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Rotational motion problem

  1. Nov 7, 2005 #1
    A spool of thin wire rotates without friction about its axis. A man pulls down on a wire connected to an inner radius with force of 16.3 N. How long does it take to increase the angular velocity of the spool from 15.0 rad/s to 33.0 rad/s? Icm = 0.550 kgm2 Inner radius, r = 0.300 m Outer radius, R = 0.560 m.
    I got this part, the answer was 2.02 s.
    For the second part, it says, Consider the same spool as in the previous problem. Now, instead of pulling on the wire, the man attaches the wire to a mass of 0.900 kg. The man lets go of the mass, and the spool starts to turn. What is the speed of the mass after it has travelled downward a distance of 0.650 m? (Assume there is no energy lost to friction.)
    Here's what I did
    Conservation of energy: [tex] (1/2)Iw_f^2 - (1/2)Iw_i^2 = W [/tex]
    (1/2)(.550)(33.0)^2 - (1/2)(.550)(15)^2= 237.6 J
    [tex]W= \tau (\Theta_f - \Theta_i) [/tex]
    [tex]237.6 = \tau (.650) [/tex]
    [tex]\tau=109.5 [/tex]
    [tex]\tau= I \alpha[/tex]
    [tex]\alpha= 199.1 [/tex]
    I don't think I'm doing this the right way.. can someone please help me?
  2. jcsd
  3. Nov 7, 2005 #2
    The mass is being pulled down by gravity but is being 'pulled up' by the rope. Gravity has to pull down the mass as well as turn the spool of string.
  4. Nov 7, 2005 #3
    So how would I get that to come into play?
  5. Nov 7, 2005 #4
    Gravity pulls the mass down with force F. The force F is then transferred to the spool of string and it experiences a _____ because it is being pulled off center.
  6. Nov 7, 2005 #5

    so I would use [tex] /tau = r x F [/tex] where F=mg, this would be the pull from gravity..
    Isn't there a normal force acting on it too?
  7. Nov 7, 2005 #6
    Where is this normal force coming from? The spool isnt on the ground.
  8. Nov 7, 2005 #7
    Ok so I found the torque to be 2.67 Nm.
    Was my inital idea of using conservation of energy to solve it correct?
    Having so many "given" variables confuses me on where to go from here.
  9. Nov 7, 2005 #8
    Well now you've got your ball's torque and its moment of inertia. Find the angular acceleration, then find the linear acceleration at the edge. Then find out how fast the mass is dropping once the string has stretched 0.65m
  10. Nov 7, 2005 #9
    Ok so
    [tex] /tau= I /alpha [/tex]
    so [tex] 2.67 = .550 * /alpha [/tex] = 4.85 rad/s^2
    [tex] /alpha= ar [/tex]
    a= 4.85(.300)
    a= 1.46
    x= v_init * t + (1/2)at^2
    .65= 2.02v + (1/2)(1.46)(2.02)^2
    solving for v gave me -1.15 m/s...
    which wasn't right...
    Am i using the right radius?
  11. Nov 8, 2005 #10
    Where did you get your value for t?
    You were fine up to a = 1.46m/s^2

    From there try [itex] v_f^2 = v_i^2 + 2ad [/tex]

    I get v = 1.9m/s towards the ground.
  12. Nov 8, 2005 #11
    I was using the t that I got from the first part of the problem. I don't think that is the right time to use in this part though, since the man was pulling on it before and now a mass is.
    I tried using the equation you used, and the answer was still wrong..
    Should I try energy conservation?
  13. Nov 8, 2005 #12
    I really don't know how to do this problem and if I am even using the right radius. Can someone please help? Thanks
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