Question

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?

Answer 1

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

Answer 2

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

Answer 3

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)

Answer 4

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.