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Rotational kinectic energy

  1. Jul 2, 2006 #1
    a trebuchet of negligible mass, 3m long joining particles of mass 60 kg and 0.12kg at it's ends. it can turn on a frictionless horizontal axle perpendicular to the rod and 14cm from the largemass particle. the rod is relaesed from rest in a horizontal orientation.find the max. speed that the small mass object attains.

    there the problem...i dont know how to start it....if some1 could jus tell me how to struture the energies, then i may be able to solve the problem..

    regards
    vijay
     
  2. jcsd
  3. Jul 2, 2006 #2

    Andrew Mason

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    What is the change in energy of the two masses when the rod moves from from horizontal to vertical? What is the ratio of speeds of the heavy mass to the light mass?

    AM
     
  4. Jul 2, 2006 #3
    i dont know what you mean....i calculated the moment of inertia of the object and also found out the rotaional kinectic energy of the whole system. but i dont know how to apply that concept in findind out how fast the smalller can go.....plz help..
     
  5. Jul 2, 2006 #4
    What's the formula for kinetic energy (rotational KE in this case, but if you know the equation for linear KE then you can work it out from there) in terms of some other useful variable?
     
  6. Jul 2, 2006 #5
    i dont think linear ke is effect in solving this problem....this is because the system run not only on translational motion but also in rotational motion.so linear kinectic energy can be used in some parts of solving but the main part of it requires rotaional ke.
    the formula of rotaional ke is 0.5I(omega)(squared).were I is the moment of inertia and omega is angular velocity. so could find the rotational ke but i still cant relate it with the translational part
     
  7. Jul 2, 2006 #6

    Doc Al

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    There's nothing wrong with using rotational KE to analyze this problem. If you found the rotational KE (how did you find it?) and the rotational inertia, then you should be able to find the maximum angular speed. Use that to find the linear speed.
     
  8. Jul 2, 2006 #7
    Apologies, I more meant that if you know [itex]KE\sub{linear} = \frac{1}{2} m v^2[/itex] then you'd be able to figure out how rotational KE worked from that. As Doc Al said, if you have the rotational KE and the moment of inertia you can find angular velocity, and from there you can find linear velocity.
     
  9. Jul 2, 2006 #8

    Andrew Mason

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    Since rotational KE is the same as the sum of the linear KE of each mass, it doesn't matter whether you use rotational KE and linear KE. The analysis is the same. The essential part of this problem is to determine what that KE is. What is the source of the kinetic energy?

    AM
     
  10. Jul 2, 2006 #9
    yea, i think you guys are right.
    but once you find the kinetic energy, how is it possible to link them up to form the final equation.
    i mean, is rotaional ke=gravitational ke+.5mv(sqaured)....
    yea thats jus an example, but how do you link it up for this problem?
    regards
    vijay
     
  11. Jul 2, 2006 #10

    Doc Al

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    Either think of the two masses as a rotating system with rotational KE or as two masses with their own linear KE (as Andrew Mason has explained). Pick one or the other method (not both!) since the two methods give identical results.

    Realize that the motion of the two masses is linked by their attachment to the rod, no matter which method is used.
     
  12. Jul 3, 2006 #11

    Andrew Mason

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    The change in potential energy is equal to the kinetic energy:

    [tex]\Delta U = Mg\Delta h_1 + mg\Delta h_2 = KE = \frac{1}{2}Mv_1^2 + \frac{1}{2}mv_2^2 (= \frac{1}{2}I\omega^2)[/tex]

    You can work out the ratio of v2 to v1 easily enough from their respective radii.

    AM
     
  13. Jul 3, 2006 #12
    thanks for the equation mason...the ans. i got is correct....
     
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