The center of mass is the center of a system of masses, located at a point that is central in relation to the masses of all of the objects within the system. To describe the position of the center of mass, we use the equation$$\vec{r}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{r}_i}{\sum\limits_{i=0}^n m_i}$$$$\vec{r}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{r}_i}{M_{total}}$$
The velocity of the center of mass is the same thing, except finding velocity instead of position.$$\vec{v}_{CM} = \frac{\sum\limits_{i=0}^n m_i \vec{v}_i}{M_{total}}$$
If the velocity is very small compared to the speed of light, $\gamma = 1$ we can say that the total velocities and masses of the system are equal based off of the equation for momentum. This helps us find the momentum of the center of mass:$$\vec{p} = \gamma\times m\vec{v}$$$$\vec{p}_{sys} = M_{total}\vec{v}_{CM}$$$$\vec{p}_{sys} = \sum\limits_{i=0}^n m_i \vec{v}_i$$
Showing posts with label mass. Show all posts
Showing posts with label mass. Show all posts
Sunday, March 17, 2013
Monday, March 4, 2013
Momentum
In real life, it seems obvious that stopping a baseball traveling at high speed is much easier than, say, a refrigerator hurdling towards you at high speed. The mass, or weight, of the refrigerator is much greater than the baseball! Everyone knows that it becomes increasingly difficult to move or change something as it becomes heavier, but how are we to describe this phenomenon using physics?
In order to represent the combination of an object's mass ($m$) and velocity ($\vec{v}$), a vector quantity called momentum, $\vec{p}$, is defined. Notice that mass is a scalar and velocity a vector. In our visible, daily lives, this equation exists simply as $\vec{p} = m \times \vec{v}$. However there is another important factor that we must account for.
Certain experiments have shown that as particles travel closer and closer to the speed of light ($c = 3 \times 10^8$ m/s), the amount of interaction required in order to create an increase in velocity becomes increasingly large. This is Einstein's "relativistic" definition of momentum, defined below.
In order to model the disparity between low speed velocity and velocity that approaches the speed of light, we introduce the scalar proportionality factor, gamma $\gamma$, which is equal to: $$\gamma = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}}$$
Using this proportionality factor, we can now write the correct equation to model momentum at all speeds: $$\vec{p} = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}} m \vec{v}$$ $$\vec{p} = \gamma \times m \vec{v}$$
Generally, gamma is approximately exactly 1 at low speeds, and is only effective at very high speeds. We can simplify our momentum equations in low speed situations for the sake of ease down to: $$\vec{p} = 1 \times m \vec{v} = m \vec{v}$$
Momentum is the integral of the net force ($\vec{F}_{net}$)$$\vec{p} = \int \vec{F}_{net}$$Using the definition of acceleration and its relation to velocity ($\vec{a} = \frac{\delta \vec{v}}{\delta t}$), we can define the approximate rate of change of momentum as$$\frac{\delta \vec{p}}{\delta t} = m\vec{a}$$$$\vec{F}_{net} = (\frac{\delta \vec{p}}{\delta t} = m\frac{\delta \vec{v}}{\delta t}) = m\vec{a}$$
In order to represent the combination of an object's mass ($m$) and velocity ($\vec{v}$), a vector quantity called momentum, $\vec{p}$, is defined. Notice that mass is a scalar and velocity a vector. In our visible, daily lives, this equation exists simply as $\vec{p} = m \times \vec{v}$. However there is another important factor that we must account for.
Certain experiments have shown that as particles travel closer and closer to the speed of light ($c = 3 \times 10^8$ m/s), the amount of interaction required in order to create an increase in velocity becomes increasingly large. This is Einstein's "relativistic" definition of momentum, defined below.
In order to model the disparity between low speed velocity and velocity that approaches the speed of light, we introduce the scalar proportionality factor, gamma $\gamma$, which is equal to: $$\gamma = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}}$$
Using this proportionality factor, we can now write the correct equation to model momentum at all speeds: $$\vec{p} = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}} m \vec{v}$$ $$\vec{p} = \gamma \times m \vec{v}$$
Generally, gamma is approximately exactly 1 at low speeds, and is only effective at very high speeds. We can simplify our momentum equations in low speed situations for the sake of ease down to: $$\vec{p} = 1 \times m \vec{v} = m \vec{v}$$
Momentum is the integral of the net force ($\vec{F}_{net}$)$$\vec{p} = \int \vec{F}_{net}$$Using the definition of acceleration and its relation to velocity ($\vec{a} = \frac{\delta \vec{v}}{\delta t}$), we can define the approximate rate of change of momentum as$$\frac{\delta \vec{p}}{\delta t} = m\vec{a}$$$$\vec{F}_{net} = (\frac{\delta \vec{p}}{\delta t} = m\frac{\delta \vec{v}}{\delta t}) = m\vec{a}$$
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