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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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keshavsardaBest ResponseYou've already chosen the best response.0
m*g*h=1/2*MI*\[\omega\]^2
 one year ago

meera_yadavBest ResponseYou've already chosen the best response.0
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
 one year ago

CarlosGPBest ResponseYou've already chosen the best response.0
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)
 one year ago

dianahla12Best ResponseYou've already chosen the best response.1
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.
 one year ago
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