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

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aaronqBest ResponseYou've already chosen the best response.1
RCF=[(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
 one year ago

clarkexoBest ResponseYou've already chosen the best response.0
using the nomograph I'm not getting the same answer?
 one year ago

aaronqBest ResponseYou've already chosen the best response.1
what are the values for r and RCF
 one year ago

clarkexoBest ResponseYou've already chosen the best response.0
7 cm for the radius and 600 RCF
 one year ago

aaronqBest ResponseYou've already chosen the best response.1
i got rpm = 2767.64 are you sure everything is in the same units?
 one year ago

clarkexoBest ResponseYou've already chosen the best response.0
Can 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
 one year ago

aaronqBest ResponseYou've already chosen the best response.1
RCF=[(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
 one year ago

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

aaronqBest ResponseYou've already chosen the best response.1
no prob! glad to be of help
 one year ago
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