HW1_prob1_GPS01_G04

HW1_prob1_GPS01_G04 - d p dt = F , (5) where p is the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 1.4 (3 rd ed. 1.1) Start with Newton’s 2nd Law for a particle with constant mass m , m d v dt = F . (1) Next, take the dot product of Eq.(1) with the particle velocity v to get m v · d v dt = F · v . (2) Now expand the left dot product of Eq.(2) as m v · d v dt = m ( v x dv x dt + v y dv y dt + v z dv z dt )= m 2 d dt ( v 2 x + v 2 y + v 2 z )= m 2 dv 2 dt = dT dt , (3) where T is the kinetic energy of the particle ( mv 2 2 ), to obtain the desired result, dT dt = F · v . (4) When the mass is not constant, we use Newton’s 2nd Law in the form,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d p dt = F , (5) where p is the particle momentum vector. Take the dot product of this equation with p to obtain p d p dt = F p , (6) which can be rearranged in a fashion similar to that of Eq.(3) to give 1 2 dp 2 dt = F p . (7) Since p 2 = 2 mT , Eq.(7) can be rewritten in the desired form d ( mT ) dt = F p . (8) 1...
View Full Document

This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

Ask a homework question - tutors are online