1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on rotational dynamics

  1. Oct 31, 2004 #1
    Here is the question (it's part of a lab):

    "From your lectures in mechanics you should be able to derive:
    [tex]\alpha = \frac{mgr}{I+mr^2}[/tex]
    This equation ignores the effect of the pulleys and the mass of the connecting string"

    The lab was basically setting up a mass on a string over a pulley, connected to a rotation horizontal disc which would oscillate back and forth (if this information is needed).

    Don't know where to start - I thought maybe conservation of energy, PEi + KEi = PEf + KEf, but I can't get anywhere.
     
  2. jcsd
  3. Oct 31, 2004 #2
    What is the question? Do they want you to derive that equation?
     
  4. Oct 31, 2004 #3
    Yes, that was the whole question. Don't know what to do..
     
  5. Nov 1, 2004 #4
    Anyone have an idea?
     
  6. Nov 1, 2004 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Since you haven't bothered with the simple courtesy of defining your terms, nope.
     
  7. Nov 1, 2004 #6
    If you dont know that I is moment of inertia and m is mass, you probably can't answer it anyway.
     
  8. Nov 1, 2004 #7

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I of what?
    m of what?
    r of what?
     
  9. Nov 1, 2004 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    For what it's worth, here's your solution:
    The magnitude of yhe string's acceleration is related to the disk's angular acceleration by:
    [tex]|a|=r\alpha[/tex]
    where the ang. acc. is considered greater than zero.
    Hence, the tension in the rope is given by:
    [tex]T=mg-mr\alpha[/tex]
    Hence, the momwnt-of momwntum wquation about the disk's center is:
    [tex]mgr-mr^{2}\alpha=I\alpha[/tex]

    Next time, show a proper measure of respect towards your readers than some muddøed half-baked account of an experiment..:grumpy:
     
  10. Nov 1, 2004 #9
    I was just reading the problem straight from the book, verbatim. Thanks for the answer though :blushing:
     
  11. Nov 1, 2004 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    All right, just an advice:
    Next time you relate an EXPERIMENT you've done, you need to remember that others can't read your mind. As you very sketchily described the experimental set-up, it was liable to several interpretations, not just one.

    In particular, the role of the horizontal disk is most unsatisfactorily described!
    It took some time to figure out the real experiment behind your words..:wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question on rotational dynamics
Loading...