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Angular velocities of drums

  1. Aug 13, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-8-13_19-37-28.png upload_2017-8-13_19-38-21.png



    2. Relevant equations


    3. The attempt at a solution

    Since there is no net torque or net force acting on the system( which consists of the system given in the picture), I applied conservation of angular momentum and energy.
    I took on the L.H.S. the angular momentum or energy of the system at time t and on the R.H.S, the corresponding quantity at time t = 0.
    I took the expression for ##ω_b (t) ## from the conservation of angular momentum equation as a function of ##ω_a (t) ## and substituted in the energy eqn. to get ##ω_a (t) ## as a function of t.
    But, the calculation was so much that I felt first to become sure that I am on the right path.
    So, am I correct so far?

    About applying conservation of energy, I have a doubt.
    If the total force or torque acting the system is 0, is it compulsory that the total work done is also 0?
    Total force being 0 means if there is a force ## \vec F ##, then there exists either an opposite and equal force or of sum of the rest of the forces is opposite or equal to this force. So, the work done by one force gets cancelled by the other \ sum of rest of all others. Hence, the answer to the above question is yes.A
     

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  3. Aug 13, 2017 #2

    Orodruin

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    I assume you mean external force/torque. The total energy in the system is conserved, but it can be redistributed to other degrees of freedom (such as heat). In this case, compare to an inelastic collision where there are no external forces, yet the total kinetic energy is not conserved.
     
  4. Aug 13, 2017 #3
    Why is the total kinetic energy not conserved?
    There is no net external torque or force is acting on the system. So, there is no net work being done on the system. Hence, the total mechanical energy of the system will remain constant.
    Now, the net potential energy of the system (which is here gravitational potential energy) is constant as the centre of mass remains constant.
    Hence, the net kinetic energy has to be conserved.
     
  5. Aug 14, 2017 #4

    Orodruin

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    No. Again you are ignoring the fact that forces within the system are affecting the internal degrees of freedom. No energy dissipated out of the system, but its kinetic energy may be transformed into other internal forms of energy - consider two lumps of clay colliding and sticking together.
     
  6. Aug 14, 2017 #5
    So, validity of the above statement assumes that the internal interactions in the system doesn't change mechanical energy of the system.
    Attention: So, before applying conservation of energy, I have to see whether the internal interactions causes the change in mechanical energy or not. Right ?
    This assumption is valid in case of 1) one particle system, 2) a system of particles and 3) rigid body.
    In case of clay, the assumption is not valid.
    But, in this problem, the assumption is valid as the system is a combination of rigid bodies ( two drums) and particles( sand). Isn't it?
     
  7. Aug 14, 2017 #6

    Orodruin

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    You have to be very careful here. An object that may be considered rigid for some purposes may not actually be a rigid body. A perfectly rigid body would not allow internal degrees of freedom but your drums generally will. If kinetic energy would be conserved, the sand would not stick to the outer drum. There are inelastic collisions in this problem every time a grain hits the outer drum.

    The assumption of the samd sticking to the outer drum is in direct conflict with your assumption of conservation of mechanical energy.
     
  8. Aug 14, 2017 #7
    You mean that the sticking of a sand- particle to the outer drum implies an inelastic collision between the sand-particle and drum. And this implies change in mechanical energy due to internal interactions.
    So, the conservation of mechanical energy cannot be applied to this system.
    I need to verify it.
    The problem could be modeled as a collision between two particles A of mass m and B of mass M. Initially A is moving with speed v and B is at rest. After the collision, the two moves together with speed V. I have to check whether the collision is elastic or inelastic.
    Conservation of momentum gives ## V = \frac m {m +M} v ##
    Change in kinetic energy gives ## K_f - K_i = \frac 1 2 \{ (m+M) { ( \frac m {m +M} v)^2} - mv^2 \} = ~ ~\frac{ -1} 2 mv^2 \{\frac M {M+m} \} ≠ 0##
    This means that whenever sticking happens between two colliding particles, the collision is inelastic. Doesn't it ?

     
  9. Aug 14, 2017 #8

    Orodruin

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    Yes. In a collision between two particles, it is easily verified that the relative speed (note: speed, not velocity) between the particles is conserved by looking at the problem in the centre of mass system. Thus, if they stick together the collision is inelastic.
     
  10. Aug 14, 2017 #9
    Solving the OP's problem,
    The unprimed quantity is measured at time t = o and the primed quantity is measured at time 't'.
    Conservation of Angular momentum about the axis of the drums gives,
    ##M_a ω_a a^2 = M'_a ω'_a a^2 + M'_b ω'_b b^2 ##
    I need one more eqn. to get the answer.
    I cannot apply conservation of momentum (and energy). So, what should I do?
     
  11. Aug 14, 2017 #10

    Orodruin

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    How much angular momentum does the sand carry away from the inner drum?

    Does the sand leaving change the angular velocity of the inner drum?
     
  12. Aug 14, 2017 #11
    Thanks for this insight.
    In center of mass frame , the momentum of the system is 0.So, if the particles stick together, their final speeds and so kinetic energy becomes 0 in this frame
    (While the initial kinetic energy is non-zero showing that the collision is inelastic).
    The above question is the key to the solution. Thanks for it.
    The leaving sand has angular velocity same to that of inner drum. So, it doesn't change the angular momentum of the inner drum.
    Angular momentum carried by the sand in time t = Angular momentum of bigger drum with the sand on it
    ##λtω_a a^2 =( M_b + λt)ω_b b^2 ##
    ## ω_b = \frac {λta^2 ~ω_a} {( M_b + λt) b^2} ##
    ## ω_a = ω_a(0)##
     
  13. Aug 14, 2017 #12

    Orodruin

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    Yes. That is how I would have solved the problem.
     
  14. Aug 14, 2017 #13
    Thanks
     
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