When 12000 J of work is done on a flywheel,i | vedclass

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(B) Given: Work done $W = 12000 \ J$,Initial frequency $f_1 = 10 \ Hz$,Final frequency $f_2 = 20 \ Hz$.Angular velocity is given by $\omega = 2\pi f$.

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$2 \ kg \cdot m^2$

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(B) Given: Work done $W = 12000 \ J$,Initial frequency $f_1 = 10 \ Hz$,Final frequency $f_2 = 20 \ Hz$.Angular velocity is given by $\omega = 2\pi f$.Initial angular velocity $\omega_1 = 2\pi(10) = 20\pi \ rad/s$.Final angular velocity $\omega_2 = 2\pi(20) = 40\pi \ rad/s$.According to the work-energy theorem,the work done equals the change in rotational kinetic energy:$W = \Delta K = \frac{1}{2} I \omega_2^2 - \frac{1}{2} I \omega_1^2 = \frac{1}{2} I (\omega_2^2 - \omega_1^2)$.Substituting the values:$12000 = \frac{1}{2} I [(40\pi)^2 - (20\pi)^2]$.$12000 = \frac{1}{2} I [1600\pi^2 - 400\pi^2]$.$12000 = \frac{1}{2} I [1200\pi^2]$.Using $\pi^2 = 10$:$12000 = \frac{1}{2} I [1200 	imes 10]$.$12000 = I [6000]$.$I = \frac{12000}{6000} = 2 \ kg \cdot m^2$.

When $12000 \ J$ of work is done on a flywheel,its frequency of rotation increases from $10 \ Hz$ to $20 \ Hz$. The moment of inertia of the flywheel about its axis of rotation is (take $\pi^2=10$):

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