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    Hi all,

    The question is from my revision guide.
    For those of you who can't be bothered with the whole post, what answer do you get for the bit in bold.

    " A toy train travels round a circular track at a steady speed. It completes three orbits in 1.0 minutes. The radius of curvature of the track is 40cm"
    (a) calculate the speed of the train


    So heres what I do:
    V = 2 * pi * radius / T (period)

    T = 60/ 3 = 20 seconds

    V = 2 * pi * 0.4 / 20 = pi/25 = 0.126(3sf) ms^-1

    However the answer in the book is 0.042 ms-1. Can you guys try and see what you get?

    Part B 'Calculate the angular velocity of the train'
    Well it moves 6pi radians in 60 seconds so angular velocity = 0.314 (the book says 0.31 2sf i think).

    This leads me to believe that either for A i'm being incredibly dumb, or the book is wrong.

    To double check I used V = r * angular frequency
    V = 0.4 * 0.314
    V = 0.1256 which matches my answer for A but not theirs.

    So what do you all get for A?

    Thanks
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    pi x d = 0.8pi m = circumferance.

    1 orbit / 20 seconds
    1/20 orbit / second

    1/20 x 0.8pi m/s
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    (Original post by Robbie10538)
    pi x d = 0.8pi m = circumferance.

    1 orbit / 20 seconds
    1/20 orbit / second

    1/20 x 0.8pi m/s
    Excellent we got the same.
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    (Original post by rather game than revise)
    Hi all,

    The question is from my revision guide.
    For those of you who can't be bothered with the whole post, what answer do you get for the bit in bold.

    (a) calculate the speed of the train


    So heres what I do:
    V = 2 * pi * radius / T (period)

    T = 60/ 3 = 20 seconds

    V = 2 * pi * 0.4 / 20 = pi/25 = 0.126(3sf) ms^-1
    I got the same answer. odd. mark schemes can be wrong you know.
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    i think we use a =v^2/r and v = rw = 2*pi*r/t
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    v = dtheta/dt
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    I'm doing another question now and I think that it's wrong as well... mmm.
    If something completes 400 oscillations in 6 seconds - then isn't it's frequency 400/6 Hz?
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    (Original post by rather game than revise)
    Hi all,

    The question is from my revision guide.
    For those of you who can't be bothered with the whole post, what answer do you get for the bit in bold.

    " A toy train travels round a circular track at a steady speed. It completes three orbits in 1.0 minutes. The radius of curvature of the track is 40cm"
    (a) calculate the speed of the train


    So heres what I do:
    V = 2 * pi * radius / T (period)

    T = 60/ 3 = 20 seconds

    V = 2 * pi * 0.4 / 20 = pi/25 = 0.126(3sf) ms^-1

    However the answer in the book is 0.042 ms-1. Can you guys try and see what you get?

    Part B 'Calculate the angular velocity of the train'
    Well it moves 6pi radians in 60 seconds so angular velocity = 0.314 (the book says 0.31 2sf i think).

    This leads me to believe that either for A i'm being incredibly dumb, or the book is wrong.

    To double check I used V = r * angular frequency
    V = 0.4 * 0.314
    V = 0.1256 which matches my answer for A but not theirs.

    So what do you all get for A?

    Thanks
    w = radians / time
    w * radius = speed

    radians * radius / time = speed
    6pi * 0.4 / 60 = 0.12566 m/s

    How I would of done it, sounds like the book is wrong. They have infact done the time taken and applied it to one orbit, i.e:

    radians * radius / time = speed
    2pi * 0.4 / 60 = 0.04188 m/s

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    (Original post by rather game than revise)
    I'm doing another question now and I think that it's wrong as well... mmm.
    If something completes 400 oscillations in 6 seconds - then isn't it's frequency 400/6 Hz?
    Yup

    Apologies for double post
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    (Original post by rather game than revise)
    I'm doing another question now and I think that it's wrong as well... mmm.
    If something completes 400 oscillations in 6 seconds - then isn't it's frequency 400/6 Hz?
    Indeed

    400 osc per 6 seconds
    400/6 osc per 6/6 seconds
    400/6 osc per second
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    Guys can I pick your brains with another question? If this question is wrong as well I'm gonna ditch this revision guide lol.

    A mass between two springs oscillates between 2 points 5.0cm apart, and completes 40 oscillations in 1 minute. What is its maximum speed?

    So I will use the formula V(max) = 2*pi*frequency*Amplitude

    frequency = 40/60 = 2/3
    Amplitude = I'm not sure what the amplitude would be. I know it's the maximum displacement from the equilbrium position but I don't know where the equilbrium position is? If I assume it is in the middle of the two points then A= 2.5cm so 0.025 m

    So vmax= 2*pi*(2/3)*0.025 = pi/30 = 0.105 ms-1

    The book says 0.44 ms-1

    Cheers
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    i get the same answer as you!!!!
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    Ditch the revision guide mate lol
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    Yeah it's getting ditched lol
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    (Original post by rather game than revise)
    Yeah it's getting ditched lol
    What revision guide is this **** lol?
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    happiness runs in a circular motion...
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    What guide is this mate?
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    its the 'OCR Revise Physics A2'
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    (Original post by rather game than revise)
    its the 'OCR Revise Physics A2'
    That the Physics 2 book in blue with the glass square based prism on the front?
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    It's got pink at the top, and then a picture of like a pink star/galaxy on a star background. It's for the New Spec OCR Physics A
 
 
 
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