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Rotation of a Rigid Body

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 14cm in diameter and has a mass of 2.4 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.54 Nm .

    12.P70.jpg

    2. Relevant equations
    so m1 = 4.0kg
    m2= 2.0 kg
    M = 2.4kg
    r= 0.07m
    Tf (torque due to friction) = 0.54 Nm
    ƩFy (for mass 1)= T2 - m1*g
    m1*ay1 +m1*g= T2 so T2 = m1(ay1+g) this becomes T2= m1(ay + g)

    ƩFy (for mass 2)= T1 - m2*g
    m1*ay2 + m2*g = T1 so T1 = m2(ay2 + g)
    but since ay1 = -ay2 = ay this becomes T1 = m2(g - ay)

    Ʃτ= T2*R - T1*R - Tf

    3. The attempt at a solution

    Using what i put up there i get the following formula

    τnet = R(T2 - T1) - Tf = R(m1(ay+g) - m2(g-ay)) - Tf .... equation 1
    since τ= Iα, I= 1/2 MR^2 and since α= -ay/R

    equation 1 becomes..

    1/2MR^2 * (-ay)R = R(m1(ay+g) - m2(g-ay)) - Tf ..... solving for ay with the given data i found ay=-1.65 and thus using the kinematic equation i found Δt= 1.55s but this is wrong.. :( i would appreciate if anyone points out my error or mistake, thanks
    EDIT: sorry about this, they are asking us to find the time it takes for mass 1 to hit the floor, starting at rest.
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    Redbelly98

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    Well, you never said what the actual question is. Looks like you are to figure out how long it takes the 4 kg mass to hit the floor? And everything is initially at rest?

    I agree with the 1.65 m/s2 acceleration, so your mistake is probably in applying the kinematic equations. Can you show that work?

    p.s. It's kind of confusing that you chose T2 for the rope holding m1, and T1 for the rope holding m2. Nonetheless you did find the correct acceleration.
     
  4. Nov 6, 2011 #3
    thanks! and sorry about that! Yes it is kinda confusing but im happy to know that someone else got the same acceleration. Well, I tried applying the kinematic formula:

    yf = yi + Vi*t + 1/2 a t^2 ... t = delta t... and i chose the initial point to be 0, so the final point would be -2m since it goes down... also, the initial velocity is zero since it starts from rest... so..

    -2.0m = 0 + 0 * t + 1/2 * (-1.65m/s^2) * t^2
    thus...

    -2.0 m / -1.65 m/s^2 = t^2.... applying square root i get 1.1s......which seems to be the right answer... what the hell was i doing wrong?? oh wow i think i was not dividing properly... lol so all this time i actually had the right answer before my eyes but never realized that i was making a silly mistake... my mistake was applying the kinematic formula.. well thanks for your help have a nice day :)
     
  5. Nov 7, 2011 #4

    Filip Larsen

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    [STRIKE]I get an acceleration of around 3.7 m/s and around 0.7 s.[/STRIKE] The only thing I can spot is that you multiply with R on the left-hand side of you last equation in your first post instead of dividing, but that just looks like a typo.

    Edit: Just for the record, I made a sign error. Doing it properly I too get 1.7 m/s2 and 1.1 s.
     
    Last edited: Nov 7, 2011
  6. Nov 7, 2011 #5

    Redbelly98

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    I get 1.1 s as well. :smile: Be careful, it actually drops 1.0 m in 1.1 s, not 2.0 m.
     
  7. Nov 7, 2011 #6
    wait, how come it only drops 1 meter? i thought it was gonna hit the ground? are you saying that it doesnt actually hit the floor?..... how come i got the same answer then? sorry but that made me confused...

    EDIT! OH OH! sorry i got 2m because i multiplied both sides of the last equation by 2 so i get rid of the 1/2 on the right side of the equation. ofc its gonna drop 1m cause thats the height of the mass1, i think i was looking at a different problem... lol thanks! you guys rock!
     
    Last edited: Nov 7, 2011
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