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Angular Speed and Kinetic energy

  1. Jun 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Ok I have the answers to parts a and b, which I am 100% confident in since they are simple computations
    Question is in attachement
    a)74.9 rad/s
    b)168.08 rad

    My issue comes in when I have to get parts c
    The mass of the cylinder is 8.7kg,
    its diameter is .18m,
    its .35m long and
    angular acceleration of 33.2m/s^2


    **note please ignore the markings on the image.

    2. Relevant equations
    w=theta/t
    w=v/r

    3. The attempt at a solution
    for part c) angular speed=angular velocity, w=theta/t
    w=15.5rad/s
    I am not sure about how to get the angular speed at 35 radians
     

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    Last edited: Jun 15, 2016
  2. jcsd
  3. Jun 15, 2016 #2

    CWatters

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    It's not correct. Please show your working. What formula did you use?

    i is the "moment of inertia" not just "inertia". Check the equation for the moment of inertia of a cylinder. I don't recognise i=S*r^2*d*m where S=r*theta
     
  4. Jun 15, 2016 #3
    For part c I did 35rad/2.25s=15.5, since angular speed =angular velocity wouldn't they have the same formula? Unless there is an angular speed equation that I am not aware of?
    And for part D Ill hold off on it for now, since I edited my post an already deleted it.
     
  5. Jun 15, 2016 #4

    CWatters

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    I'm off to bed as soon as I post this because it's after midnight here.
    That would be correct if it was rotating at a constant angular velocity but it's not, it's accelerating.

    If a car accelerated from rest and covered 100m in 10 seconds it's final velocity wouldn't be 10m/s. That would be the average. It would be slower at the start and faster at the finish. You would use the SUVAT equations to find the final velocity.

    PS Check your answer for b). I got a different answer.

    Back tomorrow.
     
  6. Jun 15, 2016 #5
    For part b all I did was multiply the angular velocity by the time to get the Angle in radians. so 74.7*2.25=168.08rad

    I am getting 48.2rad/s for part c, which SUVAT formula did you use? Please tell me?

    And for part D I found out that since the object can rotate or accelerating(translating) in a linear direction(PLEASE READ THE PROBLEM BACK CAUSE IM NOT SURE IF ITS BOTH, just to confirm with me) I need to add the kinetic energy of
    KElinear=(1/2)*m*v^2 and KErotate=(1/2)*i*w^2
    where i is the inertia, an its equal to i=(1/2)*m*r^2 because its a solid cylinder I checked equation online
    V is Vcylinder=√((4/3)*g*h) and
    w=v/r.

    For v=(√((4/3)* 9.8* .35))=2.138
    KElinear=.5*(8.7)*(2.138^2)=19.88joules

    For r= .09m converted cm to m i=.5* 8.7* (.09^2)=.035 w=(2.138/.09)=23.7
    KErotate=.5* .035* (23.7^2)=9.82Joules

    So in summary, I need to know what formula you used for part c, then I would like to confirm if this entire problem has just rotational kinetic energy or both, rotational and translation, and if I missed anything please let me know. I won't be able to respond when you wake up because you are 5 hours ahead of me, so please put in as much info as you can cause I need to master this problem before I go into my class, Thank you.
     
    Last edited: Jun 15, 2016
  7. Jun 16, 2016 #6
    Sorry for double post, but did you get around to look at this CWatters?
     
  8. Jun 16, 2016 #7

    CWatters

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    The velocity isn't constant so that doesn't work. Have another go using the equations for motion under constant acceleration.
    I got the same answer. Forum rules don't allow us to just give the answers or do too much of the work for you. If you show me your working I can give hints and point out where you have made a mistake if any.
    The problem mentions a flywheel. These normally rotate on a fixed axis. So I think this problem only involves rotation not translation.
    Yes that is correct.
    Yes that is correct
    That's not correct.
    ω is the angular velocity. You calculated the angular velocity after 2.25 seconds as 74.9 rad/s in question a).
     
  9. Jun 16, 2016 #8

    CWatters

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    Useful reference (better version)..
    SUVAT Table.jpg
     
    Last edited: Jun 16, 2016
  10. Jun 16, 2016 #9
    For part c, the angular speed is an inital or final speed? meaning is it w0 or w?
     
  11. Jun 16, 2016 #10

    CWatters

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    The question asks for the angular velocity after turning. So it is the final angular velocity you need to calculate.
     
  12. Jun 16, 2016 #11
    since theta=35rad, w0=74.7rad/s, T=2.25s, A=33.2,
    I would use V = w0 + A*T correct?, ok Ived tried a few of them with V in them and none are giving me 48.2. I used, for example tried theta=((w0+v)/2)*t and V^2=w0^2+2*A*theta, can you please tell me what I am doin wrong?
     
    Last edited: Jun 16, 2016
  13. Jun 16, 2016 #12

    CWatters

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    Sorry I had to delete a few posts. I'm rushing too much. The equation you need is..

    ωf2 = ωi2 + 2αθ
    which is similar to
    V2 = U2 + 2as

    ωi = 0 (starts from rest)
    α = 33.2 rad/s2 (from the problem)
    θ = 35 rads
     
    Last edited: Jun 16, 2016
  14. Jun 16, 2016 #13
    Thank you. And for part b did you get an answer around 84 point something, casue I used the theta=w0(t)+.5*a*t^2 equation?
     
  15. Jun 16, 2016 #14

    CWatters

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    Yes 84.04 rads.
     
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