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RaphaelFilgueiras

  • 3 years ago

A spherical uniform shell rotates in a vertical axis without friction.One string without mass passing through sphere equator and a sheave has a small object at its end.what is the velocity of object,when he down a distance H.

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  1. RaphaelFilgueiras
    • 3 years ago
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  2. RaphaelFilgueiras
    • 3 years ago
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    Use work energy theorem

  3. celos
    • 3 years ago
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    rotational kinetic energy of spherical shell + rotational K.E. of pulley + K.E. of object = change in P.E. of object

  4. celos
    • 3 years ago
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    since string is not stretchable: velocity of object = angular speed of shell* Radius of shell = angular speed of pulley * radius of pulley. now put the known values and get the answer.

  5. CarlosGP
    • 3 years ago
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    \[1/2mv^2+1/2I_{sphere}\omega_{sphere}^2+1/2I_{pulley}\omega_{pulley}^2=mgH\]

  6. RaphaelFilgueiras
    • 3 years ago
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    @CarlosGP ,but we dont have w,and v yet

  7. celos
    • 3 years ago
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    w= v/r

  8. celos
    • 3 years ago
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    you will get a equation in one variable and will be able to find solution also

  9. RaphaelFilgueiras
    • 3 years ago
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    @carlos i get it i will try

  10. CarlosGP
    • 3 years ago
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    \[I_{sphere}=2/5MR^2\] then, for the sphere \[1/2(2/5MR^2)w_{sphere}^2=1/5Mv^2\] and for the pulley: \[1/2Iw_{pulley}^2=1/2I(v/r)^2\] with the data given in the drawing: \[mgH=v^2[m/2+M/5+I/(2r^2)]\]

  11. RaphaelFilgueiras
    • 3 years ago
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    Isphere=2/3MR²

  12. RaphaelFilgueiras
    • 3 years ago
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    @CarlosGP thanks

  13. CarlosGP
    • 3 years ago
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    @RaphaelFilgueiras you are welcome. Isphere is 2/5MR^2

  14. RaphaelFilgueiras
    • 3 years ago
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    @CarlosGP it's just a shell

  15. CarlosGP
    • 3 years ago
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    You are right: I=2/3MR^2 for the sphere

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