A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the
mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710m, through how many revolutions will the rotor turn before coming to rest, and how long will it take?
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
The coefficient of expansion is 13 * 10^-6 m per meter length.per oK The temperature difference = 42 - - 8 = 50 oC delta T = (42 + 273) - (-8 + 273) = 50 oK delta L = L * 13* 10^6 m/oK oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m So we need to find the length and it is computed by: 0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m