Momentum vector always parallel to velocity

[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?

 

[Simon Bridge]

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.

 

[negation]

... 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.

 

[Simon Bridge]

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.