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Angular velocity and acceleration

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Using your data from trial one, calculate angular velocity w, and angular acceleration, a
    My data from trial one is period of rotation=.791s

    2. Relevant equations
    w=(change in angle/change in time)
    a=(change in angular velocity/ change in time)

    3. The attempt at a solution

    avg w=360deg/.791s= 455deg/s

    avg a=(455deg/s)/.791= 575deg/s^2

    I believe this is correct but I thought I would ask anyways. I asked my teacher several days ago whether the question was asking for average or instantaneous velocity/acceleration and he hasn't responded and it is past due so if anyone would like to comment on what they think about that.( The question I posted is word for word) also I am having trouble understanding the concept of instantaneous angular velocity because as the time is approaching zero then so is the angle so I dont see the point. thanks in advance for any help.
     
  2. jcsd
  3. Nov 26, 2008 #2
    in order to have an angular acceleration there must be a torque involved somehow. How has the object changed states? did it start from rest? we need another period of rotation measurement to get the average acceleration.
     
  4. Nov 26, 2008 #3
    This was an experiment I performed. I tied a rubber stopper to one end of a cord then the cord was threaded through a small plastic tube about the size of an average pen. then the other end was tied to a bunch of metal washers. so holding onto the tube I spun the stopper until I had it as close to equilibrium as i could get it then I timed twenty rotations. so it was already moving when I began timing. do you need any more information?
     
  5. Nov 26, 2008 #4
    hmm, can't quite visualize it. can you find an image of these materials online?

    your measurement of the period of rotation does give you the average velocity over that period of time. maybe someone else on the forums is familiar with this lab, cause now I'm interested
     
  6. Nov 26, 2008 #5
    exactly like this>>>
    http://www.thesciencefair.com/Merchant2/graphics/00000001/RubberStopper2Hole_M.jpg


    sort of like the middle tube but with a smaller diameter and it was grey:)
    http://img.alibaba.com/photo/51231264/Colored_Plastic_Tube.jpg

    washers>>
    http://www.boltdepot.com/images/Chrome/chrome-flat-washers.jpg

    basically I was just swinging something tied to string over my head. the washers were acting as a counterweight so when the system isnt moving up or down in the tube I know that the force due to gravity acting on the washers is equal to the force pulling the rubber stopper outwards. of course they would only be equal if friction did not exist and the angle of the string swinging around was 90 degrees with the hanging mass which it definitely would not be.

    I have to go do something I will try and explain it better in awhile if I need to
     
  7. Nov 26, 2008 #6
    [​IMG]

    thanks for giving me an excuse to play around with sketchup ;).

    So we set the forces on each of the masses equal to one another. However in order to get the centripetal acceleration I think we need to know r or theta somehow... where there any measurements of that kind?
     
  8. Nov 26, 2008 #7
    my theoretical centripetal force was 0.604 n, calculated using the equation Ftc=M(4(pi^2)R/T^2), where Ftc= theoretical centripetal force in n, (rubber stopper)M=0.013kg, R=.736m, T= .791s. my "experimental" centripetal force was 0.441n, Fec=MG where Fec= Experimental Centripetal force in n, (washers)M=0.45kg, G= acceleration due to gravity. part of the experiment is explaining the error here because it is supposed to be negligible. If the angle is 90°, then the tension in the string (and therefore the centripetal force) is just equal to Mg however it is not possible for me to maintain 90degrees. BUTTT!!!!!!!! is this really necessary to calculate angular acceleration Cant I just divide change in angular velocity by change in time.??
     
  9. Nov 27, 2008 #8
    [itex]
    a= \frac{ \Delta v }{ \Delta t}
    [/itex]

    so you always are looking at 2 velocities and how they change in time.

    wait so you do know the radius? did you measure the length of string coming out of the top of the tube or something?
     
  10. Nov 27, 2008 #9
    Yes that is exactly what I did. For the other trials I simply changed the hanging mass by adding washers, and changed the radius. by that's not really important for this question I don't think so anyways. I have already calculated centripetal force and acceleration.
     
  11. Nov 27, 2008 #10
    The length of the string was assumed to be the radius because the angle would be difficult to measure.
     
  12. Nov 27, 2008 #11
    If I was spinning the rubber stopper at what was supposed to be a constant angular velocity then wouldn't the angular acceleration actually be zero since the change in angular velocity would have been zero. I think the fact that you didn't know how to help me forced me to figure it out on my own. did you do that on purpose? I think I will have to answer angular acceleration is equal to zero unless you can explain otherwise before I get off work tonight. I really need to hand this in asap its 10:47 my time so I start at 12 and I am off at 8 and I want to hand it in about 9-930 my time anyways thnx for making me think it through.
     
  13. Nov 28, 2008 #12
    Sorry for the mixed messages, I'm a bit rusty atm. And the question of the angular acceleration is a bit wierd since it seems to be zero.


    I thought the problem through yesterday, and it simplified a bit. The force exerted by m2 has to equal the total force exerted on m1. There are two forces on m1, one due to gravity and the other due to centripetal acceleration.

    [itex]
    m_2g=\sqrt{ (m_1g)^2 + \left( \frac{m_1 v^2}{r} \right)^2}
    [/itex]

    so on the right is the magnitude of the force on m1. We know the value of all the variables except r. So solve for r and plug it in to your equation

    [itex]
    F_{centripetal} = \frac{m_1 v^2}{r}
    [/itex]
     
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