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Time required for disk to reach angular speed?

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A uniform disk of mass M = 3 kg and radius r = .22 meters is mounted on a motor through its center. The motor accelerates the disk uniformly from rest by exerting a constant torque of 1 n * m

    What is the time required for the disk to reach an angular speed of 800 rpm

    2. Relevant equations
    tor = i a
    i = rotational intertia
    a = acc

    3. The attempt at a solution
    So I convert 800 rotations / min to 80/6 rotations a second.

    since angular acceleration = Torque / rotational interta, and I have torque is 1 n * m and rotational intertia = (1/2) m r^2

    so I have (1) / (1/2)(3)(.22)^2 = 13.77

    so that should be my angular acceleration right?

    Now to find time all I have to do is use euqation
    [itex] W_0 + ax t = W_x [/itex]

    w_o = 0 so

    t = (W_x)/(a_x)

    t = 80/(6)(13.77) = .9682 s

    but my books answer was 6.08 seconds.

    Does anyone know where I went wrong? I have a feeling I calculated acceleration wrong :/
     
  2. jcsd
  3. Apr 29, 2017 #2

    kuruman

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    Wrong. That looks like the moment of inertia.
     
  4. Apr 29, 2017 #3
    Wow.. you are right.

    So angular acc = I / Torque
    so acc = (1/2)mr^2 = (1/2) (3) (.22)^2
    But doing this I got angular acc = .0726 rad /s^2

    Now using w_0 + ax t = wx

    I get t = wx/ax = (80/6) (1/.0726) which = 183.65 which is still the wrong answer??
     
  5. Apr 29, 2017 #4

    kuruman

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    Incorrect. It would help if you got into the habit of appending units to any number you write down. What are your units in this case?
     
  6. Apr 29, 2017 #5
    Sorry I will do that from now on.

    https://www.boundless.com/physics/t...een-torque-and-angular-acceleration-319-6062/

    This page makes the statement "Torque is equal to the moment of inertia times the angular acceleration."

    moment of intertia = (1/2)MR^2 and torque = 1 N m

    Why can't I set angular acceleration = Torque / moment of intertia?
     
  7. Apr 29, 2017 #6

    kuruman

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    You can. What you had before is the inverse of that. You have to be careful because small things like that will trip you up. Now can you finish the problem?
     
  8. Apr 29, 2017 #7
    I'm not sure :(

    So I can state "acceleration = Torque / moment of intertia"

    torque was given to be 1 N * m
    moment of intertia given by my book is (1/2)MR^(2) = (1/2)(3kg)(.22m)^2 = .0736 kg m^2

    so 1 N / .0736 kg m = 13.77 N/ (kgm) = acceleration

    What am I missing? Why is this still the wrong acceleration value?? I'm literally plugging the values right into the equation I dont understand how it could be wrong..
     
  9. Apr 29, 2017 #8

    kuruman

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    You are missing the correct units for the acceleration. You list them as N/(kg m). Is that what they are? What does your textbook say they ought to be?

    Edit: The number 13.77 is correct. It's the units you need to understand because that matters for finishing the problem correctly.
     
  10. Apr 29, 2017 #9
    I see. I took the following steps and got 13.77 rad/s^2

    ( (1kg m^2)/(s^2) ) / ( (3kg/2)(.22m)^2 )

    = (2) / ( 3 * (.22)^2 ) s^2

    = 13.77rad s^(-2)
    So is my value for acceleration correct??
     
  11. Apr 29, 2017 #10

    kuruman

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    Yes it is. Now proceed to find the required time.
     
  12. Apr 29, 2017 #11
    Okay now that I found angular acceleration = 13.77 rad s^(-2)

    I use equation

    W_o + (angacc)t = W_final

    Well I know W_final = 800 rpm (I'm going to assume this to be rotations per minute not radians per minute) so 13.33 rotations per second, 1 rotation = 2pi so

    2pi(13.33) rad/s

    okay I'm just going to stop here.

    I completely get the point you were trying to get to me here.

    I got the answer that was written in my book... 6.08 seconds..

    I see now. Write down your units. and follow them. Each. And. Every. Time.

    Thank you for your patience with me.
     
  13. Apr 29, 2017 #12

    kuruman

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    Good job! :smile:
     
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