General Definitions of Impulse and Work

[motion_ar]

In classical mechanics, if we consider the motion of a particle of mass [itex]m[/itex], then



The mass [itex]m[/itex] is [itex]constant[/itex]


The vector [itex]\vec{c}[/itex] can be: [itex]\ldots[/itex] or [itex]\vec{r}[/itex] or [itex]\vec{v}[/itex] or [itex]\vec{a}[/itex] or [itex]\vec{j}[/itex] or [itex]\ldots[/itex]

[itex]\vec{c}_1 = d\vec{c} / {dt}[/itex]

[itex]\vec{c}_2 = d^2 \, \vec{c} / {dt^2}[/itex]



Definition of Impulse [itex]\vec{c}[/itex] [itex]\, ( \vec{I}_{\vec{c}} )[/itex]

[tex]\vec{I}_{\vec{c}} \; = \int_a^b m \, \vec{c}_1 \, dt \; = \Delta \; m \, \vec{c}[/tex]
[tex]\vec{I}_{\vec{c}} \; = \Delta \; \vec{P}_{\vec{c}}[/tex]
where:

[tex]\Delta \; \vec{P}_{\vec{c}} \; = \Delta \; m \, \vec{c}[/tex]

If [itex]\; \vec{c}_1 = 0[/itex]

[tex]\rightarrow \; \; \Delta \; \vec{P}_{\vec{c}} \; = 0[/tex]
[tex]\rightarrow \; \; \vec{P}_{\vec{c}} \; = constant[/tex]


Definition of Work [itex]\vec{c}[/itex] [itex]\, (W_{\vec{c}})[/itex]

[tex]W_{\vec{c}} \; = \int_a^b m \; \vec{c}_2 \cdot d\vec{c} \; = \Delta \; {\textstyle \frac{1}{2}} \, m \; \vec{c}_1^2[/tex]
[tex]W_{\vec{c}} \; = \Delta \; T_{\vec{c}_1} + \Delta \; V_{\vec{c}} \; = \int_a^b m \; \vec{c}_{2n\vec{c}} \cdot d\vec{c}[/tex]
where:

[tex]\Delta \; T_{\vec{c}_1} = \Delta \; {\textstyle \frac{1}{2}} \, m \; \vec{c}_1^2[/tex]
[tex]\Delta \; V_{\vec{c}} = - \int_a^b m \; \vec{c}_{2\vec{c}} \cdot d\vec{c}[/tex]
[tex]\vec{c}_2 = \vec{c}_{2\vec{c}} + \vec{c}_{2n\vec{c}}[/tex]
[tex]\vec{c}_{2\vec{c}} \; \; \; is \; \; function \; \; of \; \; \; \vec{c}[/tex]
[tex]\vec{c}_{2n\vec{c}} \; \; \; is \; \; not \; \; function \; \; of \; \; \; \vec{c}[/tex]

If [itex]\; \vec{c}_{2n\vec{c}} = 0[/itex]

[tex]\rightarrow \; \; \Delta \; T_{\vec{c}_1} + \Delta \; V_{\vec{c}} \; = 0[/tex]
[tex]\rightarrow \; \; T_{\vec{c}_1} + V_{\vec{c}} \; = constant[/tex]

If [itex]\; \vec{c}_2 = 0[/itex]

[tex]\rightarrow \; \; \Delta \; T_{\vec{c}_1} \; = 0[/tex]
[tex]\rightarrow \; \; T_{\vec{c}_1} \; = constant[/tex]

 

[eljose79]

If \; \vec{c}_{2n\vec{c}} = 0

\rightarrow \; \; \Delta \; V_{\vec{c}} \; = 0
\rightarrow \; \; V_{\vec{c}} \; = constant


Hello,

Thank you for your post. It is correct that in classical mechanics, the mass of a particle is considered to be constant. This is known as the conservation of mass, which states that the total mass of a closed system remains constant over time.

In addition, you have correctly defined the vector \vec{c} as a function of time, with its first and second derivatives being denoted as \vec{c}_1 and \vec{c}_2, respectively. These derivatives play an important role in the definitions of impulse and work, as you have shown.

Impulse, denoted as \vec{I}_{\vec{c}}, is defined as the change in momentum (\Delta \vec{P}_{\vec{c}}) of a particle over a certain time interval. This can also be expressed as the integral of the product of the mass and the first derivative of \vec{c} with respect to time. This is a fundamental concept in classical mechanics, as it describes the effect of a force on a particle over time.

Similarly, work, denoted as W_{\vec{c}}, is defined as the change in kinetic energy (\Delta T_{\vec{c}_1}) of a particle over a certain time interval, or the integral of the product of the mass and the second derivative of \vec{c} with respect to time. This concept is also crucial in understanding the effects of forces on particles in classical mechanics.

Finally, you have correctly pointed out that if certain conditions are met (such as \vec{c}_1 = 0 or \vec{c}_{2n\vec{c}} = 0), then the values of impulse and work can be simplified to constants. This is important in certain scenarios where the motion of a particle is constrained or constant.

Overall, your post demonstrates a good understanding of classical mechanics and the role of mass, impulse, and work in describing the motion of a particle. Keep up the good work!