sir, at 10:08 in energy approach why we considered translation KE for point mass why dont you take it as rotational KE just like for mass of rod you have taken
I would suggest you to also see this prob. see only up to energy method for simple pendulum 7 .00 min. Plz understand, in case of applying energy method to a mass moving in a curved line. either you will apply KE as 1/2mv^2 or 1/2I(omega)^2 where I is MOI about the point of rotation. so I = mr^2. So when you put I into second expression, it will become same as first expression as (omega*r)^2=v^2. I hope it is clear, if not let me know.
Sir, I am working on a project is it possible that you can help me to derive the natural frequency of my system. it is very similar to this problem but just with one spring.
Energy method starts at 6:00
This degree of explanation is simply genius.
Sir this system is in a plane..... Because here potential energy of rod and potential energy of mass m are absent... Please clearify this.
when simplifying the equation of motion in the first method, how did the L's disappear?
thank you! so helpful
Why was angle theta not used in the equation
Why are all the forces assumed in the same direction..and when i assume one force in the spring to go a different direction the answer is different
I guess you are talking about spring forces. When the body moves in a particular direction, all spring forces act opposite of it.
Thank you I understand now.. all forces not in direction of motion of mass are negative
Thank you I understand now.. all forces not in direction of motion of mass are negative
sir,
at 10:08 in energy approach why we considered translation KE for point mass
why dont you take it as rotational KE just like for mass of rod you have taken
I would suggest you to also see this prob. see only up to energy method for simple pendulum 7 .00 min. Plz understand, in case of applying energy method to a mass moving in a curved line. either you will apply KE as 1/2mv^2 or 1/2I(omega)^2 where I is MOI about the point of rotation. so I = mr^2. So when you put I into second expression, it will become same as first expression as (omega*r)^2=v^2. I hope it is clear, if not let me know.
@@ashishpurohit9138 yes sir got it ,now i got both as same thing ...
Thank you so much for your reply ❤️
Sir this system is in a plane..... Because here potential energy of rod and potential energy of mass m are absent... Please clearify this.
sir =excellent =pls suggest book =thank u =amarjit
Awesome sir❤️❤️👍
Sir, I am working on a project is it possible that you can help me to derive the natural frequency of my system. it is very similar to this problem but just with one spring.
وعلي فهمت
Thanks 😊
Is that 8 or 16
why mg is not considered for the inertial mass during newton's method?
It cancels itself out in the beginning.
Thank you
Sir are u from odisha??
No Dear Subrat, I belongs to Central India. Bu currently in punjab.
@@ashishpurohit9138 sir I want to talk with u for 5 minutes can u give your contact no plz.....I want to talk with u regarding mtech on iiT
ashishpurohit1(a)gmail
Thanks 🙏🏼