The fluctuation of the energy of the engine flywheel will be 2722.71 J.
The device which stores kinetic energy is known as the flywheel. It supplies that energy when it is required.
An engine flywheel has a mass of 6. 5 tonnes and the radius of gyration is 2 m. If the maximum and minimum speeds are 120 R.P.M. And 118 R.P.M. Respectively.
Then the energy fluctuation will be
E = 1/2 (mr²) (ω₂² - ω₁²)
Then we have
[tex]\rm \Delta E = \dfrac{1}{2} \times 6.5 \times 10^3 \times 2^2 \left ( \dfrac{2\pi }{60} \times (120 - 118) \right)\\\\\rm \Delta E = 13000 \times 0.1047 \times 2\\\\\Delta E = 2722.71 \ J[/tex]
More about the energy of the flywheel link is given below.
https://brainly.com/question/16969408
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