Kinetic energy for single particle with constant mass post
Posted: July 14th, 2008, 9:03 pmI can't find the post anymore, but I had a question about it
Here is what it said:
Show that dT/dt=F dot v where F is force, v velocity, T kinetic energy and t time. Also show d(mT)/dt= F dot p (all same variables but p is momentum)
This is derivation 1 in Goldstein chapter 1. So I was able to get the answer, but am unsure as to why it is correct.
The second one is the most confusing. It states the same but with variable mass. It solves by this method: d(mT)/dt= d(p^2)/dt=p*dp/dt=p*F=p dot F in three coordinate space. My original idea was to try this instead and am not sure why it doesn't work while the above method does.
I tried: d(mT)/dt=1/2 d(m^2v^2)/dt which by the chain rule gives m(dm/dt)*v^2+m^2*v*(dv/dt)=mv(dm/dt*v+ma)=p(dm/dt*v+F)=p dot F + dm/dt*v which is the same as above with an extra term of dm/dt*v. What if anything did I do wrong and how do I get the two to reconcile?
Here is what it said:
Show that dT/dt=F dot v where F is force, v velocity, T kinetic energy and t time. Also show d(mT)/dt= F dot p (all same variables but p is momentum)
This is derivation 1 in Goldstein chapter 1. So I was able to get the answer, but am unsure as to why it is correct.
The second one is the most confusing. It states the same but with variable mass. It solves by this method: d(mT)/dt= d(p^2)/dt=p*dp/dt=p*F=p dot F in three coordinate space. My original idea was to try this instead and am not sure why it doesn't work while the above method does.
I tried: d(mT)/dt=1/2 d(m^2v^2)/dt which by the chain rule gives m(dm/dt)*v^2+m^2*v*(dv/dt)=mv(dm/dt*v+ma)=p(dm/dt*v+F)=p dot F + dm/dt*v which is the same as above with an extra term of dm/dt*v. What if anything did I do wrong and how do I get the two to reconcile?