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Mass in relation to Tangential Force

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    A child exerts a tangential 42.4 N force on the rim of a disk-shaped merry-go-round with a radius of 2.20 m. If the merry-go-round starts at rest and acquires an angular speed of 0.0870 rev/s in 4.50 s, what is its mass?

    2. Relevant equations
    t=I[tex]\alpha[/tex]
    I=.5mr2

    3. The attempt at a solution
    42.4=.5*2.22*.8087*m
    m=21.665

    Would the above be correct? I'm hesitant about entering it as I only have one more attempt to solve this problem. If not would converting the revolutions per second into radians per second be the right thing to do? Anyone see anything I'm missing? Any help is appreciated. Thank you.
     
    Last edited: Nov 30, 2009
  2. jcsd
  3. Nov 30, 2009 #2
    There is mistake in your formula.

    Firstly, torque formula is [tex]\tau = F.R.sin(\theta) (\theta [/tex] is angle between F and R)

    Besides, torque can be found by using this formula: [tex]\tau = I. \alpha (\alpha [/tex] is angular acceleration)

    p/s: what is '0.8087' number in your calculation? I dont get it :(
     
  4. Dec 1, 2009 #3
    Ah, thank you for that. That number was me mistyping 0.0870. Should I convert the angular speed to radians per second before I solve for angular acceleration or leave it as is?
     
  5. Dec 1, 2009 #4

    ideasrule

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    Homework Helper

    Try doing dimensional analysis. If you convert to radians, you'll get an angular acceleration in radians/s^2. If you leave it as is, you'll get an acceleration in revs/s^2. Both are correct, but the formula torque=I*alpha requires that alpha be in radians per second squared.
     
  6. Dec 1, 2009 #5
    Solved for it correctly thanks to the help of both of you. Thank you both.
     
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