Lagrange Multipliers - Force of constraint for a disk rolling down an incline

[@fizixx]

Nice run-through for these two methods!

[@andyrundquist7672]

It's interesting that we tend to need the second time derivative of the constraint equation, rather than just it by itself when solving for the equations of motion. I realized that when I was playing around and I realized that if I told Mathematica to honor the second derivative of the constraint equation it's able to integrate all the equations. If I just give it the constraint it (usually) fails. But if you give it the second derivative of the constraint you have to be really careful with your initial conditions. If you give the variables an initial velocity such that the first derivative of the constraint isn't zero, that solution will have that constraint constantly growing! That took me a while to figure out.

[@fcvgarcia]

Thanks for the great explanation!

[@williamperez-hernandez3968]

Comment on the signs. The torque is negative, so in R×friction, friction is not the magnitud but the component along the plane. That is OK, we saw it was negative by Langrange method. Care must be taken for the accelerations. Note s=R(theta) is taking both variables as positive, as was done here. But the derivatives have opposite signs, sdot > 0 while thetadot < 0. Same is true for the second derivates. Therefore sddot = - R (thetaddot).

[@nishchayy]

16:18 friction force = I *theta double dot / R square... I think you miss the square on R

[@rajvardhansinghsisodiya1095]

Why dont you find constraint in theta ??

[@sahidtenorio601]

Sir. The KE of a rotating body is T = 1/2mv^2 + 1/2Iw^2... Not theta dot

[@DotPhysics]

Sorry - I don't remember exactly what's there, but it looks like I have (1/2)*I*thetadot^2, right?

[@FULLAWESOMENESS12]

Theta dot = omega