The angular acceleration and the moment of inertia of the flywheel is 0.6 $$\frac{rad}{s^{2}}$$ and 2500 $$kg-m^{2}$$ respectively. What is the difference in kinetic energy (in
kN.-m) of the flywheel after 5 seconds and 10 seconds from the start?

Given : Angular acceleration of flywheel $$\alpha=0.6$$ $$rad/s^2$$

Moment of inertia of flywheel, $$I_f=2500$$ $$kg/m^2$$

Let $$\triangle KE=$$ Difference in Kinetic Energy of flywheel after 5 and 10 sec respectively from start.

Case 1 : When time $$t_1=5$$ sec

=> Angular velocity $$w_1=\alpha t_1$$

= $$0.6\times5=3$$ rad/sec

Case 2 : When time $$t_2=10$$ sec

=> Angular velocity $$w_2=\alpha t_2$$

= $$0.6\times10=6$$ rad/sec

$$\therefore$$ Change in Kinetic Energy $$\triangle KE=\frac{1}{2}I_f(w_2^2-w_1^2)$$

= $$\frac{1}{2}\times2500\times(6^2-3^2)$$

= $$\frac{1}{2}\times2500\times(36-9)$$

= $$33750$$ J = $$33.75$$ KJ

=> Ans - (A)