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TomLikesPhysics

What forces act on the block with mass m2 (as seen on the picture)? Gravity for sure but I guess the cylinder does also something. Perhaps Tension plays also some role?

  • one year ago
  • one year ago

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  1. TomLikesPhysics
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    • one year ago
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  2. TomLikesPhysics
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    The cylinder can rotate without friction and is fixed there. The moment of inertia of the cylinder is 1/2*M1*R^2.

    • one year ago
  3. cheesesticks
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    gravity and friction

    • one year ago
  4. cheesesticks
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    tension is like tug of war

    • one year ago
  5. TomLikesPhysics
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    Sry, I forgot to mention that there is no friction.^^

    • one year ago
  6. cheesesticks
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    o is it multiple choice?

    • one year ago
  7. TomLikesPhysics
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    Nope :D

    • one year ago
  8. Vincent-Lyon.Fr
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    Weight, normal force, tension in the rope.

    • one year ago
  9. cheesesticks
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    WAIT THATS A ROPE?

    • one year ago
  10. mahmit2012
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    |dw:1354312298270:dw|

    • one year ago
  11. Vincent-Lyon.Fr
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    That seems perfect!

    • one year ago
  12. mahmit2012
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    |dw:1354312672595:dw|

    • one year ago
  13. TomLikesPhysics
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    Do you mean with T the Tension or Torque? If you mean Tension - why is T*R=I*alpha?

    • one year ago
  14. Vincent-Lyon.Fr
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    T is tension. Torque is: \(\large \tau\normalsize=TR\)

    • one year ago
  15. TomLikesPhysics
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    So the moment of inertia of the cylinder is connected with the tension... I have never seen this formula: T=I*alpha. Where does it come from? Normally Torque = Moment of Inertia * angular acceleration but now you say it also equals a force (tension)? I think the units do not add up because normally I*alpha gives you "Nm" and not "N".

    • one year ago
  16. shamim
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    \[m _{2}gsin \theta -T=m _{2}a\]

    • one year ago
  17. shamim
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    \[R=m _{2}gcos \theta\]

    • one year ago
  18. shamim
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    R is a force acted by the surface on the body of mass m2

    • one year ago
  19. Vincent-Lyon.Fr
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    I don't think anybody wrote : T=I*alpha, where T is a tension. This is dimensionnaly wrong. Equivalent of N's 2nd law for rotating bodies is: net torque \(\Large \tau\) = \(I\;\alpha\)

    • one year ago
  20. Vincent-Lyon.Fr
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    @shamim careful, R is the radius of the pulley. The problem has already been solved by mahmit2012

    • one year ago
  21. shamim
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    ok

    • one year ago
  22. TomLikesPhysics
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    Ok, but why is T*R=I*alpha? I only saw this guy when T meant Torque so I have no clue where mahmit2012 got it from.

    • one year ago
  23. Vincent-Lyon.Fr
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    Because torque acting on the pulley is the moment of the force \(\vec T\) about the axis of the pulley. Moment of \(\vec T\) is simply \(RT\) because \(\vec T\) acts at right angles to the radius of the pulley.

    • one year ago
  24. TomLikesPhysics
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    Oh ... Torque = radius cross Force. Here the force on the cylinder is the tension. Also Torque equals moment of Inertia times alpha.

    • one year ago
  25. Vincent-Lyon.Fr
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    Exactly!

    • one year ago
  26. TomLikesPhysics
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    k, I think I get it now. Thanks a lot Vincent-Lycon.Fr and mahmit2012.

    • one year ago
  27. Vincent-Lyon.Fr
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    yw :)) goodbye!

    • one year ago
  28. TomLikesPhysics
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    A shame that I can not get you booth a "Best Response". mahmit2012 solved the thing but you (Vincent-Lyon.Fr) explained it further.

    • one year ago
  29. Vincent-Lyon.Fr
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    NP, mahmit deserves the award.

    • one year ago
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