Rotational MotionMedium
Question
A circular disk of moment of inertia It is roatating in a horizontal plane, about the symmetry axis, another disc of moment of inertia Ib is dropped coaxially onto the rotatating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ωf. The energy lost by the initially rotating disc to friction is
Options
A.
B.
C.
D.
Solution
As no external torque is applied to the system, the angular momentum of the system remains conserved.
∴ Lf = Lf
According to given problem,
Itωi = (It + Ib)ωf
or, ωf =
.......(i)
Initial energy, Ei =
Ttωi2 .......(ii)
Final energy, Ef = (It + Ib)ωf2 .......(iii)
Substituting the value of ωf from the equation (i) in equation (iii), we get
Final Energy, Ef = (It + Ib)
.......(iv)
Loss Energy. ᐃE = Ei - Ef
[Using (ii) and (iv)]
∴ Lf = Lf
According to given problem,
Itωi = (It + Ib)ωf
or, ωf =
.......(i)Initial energy, Ei =
Ttωi2 .......(ii)Final energy, Ef =
(It + Ib)ωf2 .......(iii)Substituting the value of ωf from the equation (i) in equation (iii), we get
Final Energy, Ef =
(It + Ib) .......(iv)Loss Energy. ᐃE = Ei - Ef
[Using (ii) and (iv)]Create a free account to view solution
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