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dianahla12

  • 3 years ago

There is a nonuniform rod of length 2 m and mass 2 kg. It can freely rotate about one of its ends and the moment of inertia about this point is 4 kg*m^2. The center of gravity is 1.2 m from the pivot. What is the angular velocity as it reaches the vertical position if it is let go from rest in the horizontal position?

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  1. keshavsarda
    • 3 years ago
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    m*g*h=1/2*MI*\[\omega\]^2

  2. meera_yadav
    • 3 years ago
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    hii, Methinks u can here use law of conservation of energy here. choosing potential energy equals to zero when it's at horizontal position , then initial total energy = energy at vertical position total energy at vertical position = \[I w ^{2}/2 - mgh\] initial energy =0 substitute values u';; be there

  3. CarlosGP
    • 3 years ago
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    The initial energy is given by the torque (momentum) M=mgd, where: m=mass, g=gravitational acceleration and d=distance of center of gravity. This momuntum is zero when gravitational force is aligned with the rod (vertical position) which means that has fully turned into rotational energy =1/2*I*w^2, then :w=SquareRoot(2mgd/I)

  4. dianahla12
    • 3 years ago
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    Could you get the angular velocity from finding the angular acceleration though?\[ \alpha=\tau _{net}\times I\] You could get the angular acceleration from that.

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