4. A exerted on the cord. The wheel is mounted in frictionless bearings on a hor

Question

4. A exerted on the cord. The wheel is mounted in frictionless bearings on a horizontal shaft through the center. The moment of inertia of the wheel is 4 kgm2 cord is wrapped around the rim of a flywheel 0.5 m in radius and a steady pull of 50 N is a) Compute the angular acceleration of the wheel. b) Show that the work done in unwinding 5 m of cord equals the gain in kinetic energs of the wheel. c) If a mass having a weight of 50 N hangs from the cord, compute the angula acceleration of the wheel. Should the solution be the same or different as in part (a)? Why? 50 N

Explanation / Answer

A) Writing torque equation,

Tr = i alpha

50*0.5 = 4 alpha

alpha = 25/4 = 6.25 rad/s^2 Answer

B) By work energy theorem,

Total work done by all forces = change in KE

Work done by tension = increase in KE

= Td = 50*5 =250 J

C) Let tension be T in this case and acceleration be a,

W - T = Ma

Or 50- T = (50/9.8) a

Not by torque equation, Tr = i alpha

T*0.5 = 4 alpha where alpha =a/r = 2a

T = 16a

Putting this in force equation,

50 - 16a =5.1a

21.1a=50

a = 50/21.1 = 2.37 m/s^2

Angular acceleration alpha = 2a = 4.74 rad/s^2 Answer

It should be different because tension in string is different in both case.

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