Moment of inertia of the flywheel, I = 0.62 kg m^{2}

Initial angular velocity of the flywheel, ω_{1} = \(\cfrac{2\pi N_1}{60}\)

= \(\cfrac{2\pi \times2500}{60}\) = 261.8 rad/s.

Temperature of insulated system, T_{0} = 20 + 273 = 293 K

Water equivalent of shaft bearings = 1.9 kg

(i) Initial available energy of the flywheel,

(K.E.)_{initial} = \(\cfrac12\) Iω_{1}^{2}

\(\cfrac12\) × 0.62 × (261.8)^{2 }= 2.12 × 10^{4} N.m = 21.2 kJ

When this K.E. is dissipated as frictional heat, if ∆t is the temperature rise of the bearings, we have

Water equivalent of bearings × rise in temperature = 21.2

i.e., (1.9 × 4.18) ∆t = 21.2

∆t = \(\cfrac{21.2}{1.9\times4.18}\) = 2.67°C

Hence, rise in temperature of bearings = 2.67°C.

∴ Final temperature of the bearings = 20 + 2.67 = 22.67°C.

(ii) The maximum amount of energy which may be returned to the flywheel as high-grade energy is,

The amount of energy rendered unavailable is

U.E. = (A.E.)_{initial} – (A.E.)_{returnable as high grade energy }

= 21.2 – 0.096 = 21.1 kJ.

Since the amount of energy returnable to the flywheel is 0.096 kJ, if ω_{2} is the final angular velocity, and the flywheel is set in motion with this energy, then

0.096 × 10^{3} = \(\cfrac12\) x 0.62 × ω_{2}^{2}

ω_{2}^{2 }= \(\cfrac{0.096\times10^3\times2}{0.62}\) = 309.67 or ω2 = 17.59 rad/s.

If N_{2} is the final r.p.m. of the flywheel, then

ω_{2} = \(\cfrac{2\pi N_1}{60}\) or N_{2} = \(\cfrac{17.59\times60}{2\pi}\) = 168 r.p.m.

Final r.p.m. of the flywheel = 168 r.p.m.