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TomLikesPhysics

  • 2 years ago

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?

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  1. TomLikesPhysics
    • 2 years ago
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  2. TomLikesPhysics
    • 2 years ago
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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.

  3. cheesesticks
    • 2 years ago
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    gravity and friction

  4. cheesesticks
    • 2 years ago
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    tension is like tug of war

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

  6. cheesesticks
    • 2 years ago
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    o is it multiple choice?

  7. TomLikesPhysics
    • 2 years ago
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    Nope :D

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

  9. cheesesticks
    • 2 years ago
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    WAIT THATS A ROPE?

  10. mahmit2012
    • 2 years ago
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    |dw:1354312298270:dw|

  11. Vincent-Lyon.Fr
    • 2 years ago
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    That seems perfect!

  12. mahmit2012
    • 2 years ago
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    |dw:1354312672595:dw|

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

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

  15. TomLikesPhysics
    • 2 years ago
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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".

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

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

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

  19. Vincent-Lyon.Fr
    • 2 years ago
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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 \(\Large \tau\) = \(I\;\alpha\)

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

  21. shamim
    • 2 years ago
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    ok

  22. TomLikesPhysics
    • 2 years ago
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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.

  23. Vincent-Lyon.Fr
    • 2 years ago
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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.

  24. TomLikesPhysics
    • 2 years ago
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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.

  25. Vincent-Lyon.Fr
    • 2 years ago
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    Exactly!

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

  27. Vincent-Lyon.Fr
    • 2 years ago
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    yw :)) goodbye!

  28. TomLikesPhysics
    • 2 years ago
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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.

  29. Vincent-Lyon.Fr
    • 2 years ago
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    NP, mahmit deserves the award.

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