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Kinectics of particles. impact between 2 bodies 1 of which is stationary

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data
    An engine is suddenly coupled to a rotating drum by a friction clutch. the moment of inertia of the engine is equivalent to a mass of 40kg acting with a radius of gyration of 111mm. the drum has a mass of 10kg and a radius of gyration of 100mm. the initial velocity of the engine before engagement is 12rev/s and the drum is initially at rest. find the velocity of engine and drum immediately after connection.



    2. Relevant equations
    total moment before impact = total moment after impact
    (m1.u1) +(m2.u2) = (m1.v1)+(m2.v2)


    3. The attempt at a solution
    ive seen questions like this before without the gyration part in the question & i know that i'd just plug the weight into the equation along with the velocity for example
    (40x12) +(10x0) = (40xV)+(10xv) and then just rearrange to find the end velocity. However the addition of the moment of inertia and radius gyration has thrown me. do i use the equation I=m.k² and use the answers for the engine and drum and replace the m1 and m2 with the I1 and I2 value id get from that equation?
     
  2. jcsd
  3. Apr 11, 2012 #2

    gneill

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    Staff: Mentor

    Radius of gyration, rg is related to the mass and moment of inertia of a rotating body:

    ##r_g^2 = \frac{I}{M}##
     
  4. Apr 11, 2012 #3
    yeah i've got that part but how does the moment of inertia from that equation help me with finding the impact of the engine and drum? do i replace the mass in the impact before and after with I=M.R² so it would be (I1.u1) +(I2.u2) = (I1.v1)+(I2.v2) or do i use a different formula all together? I understand the theory of why the gyration is given in the question but how do i apply all the values i've been given into finding the answer?
     
  5. Apr 11, 2012 #4

    gneill

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    Staff: Mentor

    You're dealing with an inelastic collision of the angular variety. When the bodies come together their moments of inertia combine...
     
  6. Apr 11, 2012 #5
    so from that then would it follow that (I1.u1) +(I2.u2) = (I1+I2).V2,,,,,, V2=(I1.u1) +(I2.u2)/(I1+I2). therefore, I1=40X0.111m²=0.49284 & 10x0.1m²=0.1 then v2=9.98rev/s ? Would you agree with that or am i going in the wrong direction :)
     
  7. Apr 11, 2012 #6

    gneill

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    Staff: Mentor

    Almost there... Remember that it's body 2, the motor, that's initially rotating.
     
  8. Apr 11, 2012 #7
    oh of course yes! thank you for your help :)
     
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