## anonymous 4 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. anonymous

m*g*h=1/2*MI*$\omega$^2

2. anonymous

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. anonymous

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. anonymous

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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