.
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  7. #22
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    What does \bar{a} mean, average? So you're saying there is no way to compare them? I can handle this by getting rid of 1-3 decimal places to make them have the same starting velocity but it is frustrating.

    Thanx for your help, I'll have more questions when I start to collect the data.
    Oh, sorry! Yes, \bar{a} means "average." I don't know that your procedure will or won't work. (This non-linear stuff is hard to make predictions with.) All I can say is take your data and see if it will work. The problem is that there is no way to find an explicit value for T, all that can be done is to set up the problem and find it numerically. The best thing for you to do would be to measure it since the process involves some nastily tedious Calculus. (I used to have a program that would do it, but I lost it when my computer crashed. I really need to back up more often!)

    I will say that if the standard error in your initial velocity is small enough the data points should "cluster" about the same value of T. (ie. you will have a small standard error in T also.) Or you could use the range and look at the same thing. If this doesn't happen then you have problems. So maybe you should do a few trials early on to see what the data will do.

    -Dan
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  8. #23
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    Hmmmm.... I see what you are trying to get at now.

    The problem is not so much that the initial velocities are different, but that we don't know the velocity curve. This means we don't know how to compare the averages. (It will not, unfortunately, be as simple as a linear transformation like you suggest.) What I am not able to say is if you procedure is "good enough" for a decent approximation.

    If we assume a linear resistance term, I can calculate it using Calculus:
    a = - \frac{1}{m}cv where v = v_0e^{-ct/m}
    (This is first the equation of motion and second the solution to the equation.)

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-cv_0}{m}e^{-ct/m} \,dt, where T is the time of flight.

    \bar{a} = \frac{-cv_0}{mT} \cdot \left ( - \frac{m}{c} \right ) \left ( e^{-cT/m} - 1 \right )

    \bar{a} = -\frac{v_0}{T} \left ( 1 - e^{-cT/m} \right )

    For the quadratic resistance case:
    a = -\frac{1}{m}cv^2 for v = \frac{mv_0}{v_0ct + m}

    So
    \bar{a} = \frac{1}{T} \int_0^T \frac{-c}{m} \left ( \frac{mv_0}{v_0ct + m} \right )^2 \, dt

    \bar{a} = - \frac{cmv_0^2}{T} \int_0^T \frac{1}{(v_0ct + m)^2} \, dt

    \bar{a} = - \frac{cv_0^2}{cv_0T+m}

    As you can see in both of these equations, simply scaling v0 is not going to be enough because of the T dependence. You'll just have to get the numbers and see how they turn out.

    If you wish to use these equations, the velocity solutions are from "Analytical Mechanics" 4th ed. by Fowles, pg. 49. You can say I assisted you in finding the equations for the average acceleration.

    -Dan
    what is c, m, and lower-case t represent?
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  9. #24
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    what is c, m, and lower-case t represent?
    c: coefficient of resistance (either linear or quadratic)

    m: mass of the object

    t: time

    I used "T" for the total time of flight over the range, which I presume is the time you are going to measure.

    -Dan
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  10. #25
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    c: coefficient of resistance (either linear or quadratic)

    m: mass of the object

    t: time

    I used "T" for the total time of flight over the range, which I presume is the time you are going to measure.

    -Dan
    what's the difference between T and t? t measures time of what?
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  11. #26
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    what's the difference between T and t? t measures time of what?
    Sorry, I guess I'm not being clear enough.

    T is the time it takes for your projectile to go from the launch point to hit the ground. It is a constant.

    t is an intermediate time, a parameter of the motion. It's just the usual time variable.

    -Dan
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  12. #27
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    Sorry, I guess I'm not being clear enough.

    T is the time it takes for your projectile to go from the launch point to hit the ground. It is a constant.

    t is an intermediate time, a parameter of the motion. It's just the usual time variable.

    -Dan
    so t is just the seconds variable? Ok, that clears things up
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  13. #28
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    Quote Originally Posted by topsquark View Post

    If we assume a linear resistance term, I can calculate it using Calculus:
    a = - \frac{1}{m}cv where v = v_0e^{-ct/m}
    (This is first the equation of motion and second the solution to the equation.)
    (Just for information - I know its unsuitable for pedagogical purposes)

    Last time I did the calculations for these type of experiments I concluded
    that the high Reynold's number regime applied and so you should expect to
    find drag proportional to |v|^2.

    RonL
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  14. #29
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    design change?

    hmmm, I got an interesting suggestion from my teacher on how to do the project without using a catapult or just plain dropping it.

    He suggested attaching the ball to a string (like a pendelum) and releasing it at a certain degree and then find the air resistance from how long it takes to get to a certain degree above the base (because if I waited for the ball to stop I would run into human error) and then record the height and time.

    How would I calculate air resistance? I don't need you to type the whole thing if you don't want, but I do need help seeing the general idea.

    My solution: All I could think of was that the highest point the ball reached on each side per swing and the time it takes to get there, along with the times to the base, are the things I need to record.
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  15. #30
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    Quote Originally Posted by Quick View Post
    hmmm, I got an interesting suggestion from my teacher on how to do the project without using a catapult or just plain dropping it.

    He suggested attaching the ball to a string (like a pendelum) and releasing it at a certain degree and then find the air resistance from how long it takes to get to a certain degree above the base (because if I waited for the ball to stop I would run into human error) and then record the height and time.

    How would I calculate air resistance? I don't need you to type the whole thing if you don't want, but I do need help seeing the general idea.

    My solution: All I could think of was that the highest point the ball reached on each side per swing and the time it takes to get there, along with the times to the base, are the things I need to record.
    Quick I hope you select the pendulum. The expert(s) here on applied math can provide the differencial equations that represent the motion of the pendulum. And solve them! Then you can present that to your stupid female physics teacher who does not know what a hyperbola is and scare her. I hope you starts getting afraid of you.
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