A disk and a ring are mounted on low-friction bearings on the same axle and can

Question

A disk and a ring are mounted on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. The mass is uniformly distributed throughout the objects. The disk and the ring both have a radius of 10cm and a mass of 4kg. The objects are set spinning about their central axis, the disk at 400rev/min in clockwise direction and the ring at 800rev/min in the opposite direction. They then couple together and rotate with an angular velocity of 400rev/min in counter-clockwise direction. Determine whether the angular momentum of the system is conserved, and what average torque might have acted on the system if it was applied steadily over aperiod of 1 minute. The moments of inertia of a solid disk and a ring with respect to theircentral axis are ID =MR^2/2, IR = MR^2 .

Explanation / Answer

Here,

as 1 rev/min = (2pi/60) rad/s
Talking, counter clockwise ti be negative direction -

angular velocity of system,
wd = 400 rev/min = (400*2*3.14)/60 = -41.9 rad/s
Wr = -800 rev/min = (800*2*3.14)/60 = 83.4 rad/s

Wf = -400 rev/min = (400*2*3.14)/60 = -41.9 rad/s

Moment of inertia for disc , Id
Id = 0.5mr^2
Id = 0.5 * 4 * (0.10)^2
Id = 0.02 Kg.m^2

Moment of inertia for Ring , Ir
Ir = mr^2
Ir = 4 * (0.10)^2
Ir = 0.04 Kg.m^2

Initial Angular momentum :
Li = Ir*wr + Id*wd
Li = 0.04 * 83.4 - 0.02*41.9
Li = 2.498 kg*m/s^2

Final Angular momentum :
Lf = -Inet * wf
Lf = -(0.04+0.02)*41.9
Lf = -2.514 Kg.m/s^@

Part A :

As Final Angular momentum is greater than initial momentum therefore momentum is not conserved

Part B:
Torque = Rate of change of Angular momentum / time
t = (Lf - Li)/t
t = (-2.514 -2.948) / 60
t = - 0.09 will in Counter clockwise direction

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