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clarkexo
 2 years ago
RCF=[(rpmx2pi)/60]^2 x r/980 How would I rearrange this to solve for rpm? Please show steps! Thanks :
clarkexo
 2 years ago
RCF=[(rpmx2pi)/60]^2 x r/980 How would I rearrange this to solve for rpm? Please show steps! Thanks :

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aaronq
 2 years ago
Best ResponseYou've already chosen the best response.1RCF=[(rpmx2pi)/60]^2 x r/980 (980)(RCF)/r=[(rpmx2pi)/60]^2 ((980)(RCF)/r)^(1/2)=[(rpmx2pi)/60] [((980)(RCF)/r)^(1/2)](60)=(rpmx2pi) [((980)(RCF)/r)^(1/2)](60/2pi)=rpm

clarkexo
 2 years ago
Best ResponseYou've already chosen the best response.0using the nomograph I'm not getting the same answer?

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.1what are the values for r and RCF

clarkexo
 2 years ago
Best ResponseYou've already chosen the best response.07 cm for the radius and 600 RCF

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.1i got rpm = 2767.64 are you sure everything is in the same units?

clarkexo
 2 years ago
Best ResponseYou've already chosen the best response.0Can u please go over the steps of how your calculating it? Perhaps i'm doing it wrong..the answer using the graph is aprox 2760

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.1RCF=[(rpmx2pi)/60]^2 x r/980 multiply both sides by reciprocal of r/980...which is 980/r (980)(RCF)/r=[(rpmx2pi)/60]^2 take the square root of both sides ((980)(RCF)/r)^(1/2)=[(rpmx2pi)/60] multiply both sides by reciprocal of 2pi/60... which is 60/2pi [((980)(RCF)/r)^(1/2)](60/2pi)=rpm RCF=600, r=7 [(980)(600)/7]^(1/2)[60/2pi]=rpm multiply: 980x600 [(588000/7]^(1/2)[60/2pi]=rpm divide: 588000/7 [(84000]^(1/2)[60/2pi]=rpm Take the square root of 84000 [289.827][60/2pi]=rpm multply 289.827 by 60 17389.65/2pi=rpm divide 17389.65 by 2 8694.826/pi=rpm divide 8694.826 by pi 2767.64909=rpm

clarkexo
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you so much!! I forgot to square root! You're amazing! Thanks again :)

aaronq
 2 years ago
Best ResponseYou've already chosen the best response.1no prob! glad to be of help
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