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Mechanics. Circular motion and rotation

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    An aeroplane loops the loop in a vertical circle of radius 200 m, with a speed of 40 m s^-1 at the top of the loop. The pilot has a mass of 80 kg. What is the tention in the strap holding him into his seat when he is at the top of the loop?

    Answer: 160 N.

    2. Relevant equations
    A-Level Physics, chapter 6 formulas:
    w = Angle / t
    T = 2Pi / w
    v = wr
    a = w^2 * r
    a = v^2 / r

    3. The attempt at a solution
    r = 200 m
    v / u ? = 40 m s^-1
    m = 80 kg

    Attempt 1. 1/2 mv^2 = mgr -> 64 000 = 160 000. Wrong.
    Attempt 2. F = ma -> a = w^2 * r -> v = wr -> 40 = w*200 -> w = 0.2 rad s -> a = 0.2^2 * 200 = 8 -> F = 80*8 = 640 N. Wrong.

    Thank you in advance for any suggestions.
     
  2. jcsd
  3. May 3, 2015 #2

    billy_joule

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    You need to include units. Attempt two is close, you ignored gravity. You could have got there faster using your last relevant equation.
     
  4. May 3, 2015 #3
    You are right indeed about a = v^2 / r, thank you.

    So F = ma - mg = 640 - 80*10 (for simplicity reasons) = -160 N ?
    or F = mg - ma = 800 - 640 = 160 N. Does fit the book answer but why mg first? Maybe a silly question, but still.

    And by units you mean F = ma = 80 kg * 8 m s^1 instead of 80*8?
     
  5. May 3, 2015 #4

    billy_joule

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    Units: yes, makes it easier to understand your work, and many markers will subtract marks for missing units.

    The question asks for tension, which is always positive.
     
  6. May 3, 2015 #5
    Not sure where to post a second question, I think I'll better ask it here.

    1. The problem statement, all variables and given/known data

    A particle of mass 0.30 kg moves with an angular velocity of 10 rad s^1 in a horizontal circle 20 cm inside a smooth hemispherical bowl. FInd the reaction of the bowl on the particle and the radius of the bowl.

    Answer: 6.7 N, 22 cm

    2. Relevant equations
    R sin angle = mv^2 / r
    R cos angle = mg
    tan angle = v^2 / gr
    F sin angle = mv^2 / r
    F cos angle = mg

    3. The attempt at a solution
    https://www.physicsforums.com/threads/circular-motion-particle-in-a-bowl.252962/
    I am a bit lost because in this link with a wrong (according to the book the answer is 6.7 N) 6 N answer Thanuka got 22.3 cm, a relatively correct answer... And how to get the 6.7 N answer as in the book?
     
  7. May 3, 2015 #6

    haruspex

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    The 6N at that link is for the centripetal force, which is only the horizontal component of the reaction. The 6.7 N is the entire reaction.

    [It is generally preferred that you start a new thread for each problem.]
     
  8. May 4, 2015 #7
    Thank you. But in that case horizontal F = m(w^2)*r = 0.3 kg * 10^2 rads^-1 * 0.2 m = 6 N. And vertical F = mg = 0.3 kg * 10 (simplified g) = 3 N. In that case 6 N + 3 N = 9 N and not 6.7 N. Could you elaborate a bit on how to get 6.7 N? The bowl is smooth so I consider the only vertical force is gravity (F=mg).

    UPD: and for the second part I don't quite understand why it is
    Rcos theta= 6N as centripetal force is 6N
    Rsin theta = 3N as mass of the particle is 0.3kg

    and not R sin = 6 N for hor and R cos = 3 N for ver. Everywhere in the book the horizontal is sin and vertical is cos. Plus the angle which is created by the normal reaction also shows the same.
     
    Last edited: May 4, 2015
  9. May 4, 2015 #8

    BvU

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    This give you an idea ?

    Add_vecs.jpg
    Did you make your own drawing already (to see what's this sine and cosine business) ?
     
  10. May 4, 2015 #9
    I actually made a same drawing though instead of 3 N and 6 N I had R cos angle and R sin angle. Now I get the second part: cos angle = 6 N / R -> R cos angle = 6 N and sin angle = 3 N / R -> R sin angle = 3 N.

    But I still don't quite understand how to get the 6.7 N answer from the book for the first part.
     
  11. May 4, 2015 #10

    BvU

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    Try finding how much ?? is ...
     
  12. May 4, 2015 #11
    R = 3 / sin angle
    R = 6 / cos angle

    3/sin angle = 6/cos angle
    3 cos angle = 6 sin angle
    cos angle = 2 sin angle

    ?
     
  13. May 4, 2015 #12

    BvU

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    Are you telling me you have difficulty calculating the hypothenuse of a triangle with short sides 6 and 3 ?
     
  14. May 4, 2015 #13
    3^2+6^2 = 45, root 45 = 6.7 N... I really was looking for something complicated and totally forgot the basics. Thanks a lot!
     
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