Discussion
A load coupled to a motor through a gear of ratio \( i \) presents a referred inertia of
Answer: B
Correct answer: \( J_l/i^2 \)
Referring an inertia through a reduction of ratio \( i \) divides it by the square of the ratio:
\[ J_{referred} = \frac{J_l}{i^2} \]because the load turns \( i \) times more slowly and the kinetic energies must match. Torque refers with the first power and inertia with the second, which is the single most useful pair of facts in drive sizing - a 10:1 reduction divides the load inertia by a hundred.
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