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 radia//sec`. Another indentical 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 kilojoules is :

Options

A. `2.5`
B. `3.0`
C. `3.5`
D. `4.0`

Answer

A

Explanation

(a) `I_1 = 1 kg m^2, omega_1 = 100 rad//sec`
Since the mass is doubled,
`I_2 = (M_2 R^2)/(2) = (2 M.R^2)/(2) = 2 I_1 = 2 kgm^2`
Conservation of angular momentum,
`I_1 omega_1 = I_2 omega_2` or `1 xx 100 = 2 xx omega_2`
or `omega_2 = 50 rad//sec`
`E_1 = (1)/(2) I_1 omega_1^2 = (1)/(2) xx 1 xx (100)^2 = 5 xx 10^3 J`
`E_2 = (1)/(2) I_2 omega_2^2 = (1)/(2) xx 2 xx (50)^2 = 2.5 xx 10^3 J`
Loss in `KE = E_1 - E_2 = 5 xx 10^3 - 2.5 xx 10^3 = 2.5 kJ`.