The question is incomplete. The complete question is :
In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?
Solution :
Let M.I. of disk A = [tex]$I_0$[/tex]
So, M.I. of disk B = [tex]$3I_0$[/tex]
Angular velocity of A = [tex]$\omega_0$[/tex]
So the kinetic energy of the disk A = [tex]$\frac{1}{2}I_0\omega^2$[/tex]
After coupling, the angular velocity of both the disks will be equal to ω.
Angular momentum will be conserved.
So,
[tex]$I_0\omega_0 = I_0 \omega + 3I_0 \omega$[/tex]
[tex]$I_0\omega_0 = 4I_0 \omega$[/tex]
[tex]$\omega = \frac{\omega_0}{4}$[/tex]
Now,
[tex]$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$[/tex]
[tex]$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$[/tex]
[tex]$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $[/tex]
[tex]$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$[/tex]
[tex]$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $[/tex]
Therefore, the maximum initial K.E. = 3066.67 J
