Momentum vector always parallel to velocity

In summary: In general, the momentum vector is a vector with magnitude and direction. However, in a state of rotation, the vector is "superimposed" onto the velocity vector and therefore parallel.
  • #1
negation
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For an obejcting in a state of rotation, the velocity is always tangential to the distance acceleration to the axis of rotation.
From what I read, the momentum vector, p = mv, is always parallel.
Would it then be right to state that the momentum vector, p, is nothing more than a scalar product of m and v?
In other words, the momentum vecotr is "superimposed" onto the velocity vector and therefore parallel.

In looking at this from the cross product, v x vm (where both v are vector), the product = 0 because the angle between them is zero. Sin(0°) = 0.

Am I looking at this from the right angle or is there a better way to look at it?
 
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  • #2
negation said:
For an obejcting in a state of rotation, the velocity is always tangential to the distance acceleration to the axis of rotation.
Only in uniform circular motion - the object could be increasing or decreasing it's distance from the center of rotation, then it will have a radial component to it's velocity as well as a tangential component.

If it's angular velocity increases, then it's total acceleration will no longer point towards the center either.

From what I read, the momentum vector, p = mv, is always parallel.
Would it then be right to state that the momentum vector, p, is nothing more than a scalar product of m and v?
For objects with mass ##\vec p = m\vec v## yes.
For light ##\vec p = \hbar \vec k## ... where ##\vec k## is the wave-vector - it points in the direction the light travels and has magnitude ##k=2\pi/\lambda##.

I don't know what you mean by "nothing more", this is quite sufficient.

It is also valid to say that the velocity is nothing more than the momentum multiplied by a scalar too. However, momentum is important because it is a conserved quantity.

In other words, the momentum vector is "superimposed" onto the velocity vector and therefore parallel.
It points in the same direction as the instantaneous direction of travel yes.

In looking at this from the cross product, v x vm (where both v are vector), the product = 0 because the angle between them is zero. Sin(0°) = 0.
... this is for linear momentum.
There is also angular momentum.

note:
in general $$\vec\omega = \frac{\vec v \times \vec r}{r}$$

Am I looking at this from the right angle or is there a better way to look at it?
You seem to be overthinking the concepts here.
 
  • #3
Simon Bridge said:
... this is for linear momentum.
There is also angular momentum.

From what I gathered from the lecture.

An object in a state of rotation can be decomposed into torque and angular momentum.

Let L→ be angular momentum

L→ = r→ x p→ = r→ sin Θ x mv→

Taking the derivative of L:

dL/dt = dr/dt x p + r x dp/dt

v x mv + r x F

I'm a little confused as to why you said vx mv is true only in linear momentum. I did worked out v x mv = 0 from angular momentum. I might be missing something here.
 
  • #4
I'm a little confused as to why you said vx mv is true only in linear momentum. I did worked out v x mv = 0 from angular momentum. I might be missing something here.
... Oh I think I got confused by your notation.
 
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  • #5


Yes, you are correct in stating that the momentum vector, p, is a scalar product of the mass, m, and velocity, v. This means that the direction of the momentum vector is the same as the direction of the velocity vector. In other words, they are parallel.

However, when dealing with an object in a state of rotation, the velocity is not always parallel to the distance acceleration to the axis of rotation. This is because the velocity vector is constantly changing in direction as the object rotates.

In this case, the momentum vector is still parallel to the velocity vector at any given moment, but it may not be parallel to the distance acceleration to the axis of rotation. This is because the distance acceleration is a vector that is always perpendicular to the velocity vector, and the momentum vector is not necessarily perpendicular to the velocity vector.

So while your understanding of the scalar product and cross product is correct, it is important to note that the direction of the momentum vector may not always be the same as the direction of the distance acceleration to the axis of rotation.
 

What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is the product of an object's mass and velocity. In other words, momentum is a measure of how much force is needed to stop an object from moving.

Why is momentum important?

Momentum is important because it helps us understand the behavior of objects in motion. It is also a conserved quantity, meaning that it remains constant unless acted upon by an external force. This has important implications in various fields such as physics, engineering, and sports.

What is the relationship between momentum and velocity?

Momentum and velocity are directly related. The direction of an object's momentum is always the same as its velocity. This means that if an object is moving in a straight line, its momentum vector will always be parallel to its velocity vector.

Why is the momentum vector always parallel to the velocity vector?

This is because momentum is a vector quantity, meaning it has both magnitude and direction. The direction of an object's momentum is determined by its velocity, which is also a vector quantity. Therefore, the momentum vector will always be parallel to the velocity vector.

Can momentum ever change direction?

Yes, momentum can change direction if the direction of an object's velocity changes. This can happen when an external force is applied to an object, causing it to accelerate or change direction. However, the magnitude of an object's momentum will remain constant as long as no external forces act on it.

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