# Momentum vector always parallel to velocity

1. Mar 30, 2014

### negation

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?

2. Mar 30, 2014

### Simon Bridge

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.

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.

It points in the same direction as the instantaneous direction of travel yes.

... this is for linear momentum.
There is also angular momentum.

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

You seem to be overthinking the concepts here.

3. Mar 30, 2014

### negation

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. Mar 30, 2014

### Simon Bridge

... Oh I think I got confused by your notation.

Last edited: Mar 30, 2014