Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic energy of a moving rotating rigid body

  1. Nov 11, 2006 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is the kinetic energy of a (not necessarily homogeneous) rigid body in translational motion and rotating about its CM the sum of the kinetic energy if the object was still but rotating plus the kinetic energy if the object was in linear motion but not rotating? This seems highly unintuitive and just like that, I can't spot where the key to the proof (or disproof is).

    Yet, my thermo text says that a certain molecule of mass m and momentum p has translational kinetic energy of p²/2m, as if translational and rotational kinetic energy were indeed separable.

    So may I know what the gist of this "separation thm" is please?
     
    Last edited: Nov 11, 2006
  2. jcsd
  3. Nov 11, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Rotational energy and momentum is independent of translational energy and momentum. You can find a frame of reference in which the object has no translational kinetic energy (the frame of the centre of mass), but you cannot find an intertial frame of reference where the rotational energy is 0. The angular momentum of the object is the same in all frames of reference (ie. not affected by changes in translational momentum). The translational momentum is not affected by changes in angular momentum (eg. by changes in moment of inertia).

    AM
     
    Last edited: Nov 11, 2006
  4. Nov 11, 2006 #3

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The velocity of any part of the object is
    v=wXr + V. If you calculate rho v^2, you will see that the cross term integrates to zero if r is the distance from the center of mass.
     
  5. Nov 11, 2006 #4
    Yes, this is true.

    Why? What proof do you require? It is a simple matter of superposition.

    Provided that there is no slip, the CM will translate with a linear velocity. This means there is linear momentum.

    In addition, the body will rotate with constant angular velocity. This is in the form of rotational kinetic energy.

    It's not clear to me whats not clear.
     
    Last edited: Nov 11, 2006
  6. Nov 11, 2006 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    kinetic energy is lab frame is

    [tex]\sum_i \frac{p_i^2}{2m_i}[/tex]

    Why is that also equal to

    [tex]\frac{P^2}{2M}+K_{CM}[/tex]

    where P is the "center of mass momentum", M the total mass, K_CM is the kinetic energy in the CM frame, i.e. the rotational energy.

    Maybe it's evident to you cyrusabdollahi but there must be a why nonetheless and my mind is too blank right now to see it. (All thermo and no play makes quasar987 a dull boy. All thermo and no play makes quasar987 a dull boy. All thermo and no play makes quasar987 a dull boy...:yuck:)
     
    Last edited: Nov 11, 2006
  7. Nov 11, 2006 #6

    jtbell

    User Avatar

    Staff: Mentor

    Start with the total kinetic energy of the system in the lab:

    [tex]K_{lab} = \Sigma \frac{1}{2} m_i v_{i,lab}^2[/tex]

    The velocities of each particle in the lab and in the center of mass are related by

    [tex]{\vec v}_{i,lab} = {\vec v}_{i,cm} + {\vec V}[/tex]

    where [itex]\vec V[/itex] is the velocity of the center of mass in the lab frame.

    Using the second equation, find an expression for [itex]v_{i,lab}^2[/itex]. Hint: [itex]v^2 = \vec v \cdot \vec v[/itex]. Substitute into the first equation and see what you get.
     
  8. Nov 11, 2006 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This is what I did in my head in the begining but upon the "sight" of those cross terms I dediced it was not working. Still can't see what to do with those cross terms jtbell...
     
  9. Nov 11, 2006 #8

    jtbell

    User Avatar

    Staff: Mentor

    There's only one cross term for each particle, and you can write it as a sum for all particles:

    [tex]\Sigma m_i ({\vec v}_{i,cm} \cdot \vec V)[/tex]

    Now, the dot product [itex]{\vec v}_{i,cm} \cdot \vec V[/itex] gives you the component of [itex]{\vec v}_{i,cm}[/itex] along the direction of [itex]\vec V[/itex], right?

    Suppose for simplicity that [itex]\vec V[/itex] is in the x-direction. Then the sum becomes

    [tex]\Sigma m_i v_{xi,cm} = \Sigma p_{xi,cm}[/tex]

    What's the total x-component of momentum of the system, in the center-of-mass frame? :wink:
     
  10. Nov 11, 2006 #9

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  11. Nov 12, 2006 #10
    jtbell is quite right, my appologies for not remembering about the derivation off the top of my head.

    However, from a physical standpoint, it should still be intuitive.

    You can always break down the motion into two forms. One is linear, and the other rotational. If you add the two motions, you get the total motion. This is the principle of superposition.

    Therefore, if you break down the energy into each component of motion, and then add those components back up, you should get the energy for the entire motion. Each fraction of energy for rotation and translation will be proportional to the square of the velocity and angular velocity.

    This should be more than reasonable. No?

    This works because rotation and translation are independent motions. If you want to control a system that has translation and rotation, you are going to need one controller for each independent motion along each orthognal axis to gain system stability.
     
    Last edited: Nov 12, 2006
  12. Nov 12, 2006 #11

    jtbell

    User Avatar

    Staff: Mentor

    Oops, I just noticed that there was something odd about the units in my derivation. I started out with a quantity that has units of energy and ended up with a quantity that has units of momentum! Here's the corrected version. Fortunately the final conclusion doesn't change. As before we start with the "cross term:"

    [tex]\Sigma m_i ({\vec v}_{i,cm} \cdot \vec V)[/tex]

    Now, the dot product [itex]{\vec v}_{i,cm} \cdot \vec V[/itex] gives you the component of [itex]{\vec v}_{i,cm}[/itex] along the direction of [itex]\vec V[/itex], right?

    [added] Wrong! it gives the component of [itex]{\vec v}_{i,cm}[/itex] along the direction of [itex]\vec V[/itex], multiplied by the magnitude of [itex]\vec V[/itex]. :blushing:

    Suppose for simplicity that [itex]\vec V[/itex] is in the x-direction. Then the sum becomes

    [tex]\Sigma m_i v_{xi,cm} V = V \Sigma p_{xi,cm}[/tex]

    and it equals zero because the total x-component of momentum of the system must be zero, in the center-of-mass frame.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kinetic energy of a moving rotating rigid body
  1. Rotation of rigid body (Replies: 4)

  2. Rotation of Rigid body (Replies: 2)

Loading...