Question
Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?
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Final Answer

22.5 kgm2/s22.5\textrm{ kg}\cdot\textrm{m}^2\textrm{/s}

Solution video

OpenStax College Physics for AP® Courses, Chapter 10, Problem 38 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. There's a force of 300 newtons on the crank of an antique car engine and that crank has a lever arm of 0.300 meters and the force is applied for 0.250 seconds and the question is what momentum is given to the engine through this crank? So we know that net torque is the change in momentum divided by change in time. Then we can solve for the change in angular momentum by multiplying both sides by time. So ΔL then is τ net times Δt; the torque is force times this distance— the lever arm— and we know that these two are perpendicular because the question tells us that the force is exerted such that it creates the maximum torque which means they must be perpendicular. So we have 300 newtons times 0.300 meters times 0.250 seconds which is 22.5 kilogram meters squared per second of angular momentum.