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AravindG

  • 4 years ago

consider the arrangement shown in fig The string is wrapped around a uniform cylinder which rolls without slipping .The other end of the string is passed over a massless ,frictionless pulley to a falling weight .determine acceleration of falling mass m in terms of only the mass of the cylinder M ,the mass m and g

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  1. AravindG
    • 4 years ago
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    |dw:1327757617642:dw|

  2. anonymous
    • 4 years ago
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    i consider that both of the object have the same mass. F = ma F = mg , (gravitational force which pulling the square object) F = (m+m)a ,(grouping thses 2 object into the system) F/2m = a

  3. JamesJ
    • 4 years ago
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    No, that analysis isn't right. The cylinder has a moment of inertia, \[ I = \frac{Mr^2}{2} \] where \( r \) is its radius. Hence if the cylinder has velocity \( v \), then because \( v = \omega r \) where \omega is the angular velocity. Thus the total kinetic energy is \[ KE_{linear} + KE_{rotational} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \] \[ = \frac{1}{2}Mv^2 + \frac{1}{2}Mr^2\frac{v^2}{r^2} \] \[ = Mv^2 \] Now, given that, you can use conservation of energy to find the energy of the cylinder when the weight \( m \) has fallen a distance \( h \). From that, use the Work-Energy Theorem and the fact \[ \Delta U = \Delta KE_{block} + \Delta PE_{m} = Work = \int F \cdot dy = \int ma \ dy \] to find the acceleration \( a \) on the block.

  4. anonymous
    • 4 years ago
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    james why arent u bringing tension oor gravity into the poicture?

  5. anonymous
    • 4 years ago
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    James is correct, though I believe to correct a small typo, the rotational kinetic energy should be \[ \frac{1}{2} I \omega^2 = \frac{1}{2}\left(\frac{Mr^2}{2}\right)v^2r^2 = \frac{1}{4}Mv^2\] so the total kinetic energy is \[\frac{3}{4}Mv^2\]

  6. anonymous
    • 4 years ago
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    @salini we automatically include tension when we equate the velocity of the center of mass of the cylinder with the velocity of the hanging weight, and the F in the integral happens to be the gravitational force.

  7. anonymous
    • 4 years ago
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    when we equate the velocity and acceleration*

  8. AravindG
    • 4 years ago
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    wndful

  9. anonymous
    • 4 years ago
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    And I also had yet another typo, sorry.... \[\omega^2 = \frac{v^2}{r^2} \text{ not } v^2r^2\] :)

  10. JamesJ
    • 4 years ago
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    (@jem, thanks)

  11. AravindG
    • 4 years ago
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    i hav a doubt

  12. AravindG
    • 4 years ago
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    u see in eqns of moment of inertia like mr^2,mr^2/2

  13. AravindG
    • 4 years ago
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    what is the r basically??

  14. anonymous
    • 4 years ago
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    r is the radius of the cylinder. The moments of inertia of differently shaped objects are not the same. For a disc and a solid cylinder, I = mr^2/2 (assuming we're rotating about the axis poking through the radial center), whereas for a hoop or a hollow cylinder I = mr^2.

  15. anonymous
    • 4 years ago
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    total energy is k.e+p.e right so y arent u guys including that anywhere ?

  16. AravindG
    • 4 years ago
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    |dw:1327769992982:dw|

  17. AravindG
    • 4 years ago
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    |dw:1327770030252:dw|

  18. AravindG
    • 4 years ago
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    i rotate about axis q then r is??

  19. AravindG
    • 4 years ago
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    |dw:1327770065239:dw|

  20. anonymous
    • 4 years ago
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    @Aravind that's a more complicated moment of inertia, so it's not as simple to write down.

  21. anonymous
    • 4 years ago
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    change in kinetic energy=work done by net force=m*a

  22. AravindG
    • 4 years ago
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    hey actually i did that qn

  23. anonymous
    • 4 years ago
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    no. m*a is not the work done. It is the force.

  24. AravindG
    • 4 years ago
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    pls tell the sides are made of rods

  25. anonymous
    • 4 years ago
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    in ther question they have asked for a in terms of M,m.g but we have to include displacement here

  26. AravindG
    • 4 years ago
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    wat is momentr of inertia?

  27. anonymous
    • 4 years ago
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    You have to do an integral to find that. Post a new question about calculating moment of inertia if you'd like to know how.

  28. anonymous
    • 4 years ago
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    And no we don't, not for the acceleration.

  29. AravindG
    • 4 years ago
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    hey but salini gave me the answr 3mr^2

  30. anonymous
    • 4 years ago
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    when?

  31. AravindG
    • 4 years ago
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    hey we did that remember?

  32. anonymous
    • 4 years ago
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    is athat an equilateral triangle?

  33. AravindG
    • 4 years ago
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    ys

  34. AravindG
    • 4 years ago
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    made of uniform rods of length l

  35. AravindG
    • 4 years ago
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    q is centre of mass

  36. anonymous
    • 4 years ago
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    then use trignometry to find aq

  37. AravindG
    • 4 years ago
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    axis is q

  38. AravindG
    • 4 years ago
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    aq is l/root 3 then??

  39. anonymous
    • 4 years ago
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    3*m*(aq)^2

  40. AravindG
    • 4 years ago
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    that is 3 ml^2

  41. AravindG
    • 4 years ago
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    srry ml^2

  42. anonymous
    • 4 years ago
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    annu

  43. AravindG
    • 4 years ago
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    u gud in maths?

  44. AravindG
    • 4 years ago
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    salini?

  45. anonymous
    • 4 years ago
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    arayila enniku orakam verunu anywaY PARA NOOKATE

  46. AravindG
    • 4 years ago
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    http://www.twiddla.com/736142

  47. anonymous
    • 4 years ago
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    Using Forces, Tr = Mr^2 / 2 * a / 2r ----- where a is acceleration of block which will be double of cylinder. And T is Tension. => T = Ma/4 Now plugging in equation for block, mg - T = ma mg - Ma/4 = ma Therefore, a = 4mg / ( M + 4m)

  48. anonymous
    • 4 years ago
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    siddhantsharan, why is Tr = Mr^2 / 2 * a / 2r that is I*a/2r why? and how?

  49. anonymous
    • 4 years ago
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    Salini, In the fig. the string is attached to top of the sphere. From rotation dynamics we know that acceleration of the point at the top of sphere = a (COM) + (angular Acceleration) / r = a + a ( pure rolling => a = Angular acceleration / r) = 2a. Where a is acceleration of COM of sphere which is what we consider the sphere's acceleration to be during Force Concept. Now along the string Acceleration has to be equal. Hence, a ( top of sphere) = a (block) Therefore a(block) = 2a. In the answer I have assumed a to be acceleration of block and solved for a. Therefore Acceleration of sphere = a /2 Cheers. :)

  50. anonymous
    • 4 years ago
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    |dw:1327926218152:dw|@ siddhantsharan your 3rd step: acce of sphere = a (COM) + (angular Acceleration) / r

  51. anonymous
    • 4 years ago
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    Sorry. Typo error. Answer is still the same.

  52. anonymous
    • 4 years ago
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    YOu got the concept right?

  53. anonymous
    • 4 years ago
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    Tr = Mr^2 / 2 * a / 2r here, is a/2r the angular acceleration of cylinder if so T=M*a*r/4 not Ma/4

  54. anonymous
    • 4 years ago
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    Wait I'll draw it for clarity.|dw:1327878291024:dw|

  55. anonymous
    • 4 years ago
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    last doubt.....how did u get T*r ?

  56. anonymous
    • 4 years ago
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    |dw:1327880069640:dw| Hence , Force on sphere is Tension. Torque of Tension = r x F = r x T Hence Torque is Tr.

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