A centrifuge rotor has a moment of inertia of 3.25 x 10^{-2} kg \cdot m ^2. How much energy is required to bring it from rest to 8750 rpm?

This is Giancoli Answers with Mr. Dychko. The amount of energy to bring this centrifuge from rest to 8750 rpm is gonna be the change in kinetic energy— KE f minus KE i— but there's no kinetic energy initial so we just have the kinetic energy final is gonna answer our question. So we have one-half times moment of inertia which we are given— 3.25 times 10 to the minus 2 kilogram meter squared— and then we take this final angular speed and convert the rpm's into radians per second and then square that result and we get 1.36 times 10 to the 4 joules.

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A centrifuge rotor has a moment of inertia of 3.25×10−2 kg⋅m23.25 \times 10^{-2} \textrm{ kg}\cdot\textrm{m}^23.25×10−2 kg⋅m2. How much energy is required to bring it from rest to 8750 rpm?

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A centrifuge rotor has a moment of inertia of $3.25 \times 10^{-2} \textrm{ kg}\cdot\textrm{m}^2$ . How much energy is required to bring it from rest to 8750 rpm?