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Math Help - Need help with confusing proportion problem

  1. #1
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    Exclamation Need help with confusing proportion problem

    The thrust, T, delivered by a ship's propeller is proportional to the square of the propeller rotation speed, R, times the fourth power of the propeller diameter, D.

    a) What happens to the thrust if the propeller speed is doubled?
    b) What happens to the thrust if the propeller diameter is doubled?
    c) If the propeller diameter is increased by 50%, by how much can the propeller speed be reduced to deliver the same thrust?

    ================================================== =======

    I have found out the equation to this to be "T = k*R^2*D^4.

    **EDIT**
    I think I found out how to do a) and b). Since they are proportional I can just set it to T = R^2 and T = D^4 and then plug in 2 for R and D. The answer should be 4 and 16 times greater.


    I still need help with c) What would we need to do to find how much the propeller speed be reduced to deliver the same thrust?
    Last edited by florx; April 26th 2010 at 09:45 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    The thrust, T, delivered by a ship's propeller is proportional to the square of the propeller rotation speed, R, times the fourth power of the propeller diameter, D.

    a) What happens to the thrust if the propeller speed is doubled?
    b) What happens to the thrust if the propeller diameter is doubled?
    c) If the propeller diameter is increased by 50%, by how much can the propeller speed be reduced to deliver the same thrust?

    ================================================== =======

    I have found out the equation to this to be "T = k*R^2*D^4.

    What I don't understand is how to answer the above questions. If it is proportional then wouldn't the answer to a) be increase by a factor of 4? And the answer to b) be increased by a factor of 8? Also it would be greatly appreciated if you could help me with c).
    Proportional does not mean being increased by a factor of 4

    The equation you have obtained is correct.

    T = k*R^2*D^4

    where: R is the rotation speed and D is the diameter.

    Now when the speed is doubled , your rotation speed = 2R

    and

    T = k \times (2R)^2 \times D^4 = k \times 4R^2 \times D^4

    Note that D is not changed in this case.

    so the ratio of new T to the old T is:


    \frac{k \times 4R^2 \times D^4}{k \times R^2 \times D^4}

    =4

    So your T becomes 4 times as much as it was before..

    Do the same for (b)
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    The thrust, T, delivered by a ship's propeller is proportional to the square of the propeller rotation speed, R, times the fourth power of the propeller diameter, D.

    a) What happens to the thrust if the propeller speed is doubled?
    b) What happens to the thrust if the propeller diameter is doubled?
    c) If the propeller diameter is increased by 50%, by how much can the propeller speed be reduced to deliver the same thrust?

    ================================================== =======

    I have found out the equation to this to be "T = k*R^2*D^4.

    **EDIT**
    I think I found out how to do a) and b). Since they are proportional I can just set it to T = R^2 and T = D^4 and then plug in 2 for R and D. The answer should be 4 and 16 times greater.


    I still need help with c) What would we need to do to find how much the propeller speed be reduced to deliver the same thrust?
    for part(c)

    after the diameter is increased by 50%, the new diameter is (D+50% of D) = D+ \frac{50}{100} D = \frac{3D}{2}

    \therefore T = k \times R^2 \times {(\frac{3D}{2})^4}...(I)

    now , let the speed be reduced by x%. then the new speed is (R-x% of R) = R - \frac{x}{100} R = \frac{100R - xR}{100} = \frac{R(100-x)}{100}

    so T = k \times {(\frac{R(100-x)}{100})^2} \times D^4..(II)

    Now equate (I) and (II), and find x and convert it into percentage
    Last edited by harish21; April 26th 2010 at 10:17 PM.
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  4. #4
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    Thank you for your help so far.

    Just to clarify when you say equate I and II does it mean I = II and solve for x? Or is it stacking I and II and then solving for the variable x? Could you please start me off on how to solve this problem.

    Thank you so much
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Thank you for your help so far.

    Just to clarify when you say equate I and II does it mean I = II and solve for x? Or is it stacking I and II and then solving for the variable x? Could you please start me off on how to solve this problem.

    Thank you so much
    I = II

    You should end up with a quadratic equation!
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  6. #6
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    When starting this problem off I get stuck at the part (3D/2)^4. Should I leave it in that form or would I have to multiply it out?

    ie:
    (3D/2)^4
    (81D^4)/(16)
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    When starting this problem off I get stuck at the part (3D/2)^4. Should I leave it in that form or would I have to multiply it out?

    ie:
    (3D/2)^4
    (81D^4)/(16)
    Thats right.. You should have \frac{81D^4}{16}. Do the same on the right hand side, your k, D and R should be cancelled, and you will be left with a quadratic equation
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  8. #8
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    Hmm after I finished canceling out the R^2, D^4, and k, I am left with
    -x^2 = 81/16. Is that correct? And what do we do next in order to find the answer?
    Last edited by florx; April 27th 2010 at 05:19 PM.
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Hmm after I finished canceling out the R^2, D^4, and k, I am left with
    -x^2 = 81/16. Is that correct? And what do we do next in order to find the answer?
    Your calculation is incorrect. Equating I and II gives:

    k \times R^2 \times \frac{81}{16} D^4 = k \times \frac{(100-x)^2}{100^2} R^2 \times D^4
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  10. #10
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    Alright now I get x^2 - 200x -81/16 = 0. Or is it x^2 - 200x - 40625 = 0?
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  11. #11
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Alright now I get x^2 - 200x -81/16 = 0. Or is it x^2 - 200x - 40625 = 0?
    \ \frac{81}{16} = \frac{(100-x)^2}((100)^2}

    \frac{(9 \times 100)^2}{4^2} = (100-x)^2

    (\frac{900}{4})^2 = (100-x)^2

    50625 = 10000 - 200x + x^2

    solve the quadratic equation
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  12. #12
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    Solving the quadratic equation I get -125 and 325. The back of the book says the answer is 44.4%. What is the next step in solving this problem?
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  13. #13
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Solving the quadratic equation I get -125 and 325. The back of the book says the answer is 44.4%. What is the next step in solving this problem?
    Oh wait!

    sorry I misread your 3rd question...

    follow this process..

    Let R be the original speed and let r be the reduced speed

    D be the original diameter, so \frac{3D}{2} is the new diameter

    k \times R^2 \times D^4 = k \times r^2 \times (\frac{3}{2}D)^4

    R^2 = r^2 \times \frac{81}{16}

    R = r \times \frac{9}{4}

    \therefore r = \frac{4R}{9}

    (4/9) of R = 44.4% of R
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  14. #14
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    Oh there you go thank you so much for helping me with this problem as well as past problems. You are very persistent in your willingness to help me and as others and I want to thank you for that.
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