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Math Help - Stuck here--not sure how to differentiate

  1. #1
    Member Chokfull's Avatar
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    Question Stuck here--not sure how to differentiate

    Here's my problem. I have a rotating wheel, radius 40cm, and a connecting rod AP with length 1.2m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute. Find the angular velocity of the connecting rod, \frac{d\alpha}{dt}, in radians per second, when \theta=\frac{\pi}{3}.

    I have attached a diagram, and my work on the problem so far. (NOTE: later in my work, for the sake of simplicity, i replaced \alpha with "a" and \theta with "O".)

    Now, this is my problem. At the bottom, where i differentiated it, the derivative doesnt have an alpha in the equation, but i need it in the problem to solve it. I had to use the calculator to differentiate it, because i havent worked with differentiating \sin^{-1} before. I just became completely lost at this point...am i going about the problem the wrong way?..........Stuck here--not sure how to differentiate-math1.png......Stuck here--not sure how to differentiate-math2.png....(The first picture is the top half, the second is the bottom half)
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  2. #2
    A Plied Mathematician
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    You didn't differentiate with respect to the correct variable, and you didn't differentiate correctly. You have

    \displaystyle\alpha(t)=\sin^{-1}\left(\frac{1}{3}\sin(\theta(t))\right).

    What you want to compute is

    \displaystyle\dot{\alpha}(t)=\dots using the chain rule. What do you get?
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    I get \frac{d}{dt}a(t)=\frac{\cos(\theta(t))\theta(1)}{\  sqrt{9-\sin(\theta(t))^2}}

    I'm not sure about this part, either...If i replace \theta(t) with \frac{\pi}{3}, because \theta=\frac{\pi}{3} is the point at which Im trying to find the speed, I get .174*\theta(1). The correct answer is 6.563. This means that, for the equation above to be correct, \theta(1) has to equal 37.718....Right? But this makes no sense :P
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  4. #4
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    I get

    \dot{\alpha}(t)=\dfrac{\cos(\theta(t))\,\dot{\thet  a}(t)}{\sqrt{9-\sin^{2}(\theta(t))}}. I think you're on the right track.

    So you've got, correctly, that \theta=\pi/3. What is \dot{\theta}(t)? Are there any unit conversion issues you might have?
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  5. #5
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    Quote Originally Posted by Chokfull View Post
    I get \frac{d}{dt}a(t)=\frac{\cos(\theta(t))\theta(1)}{\  sqrt{9-\sin(\theta(t))^2}}

    I'm not sure about this part, either...If i replace \theta(t) with \frac{\pi}{3}, because \theta=\frac{\pi}{3} is the point at which Im trying to find the speed, I get .174*\theta(1). The correct answer is 6.563. This means that, for the equation above to be correct, \theta(1) has to equal 37.718....Right? But this makes no sense :P

    Instead of: \displaystyle \frac{d}{dt}\alpha(t)=\frac{\cos(\theta(t))\theta(  1)}{\sqrt{9-\sin(\theta(t))^2}},

    you should have: \displaystyle \frac{d}{dt}\alpha(t)=\frac{\cos(\theta(t))}{\sqrt  {9-\sin^2(\theta(t))}}{{d\theta}\over{dt}}\,.

    Alternatively you could have differentiated a preceding equation with respect to time:

    \displaystyle {{d}\over{dt}} \left( \sin(\alpha(t))\right) = {{d}\over{dt}} \left({{1}\over{3}}\sin(\theta(t))\right)

    to get:

    \displaystyle  \left( \cos(\alpha(t))\right){{d\alpha}\over{dt}} =  \left({{1}\over{3}}\cos(\theta(t))\right){{d\theta  }\over{dt}}\,.

    But then you would need to figure out what \displaystyle cos\alpha is when \displaystyle \theta = {{\pi}\over{3}}\,.
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  6. #6
    Member Chokfull's Avatar
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    Okay....If my \theta(1) is supposed to be \frac{d\theta}{dt}, then all i have to do is multiply by \frac{d\theta}{dt}. But to get the right answer, this needs to be 37.718. This is really just the rate at which the wheel turns (360 revolutions per minute, or 6 per second). Since we're working in radians and seconds, and 2\pi radians equals one full turn, it's rotating at a rate of.... 12\pi radians per second! This is exactly he answer I'm looking for. Thank you both for your help
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  7. #7
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    360 revolutions per minute is how many radians per second?
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