Acylinder with rotational inertia i l = 2.0 kg . m2 rotates clockwise about a vertical axis through its center withangular speed ml = 5.0 rad] s. = 5.0 rad/s. a second cylinderwith rotational inertia [2 = 1.0 kg . m2rotates counter clockwise about the same axis with angular speed m2 = 8,0 rad/s , if the cylinders couple so theyhave the same rotational axis what is the angular speed of the combination? what percentage of the original kineticenergy is lost to friction?

it is lost the 98.82% of the energy.  

Explanation:

for answer this, we will use the conservation of the angular momentum L:

so:

where is the moment of inertia, is the angular velocity of the cylinder 1, is the moment of inertia of the second cylinder, is the angular velocity of the second cylinder, is the moment of inertia of the cylinders couple and is the angular velocity of the cylinders couple

note: we will take the clockwise rotation as positive.

Replacing the values, we get:

solving for :

for find the kinetic energy lost we must calculate the initial Ei and the final energy Ef as:

and,

now, the percentage of the original kinetic energy that is lost to friction is calculated as:

= 0.9882 = 98.82%

that means that was lost the 98.82% of the energy.

a and b sometimes have a positive value

volts times amps 6 x 0.5 3w