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a centrifuge consists of 4 solid cylindrical containers , each of mass m at radial distance r from the axis of rotation. time t is required to bring the centrifuge to an angular velocity W from rest under a constant torque T applied to the shaft. the radius of each container is A and the mass of the shaft and arms is small compared to m. then t=.............................
ans= 2m(a2+2r2)W/T
fig:-
Dear Pranay
Plz see the pic for solution.
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
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net torque on the shaft is given...
now we can find net angular accleration of the system by
T=I(alfa) (T -torque,alfa - angura accleration and I is total moment of inertial of system )
alfa = T/I .........1
moment of inertial of any of cylender about the axis of rotation is I1
I1 = I(about its own axis) + I(about axis of rotation)
I1 = ma2 /2 + md2 ( a is radius of cylender and r is perpendicular distance from axis of rotation)
I
I posted the comlete solution but i dont know why remaining part is not appearing ...
here is the remaining solution remaining solution :
I1 = I (about axis of rotation) + I(abot its own axis)
I1 = mr2 + ma2 /2 ( Icom =ma2 /2 )
total moment of inertia is 4I1
I = 4I1 = 4mr2 + 2ma2
for constant torque we can use this eq
W =W0 + (alfa)t (alfa is angular accleration)
t=W/alfa (w0 is 0 )
t = WI/T
t = 2m(a2 + 2r2 )W/T
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