Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_1$ has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad $s^{-1}$ . Disc $D_2$ has 4 kg mass, 0.1 m radius and initial angular velocity of 200 rad$s^{-1}$. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad $s^{-1}$ ) of the system is

detailed solution

Correct option is B

Moment of inertia of disc D1 about an axis passing through its centre and normal to its plane isI1=MR22=(2 kg)(0.2 m)22=0.04 kg m2 Initial angular velocity of disc D1,ω1=50rads-1  Moment of inertia of disc D2 about an axis passing through  it centre and normal to its plane is  I2=(4 kg)(0.1 m)22=0.02 kg m2  Initial angular velocity of discD2ω2=200rads-1  Total initial angular momentum of the two discs is Li=I1ω1+I2ω2 When two discs are brought in contact face to face (one on the top of the other) and their axes of rotation coincident the moment of inertia of inertia I of the system is equal to the sum of their individual moment of inertia.I=I1+I2Let ω be the final angular speed of the system. The final angular momentum of the system isLf=Iω=I1+I2ωAccording to law of conservation of angular momentum, we getLi=Lf I1ω1+I2ω2=I1+I2ω ω=I1ω1+I2ω2I1+I2 =0.04 kg m250rads-1+0.02 kg m2200rads-1(0.04+0.02)kgm2 =(2+4)0.06rads-1=100rads-1