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    Lagrange Multipliers and Forces of Constraint

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    With reference to Figure 2, a small cylinder sits initially on top of a large cylinder of radius a, the latter being attached rigidly to a table. The smaller cylinder has mass m and radius b. A small perturbation sets the small cylinder in motion, causing it to roll down the side of the large cylinder. Assume that the coefficient of static friction u is sufficiently large that there is initially no slippage between the two cylinders. Using the method of Lagrange multipliers find the Lagrange's equations and expressions for the generalized forces of constraint. Analyze their relation with the normal force and the force of static friction. Noticing that as the cylinder rolls down the static friction will eventually not be enough to maintain it rolling without slipping, determine the angle theta for which the cylinder starts slipping.

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    © BrainMass Inc. brainmass.com December 24, 2021, 7:41 pm ad1c9bdddf
    https://brainmass.com/physics/rotation/lagrange-multipliers-forces-constraint-210366

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    Hi

    Please see the attachment.

    The energy of the motion of the cylinder's center of mass is:

    The energy of the cylinder's rotation about its own center of mass is:

    The potential energy is:

    Thus the general Lagrangian is:

    The first constraint is the fact that the cylinder rolls on the surface of the lower cylinder:

    The other is the constraint that the cylinder rolls without slipping:

    Therefore the modified Lagrangian is:

    The first equation is:

    The second equation is:

    And the third equation is:

    The three equations together:

    From the second constraint:

    Thus, from the second equation:

    Plugging it in the second equation:

    Integrating:

    Plugging this at the first equation we get an equation for the normal radial force ( ):

    The cylinder leaves the surface of the lower cylinder when the normal force is zero:

    The force of friction is the tangential force:

    The torque applied on the cylinder due to the friction force is:

    As a result:

    Then:

    We see that the friction force increases with , and it will reach its maximal value when at a certain angle:

    Beyond that angle, the friction force decreases and this when the slipping begins.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:41 pm ad1c9bdddf>
    https://brainmass.com/physics/rotation/lagrange-multipliers-forces-constraint-210366

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