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    (Original post by joostan)
    The number of days of me knowing anything about circular motion = \lim{x \to 0}\left(\dfrac{\sin(x)}{x} \right)
    Allow me to be the judge of that
    looooool :P So far your solving this the 'create an equation' way, theres another way of proceeding which I actually found less confuzzling...
    You would have to try finding co-ordinates and use velocity vectors

    Hint:
    Spoiler:
    Show
    Everything can be rescaled for different values of r. If we let r(t) be the distance between the fox and the center of the circle at time t, his (r,θ) position at time t is r(t),vt), which in (x,y) coordinates is....(_,_) The velocity vector is then....


    Sorry I can't be of too much help
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    (Original post by tigerz)
    You let your guard down tut tut
    Spoiler:
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    pssht my excuse of floppiness


    Right this is a weird one haha:

    There is a rabbit that runs in a perfect circle of radius r with a constant speed v.
    A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed v such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?
    I quite liked this But shame on you for posting a problem you can't do yourself! :naughty:

    Spoiler:
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    Let the angle swept out by the rabbit at time t be \theta \Rightarrow \theta = \dfrac{vt}{r}

    As the fox is always in line with the rabbit, the angle swept by the fox is also \theta

    Let \mathcal{R} denote the distance between the fox and the centre of the circle at time t which in polar coordinates is ( \mathcal{R}, \frac{vt}{r}) and corresponds to cartesian coordinates of ( \mathcal{R} \cos (\frac{vt}{r}), \mathcal{R} \sin ( \frac{vt}{r} ) )

    Differentiating this, we get a velocity vector of

    \left( \mathcal{R} \cdot ( - \frac{v}{r} \cdot \sin (\frac{vt}{r} ) ) + \cos ( \frac{vt}{r}) \cdot \dot{\mathcal{R}} \right) \textbf{i} + \left( \mathcal{R} \cos (\frac{vt}{r} ) \cdot \frac{v}{r} + \sin ( \frac{vt}{r} ) \cdot \dot{\mathcal{R}} \right) \textbf{j}

    \Rightarrow | \textbf{v} | = \sqrt{ \dot{\mathcal{R}^{2}} + \mathcal{R}^{2} \cdot \frac{v^{2}}{r^{2}}}

    \Rightarrow \dot{\mathcal{R}} = v \cdot \sqrt{ 1 - \dfrac{\mathcal{R}^{2}}{r^{2}}}

    \Rightarrow \dfrac{d \mathcal{R}}{dt} = v \sqrt{1 - \dfrac{\mathcal{R}^{2}}{r^{2}}}

    Separating variables, we get:

    \displaystyle\int dt = \displaystyle\int_{0}^{r} \dfrac{1}{v \cdot \sqrt{1 - \frac{\mathcal{R}^{2}}{r^{2}}}} d \mathcal{R}

    Let \mathcal{R} = r \sin \theta \Rightarrow t = \dfrac{r}{v} \displaystyle\int_{0}^{\pi /2} d \theta = \dfrac{r}{v} \cdot \dfrac{\pi}{2}
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    (Original post by tigerz)
    looooool :P So far your solving this the 'create an equation' way, theres another way of proceeding which I actually found less confuzzling...
    You would have to try finding co-ordinates and use velocity vectors

    Hint:
    Spoiler:
    Show
    Everything can be rescaled for different values of r. If we let r(t) be the distance between the fox and the center of the circle at time t, his (r,θ) position at time t is r(t),vt), which in (x,y) coordinates is....(_,_) The velocity vector is then....


    Sorry I can't be of too much help
    Vectors and calculus don't mix :cool:
    Sounds like a multivariable nightmare. I think I've made some progress.
    By noting that the Fox's circle is smaller, there needs to be a different radius, which I'll call x.
    \Rightarrow v^2 = f(t)^2 + x^2 \left(\dfrac{v}{r} \right)^2
    This new radius opens up new options, should've spotted it earlier. . .
    f(t) is some function involving x and t, I just need to work out what, it's a similar situation to before
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    (Original post by ThatRandomGuy)
    Ohhh you're talking of that paper...

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    If I remember correctly I got 72/75 when I did that paper. Dropped marks on the impulse question. I say more A level papers need to be like this one
    I see, you beat me by 2 marks , I managed to get the impulse one right but dropped all the marks in 8(c) as I didn't understand it.
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    (Original post by Felix Felicis)
    I quite liked this
    I'm resisting looking at the spoiler.
    I seem to have a problem which I can't resolve
    Do you happen to know the other component of the velocity, it's giving me gyp. Not good enough at M3/vectorsy stuff yet :cry2:
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    (Original post by Felix Felicis)
    I quite liked this But shame on you for posting a problem you can't do yourself! :naughty:

    Spoiler:
    Show
    Let the angle swept out by the rabbit at time t be \theta \Rightarrow \theta = \dfrac{vt}{r}

    As the fox is always in line with the rabbit, the angle swept by the fox is also \theta

    Let \mathcal{R} denote the distance between the fox and the centre of the circle at time t which in polar coordinates is ( \mathcal{R}, \frac{vt}{r}) and corresponds to cartesian coordinates of ( \mathcal{R} \cos (\frac{vt}{r}), \mathcal{R} \sin ( \frac{vt}{r} ) )

    Differentiating this, we get a vector of \left( \mathcal{R} \cdot ( - \frac{v}{r} \cdot \sin (\frac{vt}{r} ) ) + \cos ( \frac{vt}{r}) \cdot \dot{\mathcal{R}} \right) \mathbf{i} + \left( \mathcal{R} \cos (\frac{vt}{r} ) + \sin ( \frac{vt}{r} ) \cdot \dot{\mathcal{R}} \right) \mathbf{j}

    \Rightarrow | \mathbf{v} | = \sqrt{ \dot{\mathcal{R}^{2}} \mathcal{R}^{2} \cdot \dot{\theta}^{2}}

    \Rightarrow \dot\mathcal{R} = v \sqrt{ 1 - \dfrac{\mathcal{R}^{2}}{r^{2}}}

    \Rightarrow \dfrac{d \mathcal{R}}{dt} = v \sqrt{1 - \dfrac{\mathcal{R}^{2}}}{r^{2}}}

    Separating variables, we get:

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    t = \displaystyle\int_{0}^{r} \dfrac{1}{v \cdot \sqrt{1 - \frac{\mathcal{R}^{2}}{r^{2}}} dr


    Let \mathcal{R} = r \sin \theta \Rightarrow t = \dfrac{r}{v} \displaystyle\int_{0}^{\pi /2} d \theta = \dfrac{r}{v} \cdot \dfrac{\pi}{2}
    WOOP, you smashed it, but you can take it one step further
    I know But no worries I did make sure I would be able to explain it if required! He wanted a challenge and you guys weren't on so I had to cheat a lil' :ashamed2:

    (Original post by joostan)
    Vectors and calculus don't mix :cool:
    Sounds like a multivariable nightmare. I think I've made some progress.
    By noting that the Fox's circle is smaller, there needs to be a different radius, which I'll call x.
    \Rightarrow v^2 = f(t)^2 + r^2 \left(\dfrac{v}{x} \right)^2
    This new radius opens up new options, should've spotted it earlier. . .
    f(t) is some function involving x and t, I just need to work out what, it's a similar situation to before
    LOOL, I see...
    YAAAAY, smashed it, so now you should be able to think of an expression for v and for r'...

    P.S. Felix I know another one of this type
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    (Original post by joostan)
    I'm resisting looking at the spoiler.
    I seem to have a problem which I can't resolve
    Do you happen to know the other component of the velocity, it's giving me gyp. Not good enough at M3/vectorsy stuff yet :cry2:
    Hint
    Think about creating a function to denote the distance of the fox from the centre of the circle - you know that the angle swept out by the fox is the same as that by the rabbit - so you can exchange the fox's position vector from polar to cartesian coordinates
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    (Original post by Felix Felicis)
    Hint
    Think about creating a function to denote the distance of the fox from the centre of the circle - you know that the angle swept out by the fox is the same as that by the rabbit - so you can exchange the fox's position from polar to cartesian coordinates
    Ah thinking displacements. I should've thought of that. :rolleyes:
    So: v = \sqrt{\left(\dfrac{dx}{dt} \right)^2 + x^2 \left(\dfrac{d\theta}{dt} \right)^2} Right?
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    How do i sketch for part B??? never seen q questiion like that before
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    (Original post by joostan)
    Ah thinking displacements. I should've thought of that. :rolleyes:
    So: v = \sqrt{\left(\dfrac{dx}{dt} \right)^2 + x^2 \left(\dfrac{d\theta}{dt} \right)^2} Right?
    Yep, created a position vector and differentiated to obtain a velocity vector
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    (Original post by Felix Felicis)
    Yep, created a position vector and differentiated to obtain a velocity vector
    Yeah, that's not on OCR M1 spec - dunno why.
    Now it's a bish bosh integral - won't even bother to type it up.
    You beat me again Felix
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    (Original post by joostan)
    Yeah, that's not on OCR M1 spec - dunno why.
    Now it's a bish bosh integral - won't even bother to type it up.
    You beat me again Felix
    This took an hour tbf, this type of mechanics problem probably belongs on TPIT
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    (Original post by tigerz)
    WOOP, you smashed it, but you can take it one step further
    I know But no worries I did make sure I would be able to explain it if required! He wanted a challenge and you guys weren't on so I had to cheat a lil' :ashamed2:



    LOOL, I see...
    YAAAAY, smashed it, so now you should be able to think of an expression for v and for r'...

    P.S. Felix I know another one of this type
    I guess I brought it on myself.
    I'm off now :cool:
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    (Original post by Knoyle quiah)
    How do i sketch for part B??? never seen q questiion like that before
    tan[arg(z+6)]=1/2
    arg(z+6) = arctan1/2
    arg((x+6)+yi) = arctan1/2
    tantheta = y/x
    y/(x+6) = tan(arctan1/2)
    y/(x+6) = 1/2
    y= x/2 + 3
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    (Original post by joostan)
    I guess I brought it on myself.
    I'm off now :cool:
    yup okayss

    (Original post by Felix Felicis)
    This took an hour tbf, this type of mechanics problem probably belongs on TPIT
    LOOL, sowwy

    HOWEVER....
    Spoiler:
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    the fox is factor  \pi/2 \approx 1.5 times slower
    Spoiler:
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    to conclude while the rabbit has run a quarter circle, the fox has run half a smaller circle (half the radius)
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    I feel a little left out not being able to tackle these M3 problems :sad:
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    (Original post by tigerz)
    LOOL, sowwy

    HOWEVER....
    Spoiler:
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    the fox is factor  \pi/2 \approx 1.5 times slower
    Spoiler:
    Show
    to conclude while the rabbit has run a quarter circle, the fox has run half a smaller circle (half the radius)
    Haha, it's fine xD I'm off now, back to chem revision :ninja:
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    (Original post by Boy_wonder_95)
    I feel a little left out not being able to tackle these M3 problems :sad:
    I haven't even done M1 no worries

    (Original post by Felix Felicis)
    Haha, it's fine xD I'm off now, back to chem revision :ninja:
    LOL yup, I started this paper at like 9am :/ Keep getting distracted
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    (Original post by Boy_wonder_95)
    I feel a little left out not being able to tackle these M3 problems :sad:
    It's not really M3 tbf - it doesn't use any definitions beyond the definition of angular velocity and even that can be derived if you use speed = distance/ time and the rest is just juggling with polar/ cartesian coordinates and using the fact that v = \frac{dx}{dt}
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    (Original post by tigerz)
    I haven't even done M1 no worries
    Yet you're still able to give good explanations :lol:
 
 
 
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