Originally Posted by

**chunkylumber111** 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

Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2

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!?