Cylinder with Displaced Center of Mass Rolling Down Incline

  • #1

Homework Statement


A rigid cylinder of radius ##R## and mass ##\mu## has a moment of inertia ##I## around an axis going through the center of mass and parallel to the central axis of the cylinder. The cylinder is homogeneous along its central axis, but not in the radial and angular directions. Thus, its center of mass is displaced a distance ##a## (##0 < a < R##) from the central axis. The cylinder can roll without slipping on a plane and dissipation due to friction is assumed to be negligible. The plane is inclined on an angle ##\theta## with respect to the horizontal and the central axis of the cylinder is parallel to the horizontal plane. Find the Euler-Lagrange equations for the cylinder and discuss qualitatively the difference in motion of the cylinder for ##\theta =0## and non-zero ##\theta##.

Homework Equations




The Attempt at a Solution


The cylinder can be considered to be rotating about its instantaneous point of contact with some angular velocity ##\dot{\phi}## where ##\phi## is the angle between a line through the geometric center of the cylinder and line from the center of mass perpendicular to it. So the translational velocity of the center of mass of the cylinder is ##V_{cm} = |b| \dot{\phi}## where ##|b|## is the distance of the center of mass from the point of contact. From the cosine law this distance is ## |b| = \sqrt{a^2 +R^2 -2aR\cos{\phi}}##. So the kinetic energy is then the sum of the translational energy and rotational kinetic energy of the cylinder. The difficulty I'm having with this problem comes when I have to find the potential energy and impose the rolling constraint. Calling the displaced distance down the incline ##x## then the potential energy should be ##-\mu g x\sin{\theta}## but gravity produces a torque about the rotation axis because the center of mass doesn't lie directly above it so there is an additional term ##-\mu g (R-a\cos{\phi})##. So i don't see how to finish up finding the Lagrangian and put these pieces together. For ##\theta=0## only the second potential energy term would appear and I can solve for the motion in that case.
 

Answers and Replies

  • #2
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I would probably consider the problem as seen from the geometric axis, but your approach works as well.
So the kinetic energy is then the sum of the translational energy and rotational kinetic energy of the cylinder.
Relative to the contact point there is only rotation (but you have to find the moment of inertia I').
So i don't see how to finish up finding the Lagrangian and put these pieces together.
Which part is unclear? You still need a relation between x and ##\phi## but that shouldn't be difficult.
For ##\theta=0## only the second potential energy term would appear and I can solve for the motion in that case.
While that is true it is probably not the discussion the problem asks for.
 

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