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Momentum vector always parallel to velocity 
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#1
Mar3014, 06:12 AM

P: 819

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
Mar3014, 07:28 AM

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If it's angular velocity increases, then it's total acceleration will no longer point towards the center either. For light ##\vec p = \hbar \vec k## ... where ##\vec k## is the wavevector  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. There is also angular momentum. note: in general $$\vec\omega = \frac{\vec v \times \vec r}{r}$$ 


#3
Mar3014, 08:54 AM

P: 819

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. 


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