A baggage carousel at an airport is rotating with an angular speed of 0.32 rad/s

Question

A baggage carousel at an airport is rotating with an angular speed of 0.32 rad/s when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1300.0 kg.m^2. Ten pieces of baggage with an average mass of 24.0 kg each are dropped vertically onto the carousel and come to rest at a perpendicular distance of 1.80 m from the axis of rotation. (a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed. rad/s (b) In reality, the angular speed of a baggage carousel does not change. Therefore, what must be the external torque acting on the system if ten bags are dropped each minute? N .m

Explanation / Answer

a)

Moment of inertia of carousel, Ic = 1300 kg.m2

initial angular speed, Wi = 0.32 rad/s

So, initial angular momentum of the carousel, Mi = Ic*Wi = 416 kg.m2.rad/s

moment of inertia of each baggage, Ib = m*d^2

where m = 24 kg

d = 1.8 m

So, Ib = 24*1.8^2 = 77.76 kg.m2

So, total moment of 10 baggages and carousel, Itot = Ic + 10*Ib = 1300 + 777.6 = 2077.6 kg.m2

Now, let the angular speed be Wf

So, now, the angular momentum of the combination, Mf = Itot*Wf

But by conservation of angular momentum,

Mi = Mf

So, 416 = 2077.6*Wf

So, Wf = 0.2 rad/s <------------answer

b)

for the baggage carousel to slow down from Wi = 0.32 rad/s to Wf = 0.2 rad/s in 10 minutes, the angular acceleration is given by, A = (Wf-Wi)/t

Note : here t = 10 minutes = 10*60 = 600 s

So, A = (0.2-0.32)/600 = -1.996*10-4 rad/s2

So, the torque generated, T = Itot*A = 2077.6*(-1.996*10^-4) = -0.415 N.m

So, the external torque required for the angular speed not to change = T' = - T = 0.415 N.m <------------answer

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