Power for a turning machine

Feb 2009
11
0
I have been asked to compare the power required to turn a 80mm dia. bar at 100rpm to that of an 80mm dia. at 200 rpm. I'm not too sure how to go about this... I have made the presumption that I'm using steel which can be machined at 15cm^3/min per kilowatt.

I know that 100rpm will produce 25m/min by using:

S=(TTd/v)/1000..... so 200rpm will equal 50m/min

Just can't get my head round calculating and comparing the power........

Any help would be much appreciated.
 

Pmb

PHF Hall of Fame
Apr 2009
1,579
335
Boston's North Shore
I have been asked to compare the power required to turn a 80mm dia. bar at 100rpm to that of an 80mm dia. at 200 rpm. I'm not too sure how to go about this... I have made the presumption that I'm using steel which can be machined at 15cm^3/min per kilowatt.

I know that 100rpm will produce 25m/min by using:

S=(TTd/v)/1000..... so 200rpm will equal 50m/min

Just can't get my head round calculating and comparing the power........

Any help would be much appreciated.
It's possible to do those things with zero power since angular momentum of a closed system and if the bar has no torque on it then it takes zero power. There has to be a counter torque acting in order for the bar to exert power since power is the energy transfered per unit time. Where is the energy transfer in your question?
 
Mar 2010
105
20
Lithuania
Work done
A=Fs
s=ra,
where a is angle [rad].
Power is
P=A/t=Fra/t=Frw,
where w is angular speed [rad/s].
Or
M=Fr
P=Mw.
Asuming that M=const or F=const
P1/P2=w1/w2.
 

Pmb

PHF Hall of Fame
Apr 2009
1,579
335
Boston's North Shore
Work done
A=Fs
s=ra,
where a is angle [rad].
Ypu didn't define s, r and a. If Work done is W = F*ds where "* represents the dot product. Here where F abd ds are perpendicular so therefore F*ds = 0.