1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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 rigid body

  1. Jul 14, 2011 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    I think the problem was taken out from Landau&Lifgarbagez's book on classical mechanics.
    An inhomogeneous cylinder of radius R rolls over a plane. The mass is distributed in such a way that a principal axis is parallel to the rotational axis of the cylinder and the center of mass is at a distance "a" from the rotational axis. The moment of inertia of the cylinder about the rotational axis is I. Calculate the kinetic energy of the cylinder.

    2. Relevant equations
    [itex]T=\frac{m v_{CM} ^2}{2}+ \frac{I \omega _c ^2 }{2}[/itex].


    3. The attempt at a solution
    The center of mass suffer from a circular motion of radius a and angular velocity [itex]\omega _c[/itex].
    So that [itex]v_{CM}=a \omega _c[/itex] and thus [itex]T=\frac{\omega ^2 _c }{2} (ma^2 + I)[/itex].
    This seems wrong to me because when a tends to 0, my kinetic energy equality tells me that there's only a rotational motion and no translational motion from the center of mass, which I believe it totally wrong.
    I don't see what I did wrong though... I'd love some help to figure out what's wrong with what I did. Thanks in advance.
     
  2. jcsd
  3. Jul 15, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Rolling means both translational and rotational motion. The centre of symmetry performs translation with velocity Vc. All points of the cylinder translate with Vc and rotate around the axis of symmetry with angular frequency equal Vc/R. The velocity of a point is the resultant of the translation and rotation.

    This is not a simple problem...

    ehild
     
  4. Jul 15, 2011 #3

    fluidistic

    User Avatar
    Gold Member

    You are absolutely right, I forgot about considering the translational motion.
    If I consider rolling without slipping, I get that v of translation is worth [itex]-\omega _c R[/itex].
    Thus [itex]T=\frac{\omega ^2 _c}{2} (ma^2+R^2+I)[/itex]. Is there something I'm missing?
     
  5. Jul 16, 2011 #4
    this cant be solved like this...the velocity of cm keeps on changing during motion

    the question should give the information of initial position of cm
     
  6. Jul 16, 2011 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I suggest to read Landau&Lifgarbagez's book on classical mechanics.
    The problem is presented and solved in that book. The motion of the cylinder is considered pure rotation about the instantaneously fixed axis (the line where the cylinder touches the ground).

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinetic energy of a rigid body
  1. Rigid body kinetics (Replies: 10)

  2. Rigid Body Kinetics (Replies: 4)

Loading...