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Homework Help: Angular momentum and linear momentum

  1. Nov 11, 2006 #1
    I have gotten stuck in this question and don't know what to do.

    Show that (R X P) . (R X P) = (RP)^2 - (R . P)^2. Assume the vectors R and P lie on the xy-plane. Use this result to express the kinetic energy of a particle in terms of the momentum in the radial direction and of the square of the angular momentum.

    I have already showed that the above is true. The problem is that I don't know how to use this result for the kinetic energy equation.(I'm not even sure this is the equation that I must use since the problem doesn't say anything about how the particle moves)

    K = 1/2 mv^2 + 1/2 Iw^2.

    I'm not even sure what the momentum in the radial direction is.

    As you can see, I'm having a little trouble here, but I think it should be easy. Thanks a lot for any help.
  2. jcsd
  3. Nov 12, 2006 #2
    [tex] R \times P \cdot R \times P = (RP\sin\theta)^{2}\cos\theta = RP^{2}\sin^{2}\theta\cos \theta [/tex] . [tex] \theta = 0 [/tex] so we have [tex] RP^{2}\sin^{2}\theta [/tex]

    [tex] RP^{2} - (R\cdot P)^{2} = RP^{2} - (RP\cos\theta)^{2} = RP^{2} - RP^{2}\cos^{2}\theta[/tex]
    Last edited: Nov 12, 2006
  4. Nov 12, 2006 #3


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  5. Nov 12, 2006 #4


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    Momentum in the radial direction is just what it says. Some component of the object's motion lies along the radius and some component of the object's motion is perpendicular to the radius.

    I'm assuming that R and P must be the position vector and velocity vector, or the position vector and linear momentum vector?

    There's another way to express angular momentum.

    [tex]H=m r v sin \theta[/tex]
    where m is the mass, r the radius, and v the velocity (you could multiply the mass and velocity together to get a linear momentum vector so you don't have to worry about the mass anymore).
    theta is the angle between the radius and velocity.

    If your particle were moving in a circle, this is all you would have. The velocity vector would always be at a right angle to the position vector.

    If the velocity vector is not at a right angle to the position vector, then you must have some portion of the object's motion that lies along the radius, so you must have some linear momentum in the radial direction.

    [tex]p=m r v cos \theta[/tex]

    The cross product can get you the angular momentum, while the dot product will help you get the linear momentum along the radius.
  6. Nov 12, 2006 #5
    that's all good guys but the problem still hurts. The thing is that the question says that I have to use the validity of the expression given in the question to give the kinetic energy of a particle in terms of the angular momentum squared and radial component of linear momentum. I don't even have an idea of where or much less how to start! Thanks again.
  7. Nov 12, 2006 #6


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    Let's try going back to the beginning.

    In what follows, vectors are indicated by bold text, and scalars by regular text.

    First, we are talking about an ideal particle here. All of its mass is concentrated at a single point. It has no moment of inertia and no ω (at least not in the sense implied by your equation). It's kinetic energy is purely translational

    K = 1/2 mV² = P²/2m

    It has angular momentum relative to the origin defined by

    L = R x P

    You are given that the motion is two dimensional in the x-y plane. You could break the velocity vector or the momentum vector into two components parallel to the x and y axes, or you can choose any other pair of perpendicualr directions you want. What others have been suggesting is to choose one axis in the direction of the position vector R and one axis perpendicular to that axis. I will call that the θ direction. If you resolve R and P into components along those axes, the become

    R = Rr
    P = P_rr + P_θθ

    where r and θ are unit vectors in the "radial direction" (direction of R) and the perpendicular direction. The first term in P is what the problem is referring to as the momentum in the radial direction. The second term is the remaining component.

    The square of the momentum is the sum of the squares of these perpendicular components

    P² = (P_r)² + (P_θ)²

    The kinetic energy is therefore

    K = P²/2m = (P_r)²/2m + (P_θ)²/2m

    The first term is expressed in terms of the momentum in the radial direction. What remains to be shown is that the component of P that is perpendicular to the radial component can be expressed in terms of the angular momentum of the particle relative to the origin. Look at the geometry of the situation and the definition of L to see how (P_θ)²/2m it is related to L².
    Last edited: Nov 12, 2006
  8. Nov 12, 2006 #7
    thanks a lot dan. So, L = RP_θ, but now how do I take R out of the expression? Also, why is it that the point mass doesn't have a moment of inertia?
  9. Nov 13, 2006 #8


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    The R does not come out of the expression. It has to be there to make the term that involve L² dimensionally correct and so that the terms reduce to the expression as I left it. K = P²/2m There is no getting around that.

    Moment of inertia is a property of a rigid body, an assembly of masses held together in such a way that the distance between any two masses is constant. It is a constant for any rigid configuration of mass relative to a given axis. The definition is

    I = Σ (m_i)(r_i)²

    For a point particle, there is only one mass and the distance between any two parts of itself is zero. If the particle were at some fixed distance from some point in space, it would make sense to calculate the moment of inertia relative to that point because the calculation would yield a constant. But for a particle that is not a fixed distance from any point in space, the calculation would not be constant, and that would violate the definition.

    The equation you first wrote for K is applicable to a rigid body rotating about its center of mass as well as translating. For a particle the distance between any part of the particle and its center of mass is zero, so the moment of inertia about the center of mass must be zero.
    Last edited: Nov 13, 2006
  10. Nov 13, 2006 #9
    Thanks a lot Dan, I appreciate it. By the way, you explain things very clearly. It is a very good explanation. Thank you for the help.
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