Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular velocity of an object with radius and time

  1. Apr 3, 2014 #1
    Greetings,
    I fell upon a certain question and it seemed that no matter what I did I was unable to solve it. I wished to know how if given a radius of 0.6m and the time for one revolution: 0.3s, to find the angular velocity in radians of the object. Consider it to be a point particle. Thanks
     
  2. jcsd
  3. Apr 4, 2014 #2

    berkeman

    User Avatar

    Staff: Mentor

    When you use Google to search for applicable equations, what do you find?
     
  4. Apr 4, 2014 #3
    I get mostly websites explaining that the angular velocity is equal to the angle moved over time. I guessed that the velocity should then be equal to (2pi/t)rad/s since the angle is equal to 2pi radians. But I'm not sure because when I tried entering the answer it termed it wrong. Maybe I made a mistake as the question was asking for the rotational kinetic energy of a spree, to be more precise. The mass was 18kg so I applied the formula 18*r^2*w^2/2. Did I make a mistake here if I got the angular velocity right? Thanks
     
  5. Apr 4, 2014 #4
    The r is not squared.
     
  6. Apr 4, 2014 #5
    Isn't the formula for rotational kinetic energy 1/2Iw^2 with I the moment of inertia equal to the mass multiplied by the square of the lever arm, which in this case is the radius?
     
  7. Apr 4, 2014 #6
    Yes, my mistake. I did not realize you want the kinetic energy.

    If you show your calculations someone will figure out where is the error. If there is one.
     
  8. Apr 4, 2014 #7
    Ok. Here's the what I did: I first calculated the angular velocity by dividing 2pi by the time taken, 0.3 since the total angle covered was 2pi radians and I want the angular velocity. Then I squared my velocity, multiplied it by the square of the radius, 0.36 (r=0.6) and then by the mass 18. Finally I divided it by 2 as the formula for rotational kinetic energy is 1/2Iw^2. Did I go wrong anywhere? I checked all figures and units given in the question and they are all right. Thanks
     
  9. Apr 5, 2014 #8
    Formulas would be much better then explaining in words.
    And what are you results?
    What is your value for angular velocity? For kinetic energy?
    How do you know is not correct?

    Showing the actual problem would be a good idea too.
     
  10. Apr 6, 2014 #9
    Ok. I know it is wrong as when someone enters the value obtained for the answer it is corrected automatically by an online program.In calculating pi is 3.14159265358979
    I used colours to guide the eye: red is for numbers related to the question, orange is for the general formula, pink is for the formula I used and blue is for the answer obtained.

    Here is the question: A uniform-density sphere whose mass is 18 kg and whose radius is 0.60 m makes one complete rotation every 0.30 s. In J, what is the rotational kinetic energy of the sphere?.

    I calculated everything in SI units, so angular velocity is in rad/s.So I applied the formula ω=θ/t. In this case it results in ω=2pi/0.3. I calculated it to be:20.943951023932 rad/s. I proceeded and calculated the moment of inertia: I=mr2 which here gives I=18*0.36=6.48. Then I multiplied this value by the angular velocity twice and divided by two. That is:
    K.Erot=1/2Iω2. In this case it is equal to: K.Erot=3.24*438.64908449286 obtaining a final result of 1421.22303375687.

    When I typed that in it is marked as wrong. However I don't see any mistakes here. Did I miss something? Thanks for helping
     
  11. Apr 6, 2014 #10

    jbriggs444

    User Avatar
    Science Advisor

    I=mr2 is the formula for the moment of inertia of a point-like mass at a radius r from a chosen axis or for a hoop or cylinder of radius r around its center. The key observation for all of those is that all of the mass of the object is at radius r from the axis of rotation.

    The mass of a uniform sphere is not all at distance r from an axis through the sphere's center. All but the very outer surface right at the "equator" is closer to the axis than that. So the moment of inertia of a uniform sphere must be less than I=mr2.

    The correct moment of inertia could be determined by an integral over the volume of the sphere. But it's easier to simply Google it.

    Also, try carrying fewer significant digits in your reported results. Lots of digits are annoying and unhelpful.
     
  12. Apr 6, 2014 #11
    I see. So Actually i was supposed to use integration to find the moment of inertia. Okay that explains a lot. Thanks.
     
  13. Apr 6, 2014 #12
    It's more likely you were just supposed to look it up in a table.
    Or Google it, as jbriggs suggested already.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook