54. A cylinder with rotational inertia l1 = 2.0 kg · mrotates clockwise about avertical axis through its center with angular speed Wi5.0 rad/s.A secondcylinder with rotational inertia I2 = 1.0kg · m?same axis with angular speed W2 = 8.0 rad/sthe same rotational axis what is the angular speed of the combination? What percentage of the originalkinetic energy is lost to friction?rotates counterclockwise about theIf the cylinders couple so they have

Given information: The rotational inertia of the first cylinder, I1=2.0 kg·m2 The clock-wise angular speed of this cylinder, ω1=5.0 rad/s The rotational inertia of the second cylinder, I2=1.0 kg·m2 The counter clock-wise angular speed of this cylinder, ω2=8.0 rad/sApply the law of conservation of angular momentum Lf=LiIfωf=-I1ω1+I2ω2I1+I2ωf=-I1ω1+I2ω2 Here, the angular momentum of the first cylinder is negative due to its clockwise rotation. The final angular speed is expressed as follows ωf=-I1ω1+I2ω2I1+I2 Substitute all the known values ωf=-2.0×5.0+1.0×8.02.0+1.0=-0.67 rad/s This is the final angular speed of the combination of cylinders. Here, the negative sign indicates that the combination will rotate in a clockwise direction.The initial kinetic energy of the cylinders, Ki=12I1ω12+12I2ω22=12×2.0×5.02+12×1.0×8.02=57 J The final kinetic energy of the combination of cylinders Kf=12I1+I2ωf2=12×2.0+1.0×-0.672=0.67 J The percentage loss of the kinetic energy is calculated as follows F=Ki-KfKi×100=57-0.6757×100=98.8 % This is the percentage loss of the kinetic energyAnswer: The final angular speed of the combination of cylinders, ωf=0.67 rad/s in a clockwise direction The percentage loss of the kinetic energy is, F=98.8 %

Answered: 54. A cylinder with rotational inertia…