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anonymous

  • 5 years ago

Word problem: A particle moves around the circle x^2+y^2=25 at constant speed, making one revolution in 2 s. Find it's acceleration when it is at (3,4). How is my thinking process supposed to be here?

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  1. shadowfiend
    • 5 years ago
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    Hmm... So you're going to need the tangential acceleration, which would be the second derivative of the function. But, the function is in terms of both x and y.

  2. shadowfiend
    • 5 years ago
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    I think you'll have to find the second partial derivatives with respect to x and y, but I'm not 100% sure about that. Been a while since I've done a problem like this.

  3. anonymous
    • 5 years ago
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    You'll find the solution by converting to polar coordinates and using the given speed. It's a bit tricky, and I really suck at these kind of assignments. You'll have to treat the vector as r= x(t) + y(t) and make use of the chain rule when finding the derivatives. How would you approach it? I just find it really hard to realise all the steps without seeing them first.

  4. shadowfiend
    • 5 years ago
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    Ah. Hmm... Still a toughie for sure.

  5. shadowfiend
    • 5 years ago
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    Hm... Wait...

  6. anonymous
    • 5 years ago
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    Indeed, got a whole bunch of similar ones, luckily only 1 showed up on 2 preparation exams. So hopefully there wont be one.

  7. shadowfiend
    • 5 years ago
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    Ok, so: \[\begin{align} x(\theta) &= r \cos \theta \\ y(\theta) &= r \sin \theta \end{align}\]

  8. shadowfiend
    • 5 years ago
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    So, if we derive twice, we get: \[\begin{align} x'(\theta) &= -r \sin \theta & x''(\theta) &= -r \cos \theta\\ y'(\theta) &= r \cos \theta & y''(\theta) &= -r \sin \theta \end{align}\]

  9. anonymous
    • 5 years ago
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    Your r should just be a constant 5 here, I don't think you need to have 2 variables

  10. anonymous
    • 5 years ago
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    Correct Flying.

  11. shadowfiend
    • 5 years ago
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    Right, that's irrelevant -- we can plug that in.

  12. shadowfiend
    • 5 years ago
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    Theta, however, is a function of time. We go pi of theta in one second.

  13. shadowfiend
    • 5 years ago
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    But the time it takes for one revolution is irrelevant -- it's a distractor.

  14. shadowfiend
    • 5 years ago
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    (I think.)

  15. anonymous
    • 5 years ago
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    You're restricted to the first quadrant by the coordinates you're given, I feel like this problem just sounds difficult. It really isn't too bad though

  16. anonymous
    • 5 years ago
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    Yeah, it shouldn't really matter because your angular velocity is just the natural period of sin and cos

  17. shadowfiend
    • 5 years ago
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    Yeah, I think once you have the second derivative as r cos theta, you can find theta easily using the point you're given and the tangent, and then plug in r and theta into the two above. Then you just need to normalize it to a magnitude + direction for the acceleration vector.

  18. anonymous
    • 5 years ago
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    I think you could have left this in x and y and just done implicit differentiation since you've already been given coordinates

  19. shadowfiend
    • 5 years ago
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    Or rather once you have the two derivatives as -r cos theta and -r sin theta.

  20. anonymous
    • 5 years ago
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    solved it for y and just used specifically the top part of the circle

  21. anonymous
    • 5 years ago
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    Sounds good, the trick my teacher did was to use the points given 3 = 5cos(theta) and 4 = 5sin(theta) when you get the vector readily differentiated for acceleration. Quite neat.

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