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Find inner radius of hollow cylinder given inertia

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A roll of toilet paper ( a partially hollow cylinder with R2=7.0 cm, M=320 g, I=6.0x 10 ^(-4) kg m is mounted on an axle. initially at rest, until a child grabs the end and starts running at a constant linear acceleration.

    part a) what is the inner radius (R1)


    2. Relevant equations

    I know the equation for a hollow cylinder's moment of inertia is I=1/2 M (R1^2 + R2^2)

    3. The attempt at a solution

    I converted all my units to S.I. units

    7 cm= 0.07m
    320g= 0.320kg

    I rearranged the equation as
    R= ({(2*I)/M} -R2^2) ^(1/2)
    2 (6.0x 10^-4)=0.0012 kg m
    0.0012 kg m/.320 kg= .00375m
    0.00375 m-(0.07m)^2= -0.0015 m^2
    I would now need to take the square-root of this number, but its a negative number. I cannot seem to find what I did wrong. and I need part a) to find the other four parts of this question. Thanks to anyone who takes the time to help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 19, 2011 #2

    PhanthomJay

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    There is likely an error in the given value of I = 1.6*10^-4 kg.m^2. Bummer. Did you copy it down correctly?
     
  4. Nov 19, 2011 #3
    its straight out of the worksheet. Thanks though. I will email the professor
     
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