Professor Vandiver, thank you for another classic MIT recitation on the Cart, Pendulum and Lagrange Method and it's power impact on Engineering Dynamics.
The motion of the rod can be viewed by 2 ways. One is the rotation motion about G + the translation motion about G. The other one is the rotation motion about A. This prof. used way 1. And your thinking is way 2. These 2 ways are identical.
I know it's been a while since this was answered but maybe others will read it and get confused. Therefore: THIS IS NOT TRUE. The 2. way you mentioned will give you the wrong result for the kinetic energy, because point A is moving. Prof. Vandiver mentioned it even during the lecture at around 10:25
@@frederikrentzsch9737 What if in the term of the motion about the x axis of the entire system we include both the rod and the cart masses, so that we get (1/2)(M+m)x_dot^2 ? Can we then use the parallel axis theorem to express the rotational kinetic energy of the rod as (1/2)(1/3)mL^2(theta_dot)^2 = (1/6)mL^2(theta_dot)^2 ? I feel like this should work
You probably don't care anymore, but for anyone asking the same question: The spring applies a force when it is stretched past its natural length. So when it is unstretched, there is no force, and thus, no potential energy.
i didn't understand clearly why at 10:20 the angular momentum of H is at G but not at A....so the inertia matrix will be also with respect to rotating point A... can someone give a hand?? :D
En videos pasados, se mostró un ejemplo donde el profesor lanza un disco en el aire y menciona que cuando el cuerpo rígido se traslada y rota a la vez, entonces, el cuerpo rota y se traslada con respecto a su centro de masa, por lo que en este caso, el momento angular debe ser también con respecto a G para describir la rotación
Professor Vandiver, thank you for another classic MIT recitation on the Cart, Pendulum and Lagrange Method and it's power impact on Engineering Dynamics.
Wow! Incredible! Thank you very much. I start to see the power of the Lagrangian method!
MIT makes it look harder than what it actually is
😍😍😍Covid confinement and i am learning lagrange as if....
If you're finding v_G/O confusing, consider writing down r_G/A, and taking a derivative.
why is the kinetic energy due to rotation of the rod is calculated about G and not A ?
is it because the OvG accounts for translation and omega wrt G accounts for roation ?
The motion of the rod can be viewed by 2 ways.
One is the rotation motion about G + the translation motion about G.
The other one is the rotation motion about A.
This prof. used way 1.
And your thinking is way 2.
These 2 ways are identical.
Yes, I understood it later! Thanks
I know it's been a while since this was answered but maybe others will read it and get confused. Therefore: THIS IS NOT TRUE. The 2. way you mentioned will give you the wrong result for the kinetic energy, because point A is moving. Prof. Vandiver mentioned it even during the lecture at around 10:25
@@frederikrentzsch9737 What if in the term of the motion about the x axis of the entire system we include both the rod and the cart masses, so that we get
(1/2)(M+m)x_dot^2 ?
Can we then use the parallel axis theorem to express the rotational kinetic energy of the rod as
(1/2)(1/3)mL^2(theta_dot)^2 = (1/6)mL^2(theta_dot)^2 ?
I feel like this should work
Hi, I just wondered why didn't we account the unstrected length of the spring. Thanks for the great video by the way. :)
You probably don't care anymore, but for anyone asking the same question: The spring applies a force when it is stretched past its natural length. So when it is unstretched, there is no force, and thus, no potential energy.
@@tabhashim3887 legendary
i didn't understand clearly why at 10:20 the angular momentum of H is at G but not at A....so the inertia matrix will be also with respect to rotating point A... can someone give a hand?? :D
En videos pasados, se mostró un ejemplo donde el profesor lanza un disco en el aire y menciona que cuando el cuerpo rígido se traslada y rota a la vez, entonces, el cuerpo rota y se traslada con respecto a su centro de masa, por lo que en este caso, el momento angular debe ser también con respecto a G para describir la rotación
I think one part is missing in the solution and one of the students also raises this issue in the 24th min of the video.
GATE ME 2021 same question..
garkein drip diggah
Gate exam copied it.
Mit me board hota h