# Lecture – 4 Using the lagrangian Equation to Obtain Differential Equations(Part-I) ### 12 thoughts on “Lecture – 4 Using the lagrangian Equation to Obtain Differential Equations(Part-I)”

• January 3, 2011 at 8:16 pm

Is this taught to 1st Year IIT students?

• December 27, 2011 at 9:45 pm

very gud and clear explanation….thanks a lot to the Prof. and IITM team…. 🙂

• October 21, 2012 at 9:01 am

It is the partial derivative WRT theta. Not a proper derivative WRT "time".
eg. consider f=yx^2… df/dy = x^2 + 2yx dx/dy (via implicit differentiation)
but the partial derivative pdf/pdy = x^2, and that's it.
This is a similar scenario.

• November 14, 2012 at 12:56 am

I think it is because he is differentiating with respect theta-dot, so cos(theta) looks like a constant and stays the same just like m,l, and y-dot-s. Does that make sense?

• April 29, 2013 at 1:49 am

Why are the potentials negative?

• May 5, 2013 at 12:35 am

at 25 mins in where he defines the height of mass 2 as (l-q) shouldn't that be (l -q) – pi*r where r is the radius of the pulley, as some length of the string is wrapped around the pulley?

• May 5, 2013 at 12:51 pm

no, when he said the total length is "l", he is taking into account that part of string which is wrapped around the top half as well, in other words, the "l" he wrote, is equal to " h + R(pi) + m" where h is the distance of first string, and m is of the other, this whole thing is included in his "l". mkay?

• May 6, 2013 at 7:59 pm

Hi there, thanks for the response,

Doesn't he use the "l" to define the vertical distance of the string? which would neglect the wrapped part?
Thanks

• October 14, 2015 at 11:51 am

nice

• August 28, 2017 at 8:19 am

Superb video! Excellent explanation…

• March 27, 2018 at 1:25 am
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