<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 6.00.2800.1170" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=#ffffff> <DIV><FONT face="Courier New" size=2>Sarah,</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>If you add mass to the hammer, = keep same force on key (yes, Don, it's a torque when you multiply it by the key = radius), then the action will accelerate less.  That means less energy in = the key and wippen (we didn't change their inertias).  So if we have the = same total energy, less in the key and wippen, then MORE must have gone into the = key, not the same amount. </FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT><FONT face="Courier = New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>Don,</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" size=2>Yes.  Mostly.  Keep = in mind that angular momentum is just the integral of all the ((1/2)*m*v^2)'s, and angular velocity (av) is just = velocity/radius.  Linear and rotational motion ARE related.  If you replace v in 1/2*m*v^2 with = av*r, you get 1/2*m*r^2*av^2, which is the formula for rotational inertia for = that bit of mass.  This does not contradict the fact that that bit of = matter's instantaneous energy is 1/2*m*v^2.</FONT></DIV> <DIV><FONT face="Courier New" size=2></FONT> </DIV> <DIV><FONT face="Courier New" = size=2>-Mark</FONT></DIV></BODY></HTML>