The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 kg m^2. It is rotating with an angular velocity 100 rad s^(-1) Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilo joules is

The moment of inertia of a uniform disc about an axis passing through

1.	The moment of inertia of a uniform disc about an axis passing through its centre and perpendicular to its plane is 1 kg m^2. It is rotating with an angular velocity 100 rad s^(-1) Another identical disc is gently placed on it so that their centres coincide. Now these two discs together continue to rotate about the same axis. Then the loss in kinetic energy in kilo joules is
Answer» 2.5 `3.0` `3.5` `4.0` Solution :As no external torque is applied to the system , the angular momentum of the system remains conserved `therefore L_i= L_f` where the subscripts represent INITIAL and final , or `I_(i) omega_(i) = I_(f) omega_(f)` SUBSTITUTING the GIVEN values , we get `therefore 1 xx 100 = 2 xx 1 xx omega_(f)` or `omega_(f) = 50 rad s^(-1) "" ... (i)` Initial KINETIC energy , `K_(i)= (1)/(2) I_(i) omega_(i)^(2) = (1)/(2) xx 1 xx (100)^(2) = 5 xx 10^(3) J` Final kinetic energy , `K_(f) = (1)/(2) I_(f) omega_(f)^(2) = (1)/(2) xx 2 xx 1 xx (50)^(2)` `=2 .5 xx 10^(3) J ""` (Using (i)) Loss in kinetic energy , `Delta K = K_(i) - K_(f) = 5 xx 10^(3)-2.5 xx 10^(3) J` `=2.5 xx 10^(3) J = 2.5 kJ`