Mass multiplied by Position is...

[dtarrence]

I've been puzzled by this question lately. How do you define Mass x Position? The units would of course be kg.m , but how do you define that? What is method for defining new units of measure? How did we decide that N.m = Joule?

 

[alpha2cen]

I've been puzzled by this question lately. How do you define Mass x Position? The units would of course be kg.m , but how do you define that? kg

 

kg.m is [Mass][Length]. It is not 'Mass x Position' . The unit will be useful when we calculate an iron wire length.

L=k/M. Where L is length, M is mass and k is constant.

When we know k value, we can easily calculate the wire length from it's mass. It is not a commonly used unit for the physical problem calculation.

Edited January 1, 2013 by alpha2cen

[elfmotat]

"Position" is not a coordinate-independent (covariant) quantity. There's no such thing as a coordinate-independent "position vector," so position vectors by themselves are not physically meaningful. For example, let's say we have a position vector given by [math]\mathbf{x}=(x,y)[/math]. If we shift the origin of our coordinate system by a distance k in the +x direction, the vector (x,y) no longer corresponds to the point that it did before. Instead, our previous point now corresponds to the vector (x-k,y).

 

Consequently, the product of mass and a position vector is also not physically meaningful. Positions only become meaningful in relation to each other (i.e. when we take a difference of positions) because they represent the "distance" from one place to another. Distances obviously don't depend on your choice of coordinates.

 

"Velocity" is just a measure of change in position over time, and is therefore a coordinate-independent quantity. The product of mass and velocity is therefore physically meaningful (momentum). Similarly, acceleration and force are also physically meaningful, as are higher derivatives of position and their product with mass.

 

 

How did we decide that N.m = Joule?

 

The Joule is defined as 1 kg*m²/s² = 1 N*m. Physicists simply decided to give the unit kg*m²/s² its own name, and that name is "Joule."

L=k/M. Where L is length, M is mass and k is constant.

 

I think you mean L=kM. What you have now implies a longer wire must have smaller mass. A wire with constant cross-sectional area will have mass proportional to length, not inversely proportional.

Edited January 2, 2013 by elfmotat

[alpha2cen]

I think you mean L=kM. What you have now implies a longer wire must have smaller mass. A wire with constant cross-sectional area will have mass proportional to length, not inversely proportional.

The system can be used at the constant lever system

-----------------------------------------------------------

^ 000

const mass L M

 

In the SI unit system, L's unit is m, M's unit is kg, and k's unit is kg.m.

It is difficult to find out a general physical system to use such unit.

[elfmotat]

The system can be used at the constant lever system

-----------------------------------------------------------

^ 000

const mass L M

 

In the SI unit system, L's unit is m, M's unit is kg, and k's unit is kg.m.

Even though gravitational acceleration cancels so that you do get m₁L₁=m₂L₂, I would still regard this as a torque problem. It's more meaningful in this context to talk about (force)*(length) than (mass)*(length).

 

It is difficult to find out a general physical system to use such unit.

I agree. I'm having trouble coming up with instances where (mass)*(length) is used at all. I suppose if you wrote the center of mass equation as:

 

[math]M \bar{x}= \sum_i m_i x_i[/math]

 

Even then though, the actual quantity [math]M \bar{x}[/math] isn't really physically meaningful.

Edited January 2, 2013 by elfmotat

[alpha2cen]

This is more exact Figure.

The example of [mass. length] dimension is diffcult to make.

[swansont]

 

 

Even though gravitational acceleration cancels so that you do get m₁L₁=m₂L₂, I would still regard this as a torque problem. It's more meaningful in this context to talk about (force)*(length) than (mass)*(length).

 

 

 

I agree. I'm having trouble coming up with instances where (mass)*(length) is used at all. I suppose if you wrote the center of mass equation as:

 

[math]M \bar{x}= \sum_i m_i x_i[/math]

 

Even then though, the actual quantity [math]M \bar{x}

isn't really physically meaningful.

 

 

CoM is important, and angular momentum uses mvr. mr^2 is used when finding moments of inertia, which is crucial for rotating systems.

[michel123456]

MD can be represented as a cube.

 

 

Mass is the vertical axis.

Time is horizontal (front)

Velocity is horizontal (back)

Distance is the product of Time with Velocity, i.e. the grey surface.

The yellow surface is Momentum.

The entire cube has units of mass times distance (kg m), something like torque (or work).

[alpha2cen]

Generally, 3 dimensional graph representing

 

f(x,y) or (x, y, z)

 

This case is

(t, v(t), m)