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Homework Help: Simple question: conservation of angular momentum

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data

    small block (mass .025 kg) on frictionless horizontal surface is attached to massless cord passing through a hole in the surface. Block is revolving around the hole in the center, with initial angular speed w1 = 1.75 rad/s, 0.3 m away from the hole. The chord is pulled from below, shortening the radius of the circle to 0.15m. Treat block as a particle.

    >is angular momentum conserved?
    >what is new angular speed?
    >what is change in kinetic energy of block?
    >what is the work done by pulling the chord?

    2. Relevant equations

    k = (1/2)mv^2 + (1/2)Iw^2
    v = rw
    I = mr^2

    3. The attempt at a solution

    I'm not sure if angular momentum is conserved. I want to say no because it doesn't seem that the sum of the external force, in this case the tension in the string, sums to zero..

    But then the equation becomes K1 + Work(other) = K2. Work is W = F*dr.... but how do I solve for F? F = ma(x).... but in this case the tension in the rope is antiparallel to the radial acceleration.... a(rad) = (w^2)r. or does the force from the tension in the rope equal ma(tangtial) which is perpendicular to the force vector... where a(tan) = r*angular_accel.

    Im a little confused. Thanks for any help.
  2. jcsd
  3. Dec 3, 2009 #2
    Can a force that is always perpendicular to the velocity, change that velocity?

    This tells you a lot about the system's angular momentum as well as its energy.

    I'm afraid to approach this problem in terms of forces myself, since there may be fictitious forces at work that we don't notice (Though I may very well be wrong), so a momentum/energy approach seems like the best idea to me.
    Last edited: Dec 3, 2009
  4. Dec 3, 2009 #3
    So a perpendicular force will only change the direction of linear velocity but not the magnitude. But in this case it will change the angular velocity...

    I am also asked what the work done is by the cord so to avoid forces... isn't it the same answer to the previous question, the change in kinetic energy?

    Researching conservation of angular momentum I found in an article "If the net force on some body is directed always toward some fixed point, the center, then there is no torque on the body with respect to the center, and so the angular momentum of the body about the center is constant". So since the cord applies no *torque*, and there is no friction, then angular momentum *is* conserved. Am I correct? So in that case I can just use I_1*w_1 = I_2*w_2 ? (I = moment of inertia)

  5. Dec 3, 2009 #4

    My assumption was dead wrong. What we should have started with was the fact that force is radial, and that the axis of rotation lies on its line of action, meaning that it can apply no torque.

    [tex]\vec \tau=\frac{d\vec L}{dt}[/tex]

    So it is clear that if there is no torque acting on the mass, its angular momentum will be conserved.

    From here on out things should be fairly simple. It's always fun being proven wrong. :)

    As for your specific question regarding the quote:

    That quote is correct, though a bit poorly phrased. If the lines of action of the forces on an object cross a point P, then there can be no torque about an axis of rotation through that point P (This is immediate from the definition of the torque as [tex]\vec \tau=\vec r \times \vec F[/tex])

    This is exactly the reason I wanted to avoid dealing with forces in this question. We've made a bad assumption somewhere along the line that has led to a contradiction. If memory serves me right, the bad assumption is in that the hole can't apply any torque to the mass. I'll sink my teeth into this one and have an answer for you in a sec.

    Great question!

    I'm at a loss now myself, how can the central force perform any work on the mass if [tex]\vec F \cdot \vec {dx}=0[/tex] since [tex]\vec F \bot \vec dx[/tex] ?

    ANSWER (?) :
    The tension can perform work on the mass since the dot product [tex]\vec F \cdot \vec {dr}[/tex] is not zero since [tex]\vec F \| \vec {dr}[/tex] ?
    Last edited: Dec 3, 2009
  6. Dec 3, 2009 #5
    One definition of work is the change in kinetic energy. I wonder if it is also valid to flip that around and say that if there is a change in kinetic energy... there must have been work done? I assume (could be wrong) that since there are two questions asking about the work done and change in kinetic energy that k1 does not equal k2... so mechanical energy is not conserved but angular momentum is. Since mechanical energy isn't conserved, there must be some external force doing work (?)
  7. Dec 4, 2009 #6
    Nope, that's the work-energy theorem. The difference in kinetic energy is the conservative work done on the system.
    And just to make things clear, the work energy theorem is not a definition of work, but a consequence of the definition of work as:

    [tex]dW=\vec F \cdot \vec dr[/tex]

    Kinetic energy can't change all by itself, there always has to be a force to perform work on it.

    [tex]\Delta (\frac{1}{2}mv^2)=W[/tex]

    What I overlooked earlier, that may have mislead you, is that the radial force DOES perform work on the mass, since there is a distance over which that force performs work (The radial force performs work over the radial distance)

    Once you conclude that angular momentum is conserved (Since there is no torque) the velocity as a function of radial distance should pop out, and from there, you should be able to find the work done by the radial force in changing the kinetic energy.
    Last edited: Dec 4, 2009
  8. Dec 4, 2009 #7
    thanks, muchos =) ill work on it
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