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Homework Help: Some angular momentum problems

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    1. There is m1 on an incline connected to m2 by a string through the pulley at the top of the incline. The incline is frictionless and the string pases through the center of mass of each block. The pulley has a moment of inertia I and a radius r. (a) Find the net torque acting on the system. (b) Write an expression for the total angular momentum of the system about the center of the pulley when the masses are moving with a speed v. (c) Find the acceleration of the masses from your results for parts (a) and (b) by setting the net torque equal to the rate of change of the angular momentum of the system

    2. Relevant equations


    3. The attempt at a solution

    1. So I used Newton's 2nd law to get the acceleration as:
    (m2*g*sin[tex]\theta[/tex]- m1*g)/ (m1+m2+(1/2)M)

    However, M which is mass of the pulley is not given. I can make M=2I/r^2

    I still don't get the right answer for net torque. It should be rg(m2*sin[tex]\theta[/tex]-m1)

    Anyone know if I am on the right track ?? Any help will be appreciated
  2. jcsd
  3. Feb 24, 2009 #2


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    I'm picturing m1 on a ramp at angle theta with the horizontal so that the force on it along the ramp is m1*g*sin(theta). Also, m2 hanging straight down from the pulley, with force m2*g.
    But this would give a torque of (m2*g - m1*g*sin(theta))*r
    or rg(m2 - m1*sin(theta)).

    I must be mistaken about where the masses are?
  4. Feb 24, 2009 #3
    m2 should be on the ramp and m1 should be hanging.

    How about other parts. I think I get part a now.
  5. Feb 24, 2009 #4


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    L = Iw for the turning pulley.
    plus mvr for the moving masses.
  6. Feb 24, 2009 #5
    For part a, just to make sure, why are tensions not taken into account ?
  7. Feb 24, 2009 #6


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    The tension in the string just transfers the force from one object to another.
    If you think of the whole thing (strings and masses) being one object you don't need to bother with tension. Usually.

    If we ignore the moment of inertia of the pulley in this problem, we could think of it as one object and write
    m1*g - m2*g*sin(A) = (m1 + m2)*a

    or we could write T - m2*g*sin(A) = m2*a for the mass on the ramp and
    m1*g - T = ma for the hanging mass.
    If you add these two equations together, the T's annihilate and you get the first equation 4 lines up.

    Of course none of this helps in your problem where you have moment of inertia to deal with.
    EDIT: actually, calculating "a" this way is a big help. If we temporarily let I = 0 in the acceleration we get using the torque - angular momentum approach, it simplify to
    a = (m1*g = m2*g*sin(A))/(m1 + m2) from the equation above.
  8. Feb 24, 2009 #7
    Thanks. I could solve it now

    Another problem:

    A spaceship rotates about its longitudinal axis at 6 rev/min. The occupants wish to stop this rotation. They have small jets mounted tangentially, at a distance R=3m from the axis, as indicated, and can eject 10g/s of gas from each jet with a nozzle velocity of 800 m/s. For how long must they turn on these jets to stop the rotation ? To rotational inertia of the ship around its axis (assumed to be constant) is 4000 kg*m^2

    Any hint on how I should approach this one ?
  9. Feb 25, 2009 #8


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    the force exerted by one of those "jets" is F = d/dt(mv) = v*dm/dt
    where dm is the mass ejected in time dt and v is the exhaust velocity.
    You should be able to work out the torque from that and the radius.
    Do you have a formula relating torque and inertia (analogous to F = ma)?
  10. Feb 25, 2009 #9
    how do I calculate alpha though ?

    Should it be m*dv/dt instead of v*dm/dt ??
  11. Feb 25, 2009 #10


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    Actually it is d/dt (mv) = dm/dt*v + m*dv/dt
    Since the velocity of he exhaust is constant, the second term is zero so
    F = v*dm/dt

    To calculate alpha (angular acceleration), you need some formula analogous to F = ma relating the Torque to the angular acceleration. Do you have any formulas that say "Torque = ..." ?
  12. Feb 25, 2009 #11
    I know Torque= I*alpha.

    I have I but Toruqe and alpha are unknown so how can I calculate ?
  13. Feb 25, 2009 #12


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    You can figure out the torque caused by those jets since you know the force each one exerts.
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