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Angular momentum

  1. Nov 24, 2005 #1
    Consider this problem:
    A mass is in circular motion, held by a string, at radius Ro with angular velocity Wo. Then, the string is pulled towards the center with speed Vo, so the radius is decreasing at a rate Vo. What is the angular velocity of the mass as a function of time?

    The problem is trivial to solve using conservation of angular momentum. I'm trying to solve it without resorting to angular momentum, but I'm a bit lost. The tangential velocity is increasing, which confused me at first, but after thinking, i figured that the Vo bleeds off into the tangential velocity. Is this correct?
     
  2. jcsd
  3. Nov 24, 2005 #2

    mezarashi

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    I'd just like to say that it is odd to approach a physics problem by ignoring a law that governs it. Instead you start giving it intuitive labels like "Vo bleeds off into the tangential velocity" which I doubt will be consistent in a variety of situations. If you dismiss conservation of angular momentum, you'll be seeing some 'strange invisible' force acting on the mass as you pull it in causing its angular velocity to increase.
     
  4. Nov 24, 2005 #3
    I don't agree. I'm not ignoring a fundamental law. Conservation of angular momentum, especially as it applies here, is a direct result of newton's laws. I see no reason why applying it here would not yield correct results.

    I do agree that terms like 'bleeds off' are not precise. What I really mean to say is that the force applied to decrease the length of the string also increases the tangential velocity. The question is what the change in tangential velocity is. This is where i'm running into a problem.
     
  5. Nov 25, 2005 #4

    ZapperZ

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    Actually, Newton's laws are the results of the conservation principle, not the other way around. So what is more fundamental are the conservation laws, not newtonian laws. So you may want to reconsider.

    Zz.
     
  6. Nov 25, 2005 #5

    lightgrav

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    Of course you've realized that Work is done by the string Tension,
    a function of radius, otherwise the KE couldn't increase.

    It's kindof interesting that to use F = dp/dt for this,
    you need to have F_string be greater than mv^2/r for a while.

    I think it's a good idea to see how how scenarios treated with these
    "independent conservation principles" can be done with F = dp/dt,
    ... how often the use of the principle is for our convenience only.
     
  7. Nov 25, 2005 #6
    Very good question, asdf60! Approaching problems in different ways can certainly aid in understanding mechanics and what goes on behind the curtains of these rather interesting laws, like the law of conservation of angular momentum.
    Obviously, mezarashi's and ZapperZ's replies have provided no insight what-so-ever. So I'd like to propose a solution using Lagrangian mechanics (I hope you are familiar with Lagrangian formalism).
    The whole essence of Lagrangian mechanics is to write the Lagrangian of the system, and then use the Euler-Lagrange equation to the extract appropriate equations of motion. The Lagrangian of the system is defined by the difference between the total kinetic energy and the total potential energy of the system. So we can write
    [tex]L=T-V[/tex]​
    The kinetic energy is given by [itex]T=\frac{1}{2}mv^2[/itex], where [itex]m[/itex] is the mass and [itex]v^2[/itex] square magnitude of its velocity. Assuming the motion of the mass is constrained to lie in a plane, we can express the [itex]x[/itex] and [itex]y[/itex] coordinates of the mass, at any given time, [itex]t[/itex], in terms of its distance from the center of revolution [itex]r[/itex] and the angle, [itex]\theta[/itex] it makes with a reference horizontal, which we designate to be the [itex]x[/itex]-axis:
    [tex]x=r\cos\theta[/tex]
    [tex]y=r\sin\theta[/tex]​
    A differentiation yields the [itex]x[/itex] and [itex]y[/itex] components of the velocity of the particle.
    [tex]\dot{x}=\dot{r}\cos\theta-r\dot\theta\sin\theta[/tex]
    [tex]\dot{y}=\dot{r}\sin\theta+r\dot\theta\cos\theta[/tex]​
    ,
    where a dot over a symbol indicates the first derivative of that variable with respect to [itex]t[/itex]. Notice that since both [itex]r[/itex] and and [itex]\theta[/itex] vary with time, we make use of the product and the chain rule. We can square both components and add them to yield the velocity-square of the mass
    [tex]\dot{x}^2=\dot{r}^2\cos^2\theta+r^2\dot\theta^2\sin^2\theta-2r\dot{r}\dot\theta\cos\theta\sin\theta[/tex]
    [tex]\dot{y}^2=\dot{r}^2\sin^2\theta+r^2\dot\theta^2\cos^2\theta+2r\dot{r}\dot\theta\cos\theta\sin\theta[/tex]​
    [tex]v^2=\dot{x}^2+\dot{y}^2=\dot{r}^2+r^2\dot\theta^2[/tex][/center]
    So the kinetic energy is given by [itex]T=\frac{1}{2}mv\left(v_0^2+r^2\dot\theta^2\right)[/itex]. Notice that [itex]\dot{r}[/itex] is replaced with [itex]v_0[/itex] to maintain consistency with the notation used in the problem. Gravity is irrelevant here, so we neglect it. So, the potential of the system is zero ([itex]V=0[/itex]). Therefore, the Lagrangian of the system is
    [tex]L=\frac{1}{2}mv\left(v_0^2+r^2\dot\theta^2\right)[/tex].​
    The Euler-Lagrange equation (the equation of motion) is
    [tex]\frac{\partial L}{\partial \theta}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot\theta}\right)=0[/tex].​
    By substituting our expression for the Lagrangian, we get
    [tex]0-\frac{d}{dt}\left(mr^2\dot\theta\right)=0[/tex]​
    or
    [tex]-2mrv_0\dot{\theta}-mr^2\ddot\theta=0[/tex].​
    Remember [itex]\dot{r}=v_0[/itex]. After simplification, this yields
    [tex]\ddot\theta+\frac{2v_0\dot{\theta}}{r}=0[/tex].​
    However, [itex]r[/itex] is the distance of the mass from the center. Since we know it varies linearly with time, [itex]r=v_0t+r_0[/itex], where [itex]r_0[/itex] is the distance away from the center at time [itex]t=0[/itex]. So the proper equation of motion for the system is
    [tex]\ddot\theta+\frac{2v_0\dot\theta}{v_0t+r_0}=0[/tex].​
    The equation above is second-order ordinary differential equation, whose solution yields the angular position of the mass as a function of time:
    [tex]\theta(t)=C_1-\frac{C_2}{v_0(v_0t+r_0)}[/tex],​
    where [itex]C_1[/itex] and [itex]C_2[/itex] are integration constants which can be fitted to the initial conditions of the problem, [itex]\theta(0)=0[/itex] and [itex]\theta'(0)=\omega_0[/itex]. The solution specific to our problem becomes
    [tex]\theta(t)=\frac{r_0\omega_0t}{v_0t+r_0}[/tex].​
    From this equation, we can directly find the mathematical form of the dependence of the angular momentum by differentiating the above equation:
    [tex]\omega(t)=\frac{d\theta}{dt}=\frac{\omega_0r_0^2}{(v_0t+r_0)^2}[/tex]​
    REMEMBER, [itex]v_0[/itex] is negative to indicate that the distance between the mass and the center is shrinking.
    Check the attached graphs for plots of the position, and the angular velocity.
     

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    Last edited: Nov 25, 2005
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