Energy is to be stored in a 86.4 kg flywheel in the shape of...

Energy is to be stored in a 86.4 kg flywheel in the shape of a uniform solid disk ($I = \frac{1}{2} mR^2$) with radius $R = 1.25$ m. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 m/s$^2$. What is the maximum kinetic energy that can be stored in the flywheel? 67.5 kJ 69.2 kJ 94.5 kJ 96.9 kJ 189 kJ 194 kJ None of the above

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Energy is to be stored in an 86.4 kg flywheel in the shape of a uniform solid disk. The moment of inertia (I) of the flywheel can be calculated using the formula I = mR^2, where m is the mass of the flywheel and R is the radius of the flywheel. Given: m = 86.4 kg R = 1.25 m Using the formula, we can calculate the moment of inertia: I = (86.4 kg)(1.25 m)^2 To prevent structural failure of the flywheel, the maximum allowed radial acceleration (a) of a point on its rim is given as 3500 m/s^2. The maximum kinetic energy (K) that can be stored in the flywheel can be calculated using the formula K = (1/2)Iω^2, where ω is the angular velocity of the flywheel. To find ω, we can use the formula a = ω^2R, where a is the radial acceleration and R is the radius of the flywheel. Given: a = 3500 m/s^2 R = 1.25 m Using the formula, we can calculate ω: 3500 m/s^2 = ω^2(1.25 m) ω^2 = 3500 m/s^2 / 1.25 m ω^2 = 2800 s^-2 ω = √(2800 s^-2) Now that we have ω, we can calculate the maximum kinetic energy: K = (1/2)(86.4 kg)(1.25 m)^2(√(2800 s^-2))^2 The answer options are given in scientific notation with SI units.

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