My Math Forum  

Go Back   My Math Forum > Science Forums > Physics

Physics Physics Forum


Thanks Tree2Thanks
  • 1 Post By Benit13
  • 1 Post By J Thomas
Reply
 
LinkBack Thread Tools Display Modes
March 4th, 2015, 09:44 PM   #1
Newbie
 
Ben Hunto's Avatar
 
Joined: Mar 2015
From: Auckland, NZ

Posts: 7
Thanks: 0

Scalar Products, Derivatives and Unit Vectors

In my text there is a section on Conservation of Energy of a Point Mass. Starting with Newton's Second Law as:

F = ma = m dv/dt = mg


Gravitational field close to Earth is approximately -gz, where z is the unit vector on the z-axis.

m dv/dt = -mgz


Then take the scalar (dot) product of both sides

mv . dv/dt = - mgv . z


Now the book jumps (with a simple "This equation can be simplified") to

d/dt 1/2 mv . v = d/dt (1/2 mv^2) = -mg dz/dt

and I'm trying to shake the rust off a brain which hasn't worked with this type of math since the late 90s by filling in the gaps over which the book jumps. I know what's happening on the left, but the right side has me scratching my head.

If I recall correctly, change in position z relative to time is v, so

dz/dt = v

But then I don't know where that z goes.
Is it just that it somehow drops out with a value of 1 since it is the unit vector?
Am I just wrong that we can simply say dz/dt = v?
Do we actually need to retain that z so it should it be dz/dt = v . z?
What is the importance of that unit vector z in operational terms (I know what it means - giving the direction of the action without affecting the value - I just don't quite get how it operates)?
Is it something to do with dot products giving a scalar value?

I'm sure I'll slap my head when someone points out how simple the answer is but I can take it.

Last edited by Ben Hunto; March 4th, 2015 at 09:47 PM.
Ben Hunto is offline  
 
March 5th, 2015, 01:57 AM   #2
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,157
Thanks: 732

Math Focus: Physics, mathematical modelling, numerical and computational solutions
If $\displaystyle v = v(t)$, then using implicit differentiation,

$\displaystyle \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv \frac{dv}{dt}$.

So the reverse logic is applied to simplifying your LHS.

Also, z is a unit vector, which means that it always has a magnitude of 1. Therefore, it should only be used to indicate direction. This implies

$\displaystyle \frac{dz}{dt} = 0$.

A new scalar variable should be used, such as h, where

$\displaystyle v = \frac{dh}{dt} \cdot z$

Then integrating both sides with respect to $\displaystyle t$ will yield

$\displaystyle \frac{1}{2} mv^2 = mg \Delta h$

where we have removed the unit vectors representing up and all quantities are scalars.
Thanks from Ben Hunto
Benit13 is offline  
March 5th, 2015, 03:10 AM   #3
Member
 
Joined: Jan 2015
From: USA

Posts: 61
Thanks: 6

Quote:
Originally Posted by Ben Hunto View Post
In my text there is a section on Conservation of Energy of a Point Mass. Starting with Newton's Second Law as:

F = ma = m dv/dt = mg


Gravitational field close to Earth is approximately -gz, where z is the unit vector on the z-axis.

m dv/dt = -mgz


Then take the scalar (dot) product of both sides

mv . dv/dt = - mgv . z


Now the book jumps (with a simple "This equation can be simplified") to

d/dt 1/2 mv . v = d/dt (1/2 mv^2) = -mg dz/dt

and I'm trying to shake the rust off a brain which hasn't worked with this type of math since the late 90s by filling in the gaps over which the book jumps. I know what's happening on the left, but the right side has me scratching my head.
Their notation doesn't make sense to me either. Let's review what they're saying.

mv . dv/dt = - mgv . z

Let's call $\displaystyle V_\parallel$ the part of V that's inline with F, and $\displaystyle V_\perp$ the part that isn't inline with F.

We get a scalar, $\displaystyle |V_\parallel||F|$

d/dt 1/2 mv . v = d/dt (1/2 mv^2)

Same scalar.
v . d/dt 1/2 mv + (d/dt 1/2 mv) . v = v . m dv/dt = $\displaystyle |V_\parallel||F|$

I suppose they put the dz/dt (not dz/dt) there as a reminder that this scalar represents something in the z direction. I can see how you'd be confused by it. It doesn't make sense.
Thanks from Ben Hunto
J Thomas is offline  
Reply

  My Math Forum > Science Forums > Physics

Tags
derivatives, products, scalar, unit, vectors



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Scalar vectors : need help in proving rnck Applied Math 4 September 14th, 2012 07:25 AM
The Perpendicular Unit Vectors i and j bilano99 Calculus 2 February 8th, 2012 01:49 PM
2 unit vectors dot product onako Linear Algebra 2 September 23rd, 2010 08:48 AM
Differentiating Vectors and Gradient of a Scalar Field? cypher Linear Algebra 3 June 29th, 2010 01:40 PM
orthogonal unit vectors hrc Linear Algebra 6 March 3rd, 2008 01:59 PM





Copyright © 2019 My Math Forum. All rights reserved.