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On the acceleration of cylinder down ramp

  1. Jul 31, 2003 #1
    on the acceleration of a cylinder down a ramp

    This problem has been vexing me for some time, for the life of me I can't figure it out or figure out why I can't figure it out. It seems like gsin(Θ)=a+αr, where Θ is the angle of the ramp, a is the acceleration, α is the angular acceleration, and r is the radius of the cylinder. Which would tell me that a=gsin(Θ)/2 and α=gsin(Θ)/(2r). But then another train of thought leads me to think that a=gsin(Θ) and that α=gsin(Θ)/r, which would mean a+αr=2gsin(Θ). So the question is, given Θ, r, and maybe m if neccesary, how do you find a and α?
     
    Last edited: Jul 31, 2003
  2. jcsd
  3. Aug 2, 2003 #2

    Integral

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    Perhaps the reason there have been no responses to this post is because it is unintellagble.

    Please make a complete and consise statement of the problem, perhaps you can get some help.
     
  4. Aug 2, 2003 #3
    OKAY...

    So, given radius r, ramp angle Θ, acceleration due to gravity g, and maybe the mass m of the cylinder, how does one find the linear and angular acceleration of the cylinder?
     
  5. Aug 2, 2003 #4
    Wow, this might be our 1st 'college level help' thread, so why not move it there?

    Jonathan,
    from looking at the formulae you gave, I think I can tell that you're trying to solve this solely by means of forces, that is basically by Newton's law F = ma. Which is for point masses.
    But: This is a rigid body problem, and a problem about rotation. So here are some key words you might like to look up in your book/script, and see where that gets you. Then let's see if I can help you further:

    - conservation of energy
    - rotational energy
    - moment of inertia
     
  6. Aug 2, 2003 #5
    I remember doing this exact problem in my physics class last year. Damn if I can find the notes or remember how to do it because the teacher said we wouldn't encounter it again for awhile (seeing as this was still high school physics).

    Ah, a quick seach on google reveals:

    http://physics.arizona.edu/~varnes/Teaching/141HSpring2003/Notes/Lecture29.pdf
     
  7. Aug 3, 2003 #6
    college level?

    How is this college level? I think this is high school physics coupled with the fact that I don't have a good grasp of Newton's laws of rotational motion that made it hard. Contrary to the previous post, I don't think you should have to take the mass into account considering that I'm only interested in the acceleration, not force/torque. It should be noted that to simplify further posts, I'm only interested in a uniform cylinder.
     
  8. Aug 3, 2003 #7
    This problem is also appeared in the Mechanics paper in a Physics Olympiad Preparation Programme hosted by Toronto University. I got the paper as well as the full solution from their website (unfortunately the website currently is not working). If you (Jonathan) want to know the full solution, I'm glad to send it to you.
     
  9. Aug 3, 2003 #8

    Integral

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    Re: college level?

    You cannot neglect mass, or the moment of inertia. You have a solution.
     
  10. Aug 3, 2003 #9
    The link given by Brad_Ad23 illistrates that you can take the mass into consideration, but when finding the acceleration, the mass ends up canceling out, note: gravity accelerates all things at the same rate, no matter the mass. It seems to me that this should apply to rolling down or falling down things, but I'm not sure about the latter, that's the point of this thread.
     
    Last edited: Aug 3, 2003
  11. Aug 3, 2003 #10

    Integral

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    It is not good form in a physics problem to assume that some factor drops out. You must balance the forces, do the algebra and let the math decide on what stays or goes. It it is not entirely obvious to me that considering the moment of interia of the cylendar that all mass terms will vanish. If they do fine, but do not assume it when you start the problem.
     
  12. Aug 3, 2003 #11
    This is how my naive mind would do this.
    E = mgh + 1/2 mv2 + 1/2 Iw2.
    Now, since w = v/r, and for a full cylinder, I = 1/2 mr2,
    E = mgh + 1/2 mv2 + 1/4 mv2
    = mgh + 3/4 mv2.
    Now, conservation of energy says dE/dt = 0,
    so
    mg dh/dt + 3/4 m * 2v dv/dt = 0.
    This is where m cancels out:
    g dh/dt + 3/2 v dv/dt = 0.
    Now, dh/dt = -v sin([the]), and dv/dt = a so
    -vg sin([the]) + 3/2 va = 0
    Divide by v:
    -g sin([the]) + 3/2 a = 0
    Solve for a:
    a = 2/3 g sin([the]).

    So nothing really mystical about this, just energy conservation, OK?
     
  13. Aug 3, 2003 #12
    Hey everyone who responded to my post... thanks for the help! I just wish i could find a resource that i could search for the answers to the questions that johnathan considers to vague (i agree in their vaguenss, since it's like that becuase i'm trying to learn so much about them). Well if Bill has one thing right about his post, it's probably true that you can't find much information on magnets easily. Professors, articles, books, scientific encyclopedias... I can't find much detail at all. It's actually kinda perplexing. no wonder there's so many questions about magnetism on this board
     
  14. Aug 3, 2003 #13
    This is odd. I find it quite easy to find information on whatever I need.


    Try using www.google.com as a start and put in a bunch of key words. Just watch out for sites that give information under the pretext that all scientists are in a conspiracy or are all wrong and they know/proclaim a theory of everything that is already solved. Those generally cannot be trusted to give much of what you would be looking for.
     
  15. Aug 4, 2003 #14
    For those who has not learnt elementary calculus (like me), here is an alternative, possibly easier solution.

    The resultant force on the cylinder is

    ma = mg sin[the] - f
    but
    [tau] = fr, and [tau] = I[alpha]
    [4] fr = I[alpha]
    but I also equals mr^2/2, and [alpha] =a/r

    Now if we subistute this into the equation, and make f the subject we get:
    f = ma/2

    So we have
    ma = mg sin[the] - ma/2.
    3ma/2 = mg sin[the]
    [4] a = 2/3 mg sin[the]

    Here is an additional problem for whom may be interested:
    According to the previous problem, what is the minimum coefficient of static friction that is required for the cylinder to roll down the incline without slipping?
     
  16. Aug 4, 2003 #15
    I don't know about the coefficient, sorry.
    I misundersood Brad_Ad23's link, upon looking at it again I've come to this conclusion for a uniform cylinder:
    a=gsin(Θ)/r
    α=2gsin(Θ)/(3r)
    t=sqrt(3d/(gsin(Θ)))
    Thank you all for you help!
     
    Last edited by a moderator: Aug 4, 2003
  17. Aug 4, 2003 #16
    Can't be. Wrong unit on right-hand side.
     
  18. Aug 4, 2003 #17
    Got Confused

    Yeah, you're right, I got confused. This is what it should be:
    a=2gsin(Θ)/3
    α=2gsin(Θ)/(3r)
    t=sqrt(3d/(gsin(Θ)))
     
    Last edited by a moderator: Aug 4, 2003
  19. Aug 4, 2003 #18
    Yes. :smile:
     
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