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Rotational mechanics and Angular Momentum- help needed to check over work

  1. Nov 4, 2008 #1
    Please note, this is a series of problems, with a series of answers. I believe that I understand how to do the problems, but for some reason, my answer is wrong, and I can think of no reason why.

    1. The problem statement, all variables and given/known data
    Problem 1: http://img355.imageshack.us/my.php?image=physics1jw9.png

    i) I need to find the speed of the mass as it passes point B in m/s

    ii) I need to find the Tension of the string (Which I am sure I can do, but without a correct answer for part i, I'm not going to attempt yet)

    Problem 2: http://img186.imageshack.us/my.php?image=physics2pt6.png

    Problem 3: http://img355.imageshack.us/my.php?image=physics3el2.png
    (Note, the word that is cut off is longitudinal. I had to copy and paste parts of it into one picture, sorry it's a little shoddy <.<)

    2. Relevant equations
    All the lovely rotational mechanics equations, and their linear counterparts


    3. The attempt at a solution

    Problem 1
    i) Ok, I did not take into account the spoke on teh wheel whatsoever, since the mass of inertia was given to me, and the spokes of the wheel did not seem to have their own mass. Not sure if this would effect the problem.

    T=I(angular a)
    Angular acceleration = Torque/I

    Torque= 46 KG * 9.8 m/s^2 * 3m = 1352.4
    I= 3/4* 23 KG * 3^2 = 155.25
    Ang A= 8.7111

    a = (Ang A)*r

    a= 8.711 * 3 = 26.133

    Plugging this acceleration into the classic Xf= Xi + Vi +(at^2)*2 (assuming Vi and Xi are 0)
    I get t=2.106

    Then, using the Vf= Vi +at
    I get Vf= 55.05

    This answer is wrong though, not sure where I screwed up

    [hr]

    Problem 2
    This one seemed simple enough, just a basic torque= I * ang a

    Ang a = T/I

    I=(1/12)*(2.7)*(3^2)= 2.025

    T= rxF
    F= M*G= 9.8*2.7= 26.46
    r= 3cos(43)=2.045
    T= 54.137

    Ang a= 26.734

    It wants it in Radians/sec^2. Is it bad that I was using degrees for the cos?
    [hr]
    Problem 3
    I searched on google, and did come across this thread: https://www.physicsforums.com/showthread.php?t=17605

    However, upon trying the forumulas given there, I still got it wrong. Using the formula ShawnD had at the end of his post, I got the wrong answer.

    m= IW/rV

    I= 2950
    W=2 rev/min= .2094... rad/sec
    r=2.9
    V=755
    I got M to be .282 kg

    I then divided by .012 kg, and got that time = 21.87 s

    This answer is also wrong, not sure what happened.

    Thank you in advance for looking over this work.
     
  2. jcsd
  3. Nov 5, 2008 #2
    Slight bump, sorry, but homework's due by Thursday Night. >.>
     
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