# Linear and angular acceleration, flywheel

by alex.daciz
Tags: accleration, angular, flywheel, linear, pulley
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 P: 11 1. The problem statement, all variables and given/known data Hi, I have the following question and was wondering if anyone could help: A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds, calculate: (a) The linear acceleration of the mass (b) The angular acceleration of the wheel (c) The tension in the rope (d) The frictional torque resisting the motion 2. Relevant equations Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2 3. The attempt at a solution I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2. This would give an answer of 2.45m/sec2, unless I'm being stupid and its as straight forward as acceleration due to gravity = 9.81m/sec2? For (b) Angular velocity = linear velocity/radius = (0.5/1.5)/0.3 = 1 rad/sec Therefore, angular acceleration = change in angular velocity/time = (1-0)/1.5 = 0.6 rad/sec2 (c) Does T, tension=mg[M/(M+2m)] = 3.78 N ? (d) I'm not sure how to approach this one, I know frictional torque = total torque - accelerating torque and moment of inertia and radius of gyration must be included at some point but haven't solved it yet - help!?
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hi alex! welcome to pf!
 Quote by alex.daciz A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds … I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2.
sorry, but you're starting this completely the wrong way

find the linear acceleration from "a distance of 0.5m in 1.5 seconds"

(and your g[(2m/(M+2m)] formula would only work for a uniform disc, which this isn't)

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