Newton's Second Law: the Momentum Principle: Difference between revisions
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===A Mathematical Model=== | ===A Mathematical Model=== | ||
Recall that <math>\vec{p} = m\vec{v}</math>. Therefore, by product rule, <math>\frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}</math>. Under the assumption that mass is constant, the second term becomes 0, and the momentum principle becomes <math>\vec{F}_{net} = m\frac{d\vec{v}}{dt}</math>, or <math>\vec{F}_{net} = m\vec{a}</math>, which may be a more familiar form of Newton's second law. The form <math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math> is preferred for several reasons: | |||
<ol> | |||
<li>When a particle accumulates mass that was initially at rest, such as a snowball rolling downhill, the term <math>\vec{v}\frac{dm}{dt}</math> is not 0, and <math>\vec{F}_{net} = m\vec{a}</math> is no longer accurate, while <math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math> is;</li> | |||
<li>As a particle approaches the speed of light, <math>\vec{F}_{net} = m\vec{a}</math> is no longer accurate, while <math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math> is, as long as [[relativistic momentum]] is used; and</li> | |||
<li><math>\vec{F}_{net} = \frac{d\vec{p}}{dt}</math> has a more direct rotational analogue (that is, it will be easier to accurately learn rotational physics if you learn linear physics using this form.)</li> | |||
</ol> | |||
===A Computational Model=== | ===A Computational Model=== |
Revision as of 14:54, 17 May 2019
This page describes Newton's second law, also known as the momentum principle, which relates net force to the change in linear momentum. This principle is used to predict the effects of forces on the motion of objects.
The Main Idea
Newton's second law, also known as the momentum principle, states that [math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}_{system}}{dt} }[/math] where [math]\displaystyle{ \vec{p} }[/math] is the linear momentum of the system and [math]\displaystyle{ \vec{F}_{net} }[/math] is the net external force acting on the system from its surroundings. Often, the system in question consists of a single particle whose motion we want to predict.
The momentum principle has no derivation, as it is considered the definition of force.
A Mathematical Model
Recall that [math]\displaystyle{ \vec{p} = m\vec{v} }[/math]. Therefore, by product rule, [math]\displaystyle{ \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt} }[/math]. Under the assumption that mass is constant, the second term becomes 0, and the momentum principle becomes [math]\displaystyle{ \vec{F}_{net} = m\frac{d\vec{v}}{dt} }[/math], or [math]\displaystyle{ \vec{F}_{net} = m\vec{a} }[/math], which may be a more familiar form of Newton's second law. The form [math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}}{dt} }[/math] is preferred for several reasons:
- When a particle accumulates mass that was initially at rest, such as a snowball rolling downhill, the term [math]\displaystyle{ \vec{v}\frac{dm}{dt} }[/math] is not 0, and [math]\displaystyle{ \vec{F}_{net} = m\vec{a} }[/math] is no longer accurate, while [math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}}{dt} }[/math] is;
- As a particle approaches the speed of light, [math]\displaystyle{ \vec{F}_{net} = m\vec{a} }[/math] is no longer accurate, while [math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}}{dt} }[/math] is, as long as relativistic momentum is used; and
- [math]\displaystyle{ \vec{F}_{net} = \frac{d\vec{p}}{dt} }[/math] has a more direct rotational analogue (that is, it will be easier to accurately learn rotational physics if you learn linear physics using this form.)
A Computational Model
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