The flywheel of a steam engine of mass 1000kg has radius of gyration 1m. If the maximum and minimum speed of the flywheel is 80r.p.m and 78r.m.p respectively. Find the fluctuation of energy
"I=mr^2=1000\\cdot1^2=1000\\ (kg\\cdot m^2)"
"80\\ rpm=1.3333333\\ s^{-1}"
"78\\ rpm=1.3\\ s^{-1}"
"E_1=I\\omega_1^2\/2=1000\\cdot(2\\cdot3.14\\cdot1.3333333)^2\/2=35056\\ (J)"
"E_2=I\\omega_2^2\/2=1000\\cdot(2\\cdot3.14\\cdot1.3)^2\/2=33325\\ (J)"
"\\Delta E=35056-33325=1731\\ (J)" . Answer
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