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One for the Weekend - Pointless Problem (Just for fun) - solution Watch

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    It seems very quiet on here today, so if you are feeling bored or brave, try to find the answer to this one.



    1. What is the value of the angle theta that will cause the mass to land just at the point X?
    X and Y are the same distance from O.

    More difficult question.
    2. What value would theta need to be for the mass to be projected the furthest possible distance from O?

    Any method allowed; algebraic or numeric, programmed or spreadsheet.
    If you decide have a go, just post your answers please, not the complete solution.
    I'll give the correct answers on Sunday evening.

    If you need it, the value of g=9.8 ms-2
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    (1) \theta\approx 100^{\circ}

    (2) \theta\approx 134^{\circ}

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    Well I think this problem was a bit too difficult.
    I didn't recieve any correct replies either here or via PM.
    Next weekend's will be simpler!
    Here's the solution if you are interested.
    The algebraic method is far too complicated and it's much easier to solve this by numerical methods, either using a spreadsheet or a computer programme.
    I used a spreadsheet method.
    The formulas are entered into the sheet and a series of values for the angle theta (from 90 to 180) are entered (column 1), with the value of the range of the projectile as the final calculated value. (Column 2). The values of some of the other variables are also there to make the spreadsheet easier to set up and to enable you to check the solution more easily.
    It's then fairly simple to see where the target value of range lies and to enter more exact values for theta to match the desired answer.
    The numerical value for h was 4 and for r was 1; giving the 4 to 1 ratio.
    Any pair with a ratio of 4 would give the same final answer. I put g=10 m/s/s but the answer does not depend on g so you can use any value you want.

    Using the "Goal seek" function is also a way of doing this in Excel.

    Solution equations here.

    The value of V, the velocity of the projectile at P is found from energy considerations as there is no friction. KE = loss of PE.
    The angle for V is found from the geometry of the circle. This also gives the value of y, the height of P above the axis.
    From this, the projectile motion gives the range R, to which is added the values shown to give the total range from O to X



    Spreadsheet solution and results
    Theta is 133.96 degrees for maximum range (9.4m). (Here h=4m and r=1m)
    Theta is 171.28 degrees for the mass to just reach X. (X is at 4m from O)


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    (Original post by Stonebridge)
    Well I think this problem was a bit too difficult.
    I didn't recieve any correct replies either here or via PM.

    Spreadsheet solution and results
    Theta is 133.96 degrees for maximum range (9.4m). (Here h=4m and r=1m)
    Theta is 171.28 degrees for the mass to just reach X. (X is at 4m from O)
    Aren't there two values of theta for the mass to just reach X? You can see from your table that for theta=100 OX=3.98, and for theta=110 OX=6.57. So the other value of theta would be somewhere between 100 and 110, probably very, very close to 100.

    Is not theta=134 for the maximum value of OX a good answer?
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    (Original post by jaroc)
    Aren't there two values of theta for the mass to just reach X? You can see from your table that for theta=100 OX=3.98, and for theta=110 OX=6.57. So the other value of theta would be somewhere between 100 and 110, probably very, very close to 100.

    Is not theta=134 for the maximum value of OX a good answer?
    My apologies, jaroc.
    Yes your answers are fine. Excellent. The only correct ones received.
    There are indeed two values for the maximum range, I only quoted the second one. You have found the first one.

    What happened was I looked at your answers before I worked out the solution myself, and then forgot to re-check your answers after I had done it.

    Did you use an algebraic method? If so you are braver than I am. I decided it was easier and quicker to solve it numerically. It's my preferred method these days as I believe the technology should be put to good use.

    Again. Well done, and sorry I misread your answers.
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    (Original post by Stonebridge)
    My apologies, jaroc.
    Yes your answers are fine. Excellent.
    There are indeed two values for the maximim range, I only quoted the second one. You have found the first one.

    What happend was I looked at your answers before I worked out the solution myself, and then forgot to re-check your answers after I had done it.
    You also could have looked at the answers I posted on Friday, which were incorrect (I edited my post on Saturday)


    Did you use an algebraic method? If so you are braver than I am. I decided it was easier and quicker to solve it numerically. It's my preferred method these days as I believe the technology should be put to good use.

    Again. Well done, and sorry I misread your answers.
    Thank you So now I'm relieved. That was a really nice problem.

    Well, it seems that I used a bit more algebra in my solution than you did, but the equation I obtained was far too complex for me to solve it or find its derivative without a little help of a computer

    I'm posting my solution below, have a look if you wish.

    Spoiler:
    Show
    The equations of motion of a particle in this problem are

    x=x_0+v_{0_x}t
    0=y_0+v_{0_y}t-\tfrac{1}{2}gt^2.

    I solved the second equation for time t:

    t=\dfrac{v_{0_y}\pm\sqrt{v_{0_y}  ^2+2gy_0}}{g}.

    Of course t>0 so there is only one physical solution of the quadratic.

    I then substituted this for time in the first equation:

    x=x_0+\dfrac{v_{0_x}}{g}\left(v_  {0_y}+\sqrt{v_{0_y}^2+2gy_0} \right).

    I defined \varphi\equiv\theta-90^{\circ}.

    Then

    x_0=r\left(1+\sin\varphi \right), y_0=r\left(1-\cos\varphi \right), v_0^2=2g\left(h-y_0 \right)=2g\left(4r-r\left(1-\cos\varphi\right)\right) = 2gr \left( 3+\cos\varphi \right), v_{0_x}=v_0\cos\varphi, v_{0_y}=v_0\sin\varphi.

    I put this all into the equation. g cancelled out, so did r when I put x\equiv kr. Then I obtained the following function of \varphi:

    k=1+\sin\varphi+2\left(3+\cos \varphi \right) \sin \varphi \cos\varphi+2\sqrt{\left( 3+\cos\varphi \right)^2 \sin^2 \varphi \cos^2 \varphi +\left( 1-\cos\varphi \right) \cos^2 \varphi}.

    I then used http://wims.unice.fr/wims/en_tool~analysis~function.html to find the extremum of the function and solve the equation for k=4. I guess your way was a bit faster, and surely less prone to possible mistakes. On the other hand I chose my way because I thought it would be a bit more exact (though probably it is negligible in this case).

    Interesting thing is that I obtained the maximum value of k=9.206 which is different to your result. Also my theta for that is \theta=133.82^{\circ}, not 133.96^{\circ}, similarly there are slight differences for the values I obtained for k=4. I guess it's due to the computational errors?
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    (Original post by jaroc)
    You also could have looked at the answers I posted on Friday, which were incorrect (I edited my post on Saturday)




    Thank you So now I'm relieved. That was a really nice problem.

    Well, it seems that I used a bit more algebra in my solution than you did, but the equation I obtained was far too complex for me to solve it or find its derivative without a little help of a computer

    I'm posting my solution below, have a look if you wish.

    Spoiler:
    Show
    The equations of motion of a particle in this problem are

    x=x_0+v_{0_x}t
    0=y_0+v_{0_y}t-\tfrac{1}{2}gt^2.

    I solved the second equation for time t:

    t=\dfrac{v_{0_y}\pm\sqrt{v_{0_y}  ^2+2gy_0}}{g}.

    Of course t>0 so there is only one physical solution of the quadratic.

    I then substituted this for time in the first equation:

    x=x_0+\dfrac{v_{0_x}}{g}\left(v_  {0_y}+\sqrt{v_{0_y}^2+2gy_0} \right).

    I defined \varphi\equiv\theta-90^{\circ}.

    Then

    x_0=r\left(1+\sin\varphi \right), y_0=r\left(1-\cos\varphi \right), v_0^2=2g\left(h-y_0 \right)=2g\left(4r-r\left(1-\cos\varphi\right)\right) = 2gr \left( 3+\cos\varphi \right), v_{0_x}=v_0\cos\varphi, v_{0_y}=v_0\sin\varphi.

    I put this all into the equation. g cancelled out, so did r when I put x\equiv kr. Then I obtained the following function of \varphi:

    k=1+\sin\varphi+2\left(3+\cos \varphi \right) \sin \varphi \cos\varphi+2\sqrt{\left( 3+\cos\varphi \right)^2 \sin^2 \varphi \cos^2 \varphi +\left( 1-\cos\varphi \right) \cos^2 \varphi}.

    I then used http://wims.unice.fr/wims/en_tool~analysis~function.html to find the extremum of the function and solve the equation for k=4. I guess your way was a bit faster, and surely less prone to possible mistakes. On the other hand I chose my way because I thought it would be a bit more exact (though probably it is negligible in this case).

    Interesting thing is that I obtained the maximum value of k=9.206 which is different to your result. Also my theta for that is \theta=133.82^{\circ}, not 133.96^{\circ}, similarly there are slight differences for the values I obtained for k=4. I guess it's due to the computational errors?
    Yes there will be rounding errors in the computations depending on how many significant figures you compute to. My method uses a lot of intermediate calculations so any rounding errors will be propagated. I worked to 9 decimal places. However, we do get the same results.
    It's also interesting to note that the angle for maximum range is not 45 degrees (135 degrees) which is the case for normal projectiles. But it is very close.
    I'll take a look at that site. I've not seen it before.
    Thanks for taking part and showing an interest.
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    (Original post by Stonebridge)
    Yes there will be rounding errors in the computations depending on how many significant figures you compute to. My method uses a lot of intermediate calculations so any rounding errors will be propagated. I worked to 9 decimal places. However, we do get the same results.
    It's also interesting to note that the angle for maximum range is not 45 degrees (135 degrees) which is the case for normal projectiles. But it is very close.
    I'll take a look at that sight. I've not seen it before.
    Thanks for taking part and showing an interest.
    Very interesting problem, what is the reason for the optimum angle not being 45 degrees like for a normal projectile? I really can't think of any physical explanation. Please post another next weekend!
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    (Original post by LeeC)
    Very interesting problem, what is the reason for the optimum angle not being 45 degrees like for a normal projectile? I really can't think of any physical explanation. Please post another next weekend!
    The reason is that the 45 (135) degree angle applies to a "constant" (non variable) projectile velocity.
    By that I mean that in this case the velocity of the projectile itself depends on the angle of projection. (And of course, the range also depends on the velocity of projection) This complicates the maths. It's interesting that the angle turns out to be very close to 45, though. I tried the solution with other starting values for h and r and the angle always seemed to come close to 45 but not equal to it.
    Because the velocity of the projectile depends on the potential energy lost in the fall, reducing the angle slightly from 45 lowers the value of y and means that the velocity of projection (and range) is slightly higher. Reducing the angle also means reducing the range, however. Clearly the maths is such that for just one or two degrees reduction below 45, the effect of the increased velocity outweighs the effect of changing the angle.

    I'll try to find another for next weekend but I think this one was (or looked!) too difficult for most of the students here who are at A or GCSE level.
    I'll post a simpler one - maybe as an alternative.

    Thanks for taking an interest.
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    (Original post by Stonebridge)
    The reason is that the 45 (135) degree angle applies to a "constant" (non variable) projectile velocity.
    By that I mean that in this case the velocity of the projectile itself depends on the angle of projection. (And of course, the range also depends on the velocity of projection) This complicates the maths. It's interesting that the angle turns out to be very close to 45, though. I tried the solution with other starting values for h and r and the angle always seemed to come close to 45 but not equal to it.
    Because the velocity of the projectile depends on the potential energy lost in the fall, reducing the angle slightly from 45 lowers the value of y and means that the velocity of projection (and range) is slightly higher. Reducing the angle also means reducing the range, however. Clearly the maths is such that for just one or two degrees reduction below 45, the effect of the increased velocity outweighs the effect of changing the angle.

    I'll try to find another for next weekend but I think this one was (or looked!) too difficult for most of the students here who are at A or GCSE level.
    I'll post a simpler one - maybe as an alternative.

    Thanks for taking an interest.
    I see! Will plot this for myself using Fortran later I think. I'm a 4th year engineering student but a bit of mechanics is always fun!
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    (Original post by LeeC)
    I see! Will plot this for myself using Fortran later I think. I'm a 4th year engineering student but a bit of mechanics is always fun!
    It will work well with a simple fortran programme.
    As jaroc and I got slightly different computed values it would be interesting to know what you get and how this depends on the number significant figures the arithmetic is working at. I think the Excel spreadsheet worked to a precision of 10 but I'm not sure if that's just for the maximum displayed, or is the actual maximum in the arithmetic.
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    (Original post by Stonebridge)
    It will work well with a simple fortran programme.
    As jaroc and I got slightly different computed values it would be interesting to know what you get and how this depends on the number significant figures the arithmetic is working at. I think the Excel spreadsheet worked to a precision of 10 but I'm not sure if that's just for the maximum displayed, or is the actual maximum in the arithmetic.
    Well, for angle theta we had a difference in the third significant digit, for maximum range - in the second digit. It's quite significant, isn't it? Our ranges differ by as much as 2% - I think for such calculations it's not negligible, is it? I would have expected the results to be far more consistent. Therefore I also am very interested in finding out what result you get using another method.

    And I also am looking forward to the next weekend's problem!
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    (Original post by jaroc)
    Well, for angle theta we had a difference in the third significant digit, for maximum range - in the second digit. It's quite significant, isn't it? Our ranges differ by as much as 2% - I think for such calculations it's not negligible, is it? I would have expected the results to be far more consistent. Therefore I also am very interested in finding out what result you get using another method.

    And I also am looking forward to the next weekend's problem!
    The difference for the range does seem large.

    I'll have another look at the spreadsheet to see if I can pinpoint any places where rounding errors may have been introduced.
    I did initially round down to 8 figures for intermediate results but later removed this. It's possible I missed something.

    What is the precision of the arithmetic in the calculations on the website you used?
    Also, when you specify a limit, maximum, minimum or inflection point, what is the tolerance they use to find it?
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    How do you get so good at Physics im jealouss it was too hard lol
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    (Original post by Stonebridge)
    The difference for the range does seem large.

    What is the precision of the arithmetic in the calculations on the website you used?
    Also, when you specify a limit, maximum, minimum or inflection point, what is the tolerance they use to find it?
    To be honest, I have no idea. After a quick search I found that for finding extremes and roots of functions it uses GP/PARI Software which I don't know at all (maybe you do?)

    Here's a Wikipedia article on it: http://en.wikipedia.org/wiki/PARI/GP. It says
    PARI/GP performs arbitrary precision calculations (e.g., the significand can be millions of digits long—and billions of digits on 64-bit machines).
    I'm afraid I cannot be of more help in that matter, I don't even know where to look for relevant information on the website. There is a 'technical documentation' page but it does not open.

    EDIT: Well, when I use it to calculate something it asks me to choose "numerical precision" - is that what we are looking for? I chose "12 digits".
 
 
 
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