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Two Toilet Papers dropping-Rotational Inertia

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Okay, so I'm supposed to take two fresh rolls of toilet paper and drop them. One of which, I am supposed to let unravel while falling. The other, I just drop on its own. I'm supposed to find the height at which both will drop and hit the ground at the same time.
    2. Relevant equations
    Rotation Inertia for Toilet Papers: I= M (R-R2)^2

    Conservation of Energy: KE1 + PE1 = KE2 + PE2

    3. The attempt at a solution

    I know that

    KE1= 0 and PE2 = 0

    So

    PE1 = KE2

    mgh = (1/2) Iw^2 + (1/2)Mv^2

    Okay... this was all I wrote. At this point I realized, that I did not know the difference between the two.

    What would make the unraveling toilet paper any different from the one that falls?
     
  2. jcsd
  3. Nov 27, 2015 #2

    Doc Al

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    Staff: Mentor

    For one thing, the final PE of the unraveling toilet paper will not be zero. It stretches out as it unravels.
     
  4. Nov 27, 2015 #3
    Oh so it will be

    mgh= (1/2)(M)(Rf)((v^2)/(Rf^2)) +Mv^2 + PE?

    So what would PE equal? Would the mass change to just one sheet of paper rather than the whole roll?

    Or should I also set up Newton's 2nd Law?
     
  5. Nov 27, 2015 #4

    Doc Al

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    That's what I would do. See if you can compare the accelerations.
     
  6. Dec 2, 2015 #5
    Hate to bump this up again, but I've gotten pretty far but got jumbled on one part.

    So I did net torque = I(alpha)

    And did some substitution

    mg*R = (((1/2)(M) (R^2+R2^2) + MR^2))) (a/R)

    Thing is, would

    R^2+R2^2 be the same as the other R's?

    Sorry if I'm not being clear. But how would the formula play out? I know M cancels.
     
  7. Dec 2, 2015 #6

    Doc Al

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    This looks good. Note: R is the outer radius; R2 is the inner radius.

    Not sure what you mean. See my comment above.

    The linear acceleration will be in terms of both R and R2.
     
  8. Dec 2, 2015 #7
    @Doc Al

    Thanks for all your help, and I think I got it.

    After a bit of Algebra:

    (2/3)((g*R^2)/(R^2+R2^2))=a

    And the other a=g

    Is that all I can do?

    And then compare it?
     
  9. Dec 2, 2015 #8

    Doc Al

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    I get something slightly different, so double check your algebra.

    Once you have the two accelerations, figure out how the heights they must be dropped from have to relate for them to fall in the same time. (Hint: for a given distance, solve for the time to reach the ground.)
     
  10. Dec 2, 2015 #9
    Oh,

    So I did a bit.

    Is this correct?

    I brought the R from a/R to the other side.
    I wanted to cancel the 1/2 so I multiplied everything by 2.

    2gR^2 = (R^2+R2^2+2R^2) a?
     
  11. Dec 2, 2015 #10
    Oh I see my mistake! So this is right?
     

    Attached Files:

  12. Dec 2, 2015 #11

    Doc Al

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    This looks better. But simplify the expression within the parenthesis a bit--combine terms.
     
  13. Dec 2, 2015 #12

    Doc Al

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    Staff: Mentor

    Looks good.
     
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