Respuesta :
Answer:
a) [tex]\omega_2=41.89\ rad.s^{-1} [/tex]
b) [tex]\frac{\Delta KE}{KE_1} = 0.5[/tex]
Explanation:
Given:
rotational speed of the wheel 1, [tex]N=800\ rpm= \frac{2\pi}{60} \times 800 rad.s^{-1}[/tex]
Let the rotational inertia of the wheel 1 be, [tex]\rm I[/tex]
According to given, the rotational inertia of wheel 2, [tex]I_2=2I[/tex]
(a)
According to the law of conservation of angular momentum:
[tex]I.\omega=I_2.\omega_2[/tex]
[tex]I\times \frac{2\pi}{3} \times 40=2I\times \omega_2[/tex]
[tex]\omega_2=\frac{40\pi}{3}\ rad.s^{-1}[/tex]
[tex]\omega_2=41.89\ rad.s^{-1} [/tex]
(b)
initial kinetic energy:
[tex]KE_1=\frac{1}{2} I.\omega^2[/tex]
[tex]KE_1=0.5\times I\times 83.78^2[/tex]
[tex]KE_1=3509.54\ I[/tex]
Final kinetic energy:
[tex]KE_2=\frac{1}{2}I_2.(\omega_2)^2[/tex]
[tex]KE_2=0.5\times 2I\times 41.89^2[/tex]
[tex]KE_2=1754.77\ I[/tex]
∴the fraction of the original rotational kinetic energy lost in the process:
[tex]\frac{\Delta KE}{KE_1} =\frac{(3509.54-1754.77)}{3509.54}[/tex]
[tex]\frac{\Delta KE}{KE_1} = 0.5[/tex]
