replacing d ω ω d t we can have the last item as − − 2 r d r d t d ω ω d t = − − 2 r d r d t ( r × × a r 2 − − 2 r d r d t ω ω ) = − − 2 r 3 d r d t ( r × × a ) + 4 r 2 ( d r d t ) 2 ω ω , and we finally get