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    Struggling with this problem, if someone could help that'd be great

    F17-16
    A 20 kg sphere, with radius 0.15m rolls down an inclined plane without slipping. The plane is angled at 30 degrees to the horizontal. Determine the acceleration of the mass center of the sphere as well as the angular acceleration of the sphere.

    Here's my working:



    Higher resolution version: http://img843.imageshack.us/img843/4494/img006js.jpg

    The value for friction that I have found is far too high, can anyone see why?

    EDIT: I HAVE noticed now that I have missed a factor of 1/5 out near the bottom of the page, however once altering my working to find the angular acceleration I am infact still wrong!
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    What's the answer supposed to be?
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     \alpha = 23.4 rad / s^2

     a_o = 3.50 m / s^2
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    Hang on a minute.. it's not a coincidence that 117/5 = 23.4, let me recheck my working.
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    (Original post by Kasc)
     \alpha = 23.4 rad / s^2

     a_o = 3.50 m / s^2
    Puzzled then, since dividing 117 by 5 (which accounts for your correction to the force of 140) givs 23.4!
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    Yep, it was just me forgetting the factor of 1/5, i recalculated theta double dot wrongly :>

    Cheers anyway Ghostwalker ;-) You've helped me out a few times, I'm sure there's more questions from to follow until january
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    (Original post by Kasc)
    Yep, it was just me forgetting the factor of 1/5, i recalculated theta double dot wrongly :>

    Cheers anyway Ghostwalker ;-) You've helped me out a few times, I'm sure there's more questions from to follow until january
    :cool:
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    The book gives a solution, however it don't understand it:



    Could you briefly explain it to me please? It's the text book for this module, however the lecturer goes through problems with F=ma and M = I alpha, I assume because these apply to different situations?
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    (Original post by Kasc)
    The book gives a solution, however it don't understand it:



    Could you briefly explain it to me please? It's the text book for this module, however the lecturer goes through problems with F=ma and M = I alpha, I assume because these apply to different situations?
    OK.

    It looks like they've taken moments about the point of contact on the plane. I assume that is the A they are refering to.

    So the left hand side of your equation is the moment of the weight about that point.

    For the right hand side they've used moment of intertia times angular acceleration, BUT.

    The moment of intertia is worked out using the parallel axis theorem, so it's I + mr^2

    Then multiplying by angular acceleration and using "angular acceleration times radius = linear acceleration of the centre of gravity" gives you the RHS of the equation. The non-slipping is required to make this valid. It's possible it's required earlier when taking moments, but I'm not entirely sure on that point.
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    Ah cheers, I'm stuck again! ;>

    http://img27.imageshack.us/img27/751/imag0154w.jpg
    http://img403.imageshack.us/img403/2887/imag0156e.jpg
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    The only thing you are missing is the relationship between \ddot{x} and \ddot{\theta}. Then plug in the numbers.

    If you consider the point about which the spool is turning at any given moment, the relationship should be clear. There is a clue in the diagram.

    I presume you're familiar with "radius of gyration", if not, Google!
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    I'm really not sure :/ I've only been shown that if there is no slip then \ddot{x}=r\ddot{\theta}.

    I'm guessing you are refering to point A? If i were to take a guess, I'd say it were \ddot{x}=0.2r\ddot{\theta} but that really is a guess.
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    (Original post by Kasc)
    I'm really not sure :/ I've only been shown that if there is no slip then \ddot{x}=r\ddot{\theta}.

    I'm guessing you are refering to point A? If i were to take a guess, I'd say it were \ddot{x}=0.2r\ddot{\theta} but that really is a guess.
    Well, it's certainly not slipping a A. Or putting it another way, A is the instantaneous centre of rotation, so \ddot{x}=r\ddot{\theta}

    I don't know where you've got your second equation from; I make it \ddot{x}=0.4\ddot{\theta}
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    Oh, I understand, thank you very much ;P


    EDIT: I hope I'm not annoying you with so many questions, I must seem so clueless xD
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    (Original post by Kasc)
    Oh, I understand, thank you very much ;P


    EDIT: I hope I'm not annoying you with so many questions, I must seem so clueless xD
    If I get to the stage of thinking you're just posting rather than thinking about them, I'll stop; I may also stop for lots of other reasons as well, since I didn't do mechanics at uni.
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    (Original post by ghostwalker)
    If I get to the stage of thinking you're just posting rather than thinking about them, I'll stop; I may also stop for lots of other reasons as well, since I didn't do mechanics at uni.
    Heh.. pretty impressive you know more than me about this yet I'm the one doing a module on it at uni xD well I'm sorry to ask again about the same question but I've followed it through and come to the wrong answer.

    I'm pretty confused though, I think that the l.h.s of "M=I \alpha" is correct since \dfrac{450-Tr_i-Fr_o}{mk^2} for T = 485N gives 1.14 rad/s^2 when the correct is 1.15 rad/s^2, so I can be fairly sure the equation is right. However I have worked out the value for T wrong.


    http://img94.imageshack.us/img94/6027/imag0157u.jpg

    Know where I'm going wrong?
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    (Original post by Kasc)
    Know where I'm going wrong?
    What happened to the r_i in the denominator on the RHS?

    If you're getting 1.14 rather than 1.15, then no, it's not correct. I recall getting 1.15... when I worked it through, but that's now so long ago...
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    Oh yeah thanks!

    I must ask, do you mind helping me? I don't want to be angering you or anything
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    (Original post by Kasc)
    Oh yeah thanks!

    I must ask, do you mind helping me? I don't want to be angering you or anything
    If I didn't want to help, I wouldn't; so don't worry about it. I am pushed for time at present though, so don't expect quick replies.

    PS: the previous equation I doubted was actually correct, as the actual value of the tension is 484.56 which does give 1.15... rad s^-2 for the angular accleration.

    Edit: Managed to find my previous calculations.
 
 
 
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