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Simple problem involving pulley with mass

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    The descending pulley (disc shaped) shown in the figure have a radius 20 cm and moment of inertia 0.20 kg-m2. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1.0 kg.


    2. Relevant equations



    3. The attempt at a solution
    The moment of inertia for a disc about its CoM is: ##\frac{MR^2}{2}##. Equating this with 0.2 kg-m2 and substituting R=0.2 m, M=10 kg.

    Let T2 and T1 be the tensions in the strings. T2 on the left of disc and T1 on right of disc.

    If A is the acceleration of disc, then 2A is the acceleration of disc. Also, let ##\alpha## be the angular acceleration of disc the direction of which is anti-clockwise.

    Newton's second law on disc: ##10g-(T_1+T_2)=10A## and for block: ##T_2=2A## (mass of block is 1 kg).

    Torque about CM of disc: ##(T_1-T_2)R=I\alpha \Rightarrow (T_1-T_2)(0.2)=(0.2)\alpha \Rightarrow T_1-T_2=\alpha##.

    Next is equating the acceleration along the string. The point of contact of string with the disc on the left goes down with an acceleration of ##A+\alpha R##. Hence, ##A+\alpha R=2A##.

    I have enough equations but solving these equations gives me a wrong answer. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Nov 14, 2013 #2

    Doc Al

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    Staff: Mentor

    Your analysis seems reasonable. What did you get for the acceleration of the disc? (And what was the "right" answer?)
     
  4. Nov 14, 2013 #3
    Hi Pranav...

    It looks alright to me .What is the correct answer and the answer you are getting?
     
  5. Nov 14, 2013 #4
    Hi Doc Al and Tanya! :)

    I had ##A+\alpha R=2A \Rightarrow \alpha=5A##. Since ##T_1-T_2=\alpha## and ##T_2=2A##, we have ##T_1=7A##.

    From Newton's second law on disc: ##10g-(T_1+T_2)=10A \Rightarrow 10g-9A=10A \Rightarrow A=10g/19##.

    Hence, acceleration of block is ##2A=20g/19=10.31 \,\, m/s^2##.

    The given answer is ##5.96 \,\, m/s^2##.
     
  6. Nov 14, 2013 #5
    10.31 ms-2 is the correct answer :smile: .
     
  7. Nov 14, 2013 #6
    Thanks for the check! :)

    It doesn't look so. The acceleration of disc comes out to be 5.15 m/s^2.
     
  8. Nov 14, 2013 #7

    Doc Al

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    Staff: Mentor

    I agree with that answer.

    I don't know what that's supposed to be. Did they give a solution, or just an answer?
     
  9. Nov 14, 2013 #8
    Only the answer. :(
     
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