Question

In class we found the change of total kinetic energy (lecture notes 10/21, page 6) Ть ΣΙ Fext AT =-AU + dt (1) dt where U is the total potential energy associated with internal interaction forces and F is an external force acting on m,. Suppose that the masses m, are moving in a uniform gravitational field so that Fe -m,gy. Here, g is the acceleration of gravity and y is pointing vertically up (so that Fe is pointing vertically down) ext a) Show that in this case the second term on the right-hand side of (1) takes the form dr ΣΙά. aFex =-AUex dt (2) where AUe Mg (Y, - Y). Here, M = >m, is the total mass and Y:= =RCM.y is the 1 mF (i.e., Y RCM.y = ŘeM *ỹ). y-component of the center of mass R b) Using (1) and (2), define the total energy E of the system in such a way that E is conserved, i.e., AE = 0

Answer 1

the Fext = – mag m At madra =-M9J I am 2-mg to drem. =-mg Ito di = -mg (%6-Ta) from 0 and ① - avext ATAU = - svext DE 2 -ouext tu ext = mg AY vext a may so, E = - mgr = Ya Ya