Question

A circular disk of moment of inertia I1 is rotating in a horizontal plane,about its symmetry axis,with a constant angular speed w1.Another disk of moment of inertia I2 is dropped coaxially onto the rotating disk.Initially the second disk has zero angular speed.Eventually both the disks rotate with a constant angular speed w2.The energy lost by the initially rotating disc to friction is ?

Answer 1

initial angular momentum, $\bf{L=I_1\omega_1}$

final angular momentum , $\bf{(I_1+I_2)\omega_2}$

from law of conservation of angular momentum, as external torque is zero .
so, initial angular momentum = final angular momentum

$I_1\omega_1=(I_1+I_2)\omega_2$

$\omega_2=\frac{I_1\omega_1}{(I_1+I_2)}$

so, loss of energy is given by
$\Delta{E}=\frac{1}{2}I_1\omega_1^2-\frac{1}{2}(I_1+I_2)\left[\frac{I_1\omega_1}{(I_1+I_2)}\right]^2$

$\Delta{E}=\frac{1}{2}\left[\frac{I_1I_2}{(I_1+I_2)}\right]\omega_1^2$

Answer 2

Answer:

correct answer is (4)

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