"Value for money" makes me chortle. You have obviously practiced this material endlessly and I wish to thank you for sharing your many insights. I went through this one numerous times. I like the way it incorporates two simpler dynamic systems into one. Lacking the mental agility to do all the derivatives with great efficiency, I decided to implement the LaGrange equations with the python symbolic module called sympy. If my solutions differ greatly from yours, it has turned out to be an error in my grasp of the kinematics. You are helping me continue to enjoy my retirement despite the restraints of the pandemic-induced lockdown.
First I take the derivative, ∂L/∂θ_dot which gives me ∂L/∂θ_dot = m(L0+x)^2 · θ_dot then I need to take the time derivative of this, d()/dt. Because both x and theta_dot are functions pf time, I need to apply the product rule AND the chain rule. This gives me two terms. d (∂L/∂θ_dot) / dt = 2m(L0+x) · dx/dt · θ_dot + m(L0+x)^2 · d (θ_dot) / dt = 2m(L0+x) · x_dot · θ_dot + m(L0+x)^2 · θ_ddot)
is the Q_i term being zero at 5:45 because there are no external forces in the x direction? and in the case of the theta, there are no external torques in the theta direction? Thank you!
This is a continuation from the simple pendulum problems I worked in previous videos. In the case of a simple problem, we assume a point mass, so there is no rotational kinetic energy. I mentioned it in some of the other video, but probably should have made it clearer in this one. Typically, in the case of a simple pendulum, the rotational KE tends to be negligible and can thus be ignored unless the radius of the mass is very large compared to the length of the string.
Hello sir, When writing out the kinetic energy of the system, why didn't you split it into 2 terms, one for the translational and another for rotational to get T = 0.5*m*(xdot^2) + 0.5*I*(Theta dot^2)? "I" here is the mass moment of inertia.
Generally in the case of the simple pendulum, the rotatory kinetic effects are negligible (unless the diameter of the mass is very large or the string is very short). I had made some simple pendulum videos prior to this one and in those I believe I made the assumption that the rotatory kinetic energy could be ignored. This video follows from those, but I probably should have mentioned it.
Yes, this is true. But the application of Lagrange's Equation requires us to subtract the ∂L/∂θ term (2nd term on the left-hand side of Lagrange's Eqn has a -ve sign).
Good question. The resulting frequency/period will be a function of both the magnitude and frequency of the applied load(s). However, if you are talking about the frequencies/periods of the natural modes then these can be found by solving the eigenvalue problem.
gravitational potential energy term is [ -mg(x+l)cos0 ] that u wrote in the solution but i think it should be [ -mg(x+l)(1-cos0) ] because the height h=(x+l)-(x+l)cos0. (read 0 as theta)
This is a function of where you place the origin of your coordinate system. I have placed mine at the hinge and you have placed yours at the mass (in its equilibrium position). So the values differ by a constant. However, this makes no difference in the end because when you apply Lagrange's equations, the constant gets differentiated out - i.e. what is important is not the absolute value of the potential energy, but rather in the change in the potential energy as θ changes..
@@Freeball99 i think we can't take the ground level or origin from the bottom in elastic pendulum because the length of pendulum changes so we have to take ground level from the hinge, like u did.
@@Freeball99 Yes. Thanks for the reply! I thought setting the right hand side to 0 meant you were solving for when there was no rate of change of x and theta? How come it's 0 because there are no external loads?
@@matthewlee2023 From Lagrange's equation (#4) the right-hand sign contains the generalized load, Q for each coordinate. In this problem, there are no external loads, so Q_i is zero for each equation. If, for example, there was a moment, M(t) applied at the hinge, then this would appear on the right side of the θ equation. Similarly, if there was a force, F(t) applied at the mass in the radial direction, then this would appear on the right side of the x equation.
in the kinetic energy in the beginning why are you not deriving l (length of the pendulum) with respect to time? the length of the pendulum is l(t) so you should take it into account in the derivation, shouldn't you?
Hello, your video was very helpful with solving this problem ^^. However I have one question: how do you know that the generalized forces Qi must be equal to zero? Or, at least that's what I understood from the video
Correct. The generalized forces are the externally applied loads (can be conservative or non-conservative). Typically the internal forces (conservative) are included in the potential energy expression.
While there is definitely coupling in the motion between x and θ, x is not a function of theta. As an example, if this system is at rest (θ=0) and I give the mass a slight nudge in the vertical direction, the mass will simply vibrate up and down like the simple harmonic oscillator.
These are, in effect, the same thing. It just differs by a constant, so what you have done is assumed that zero y-location is at the mass while I have assumed that the zero y position is at the hinge. The point is this...we don't really care about the absolute value of the potential energy (and the location of zero potential energy is somewhat arbitrary). Rather, what we care about is the DERIVATIVE of the potential energy and the derivatives are identical (because you simply added a constant, l, which disappears when you differentiate it).
"Value for money" makes me chortle. You have obviously practiced this material endlessly and I wish to thank you for sharing your many insights. I went through this one numerous times. I like the way it incorporates two simpler dynamic systems into one. Lacking the mental agility to do all the derivatives with great efficiency, I decided to implement the LaGrange equations with the python symbolic module called sympy. If my solutions differ greatly from yours, it has turned out to be an error in my grasp of the kinematics. You are helping me continue to enjoy my retirement despite the restraints of the pandemic-induced lockdown.
Thank you for you kind words. This makes me smile inside.
Hi there, thanks for the video! Really helped me get comfortable with the Euler-Lagrange equations :))))
This is a perfect example for my TA class, Thanks!!!
I have a doubt at 6 min 25 sec. How you did differentiation here ? I didn't understand that step.
First I take the derivative, ∂L/∂θ_dot which gives me
∂L/∂θ_dot = m(L0+x)^2 · θ_dot
then I need to take the time derivative of this, d()/dt. Because both x and theta_dot are functions pf time, I need to apply the product rule AND the chain rule. This gives me two terms.
d (∂L/∂θ_dot) / dt
= 2m(L0+x) · dx/dt · θ_dot + m(L0+x)^2 · d (θ_dot) / dt
= 2m(L0+x) · x_dot · θ_dot + m(L0+x)^2 · θ_ddot)
❤❤nice 👍 @@Freeball99
thank you, your explanation is very simple and direct
Thanks for your work, I've been struggling with aeroelasticity and your videos have helped me
What if I want to implement a damping force so that the pendulum is both elastic and damped?
is the Q_i term being zero at 5:45 because there are no external forces in the x direction? and in the case of the theta, there are no external torques in the theta direction? Thank you!
Yes, that's correct.
This is a very helpful video. Thank you!
about the kinetic energy - dont you need to consider the Inertia energy as well? (1/2)*(I)*(omega^2)?
This is a continuation from the simple pendulum problems I worked in previous videos. In the case of a simple problem, we assume a point mass, so there is no rotational kinetic energy. I mentioned it in some of the other video, but probably should have made it clearer in this one. Typically, in the case of a simple pendulum, the rotational KE tends to be negligible and can thus be ignored unless the radius of the mass is very large compared to the length of the string.
Hello sir,
When writing out the kinetic energy of the system, why didn't you split it into 2 terms, one for the translational and another for rotational to get T = 0.5*m*(xdot^2) + 0.5*I*(Theta dot^2)? "I" here is the mass moment of inertia.
Generally in the case of the simple pendulum, the rotatory kinetic effects are negligible (unless the diameter of the mass is very large or the string is very short). I had made some simple pendulum videos prior to this one and in those I believe I made the assumption that the rotatory kinetic energy could be ignored. This video follows from those, but I probably should have mentioned it.
7:14 Don't the differential of cos is -sin?
Yes, this is true. But the application of Lagrange's Equation requires us to subtract the ∂L/∂θ term (2nd term on the left-hand side of Lagrange's Eqn has a -ve sign).
@@Freeball99 Oh, I missed that, thanks for your help.
What will be the time period for this pendulum.
Good question. The resulting frequency/period will be a function of both the magnitude and frequency of the applied load(s). However, if you are talking about the frequencies/periods of the natural modes then these can be found by solving the eigenvalue problem.
Thank you very much. Your video was helpful.
why is there no subscript i on the q for d/dq at 4:20? also thank you, really love these-- quality walkthrough with a pleasant voice!
Careless error! There should be a subscript i on the q. Thanks for catching that.
Can you please make a video on " Hamilton's equations of motions for rotating coordinate axes
Yes. I will do this eventually as soon as I can figure out how to keep the length to a minimum. Can be very lengthy.
gravitational potential energy term is [ -mg(x+l)cos0 ] that u wrote in the solution but i think it should be [ -mg(x+l)(1-cos0) ] because the height h=(x+l)-(x+l)cos0. (read 0 as theta)
This is a function of where you place the origin of your coordinate system. I have placed mine at the hinge and you have placed yours at the mass (in its equilibrium position). So the values differ by a constant. However, this makes no difference in the end because when you apply Lagrange's equations, the constant gets differentiated out - i.e. what is important is not the absolute value of the potential energy, but rather in the change in the potential energy as θ changes..
@@Freeball99 got it, thanks
@@Freeball99 i think we can't take the ground level or origin from the bottom in elastic pendulum because the length of pendulum changes so we have to take ground level from the hinge, like u did.
Thanks for your help
why when you take the derivative with respect to theta and x does it equal 0?
Are you asking why the right hand side of equations 5 & 6 are zero? If so, this is because there are no external loads in the x or θ directions.
@@Freeball99 Yes. Thanks for the reply! I thought setting the right hand side to 0 meant you were solving for when there was no rate of change of x and theta? How come it's 0 because there are no external loads?
@@matthewlee2023 From Lagrange's equation (#4) the right-hand sign contains the generalized load, Q for each coordinate. In this problem, there are no external loads, so Q_i is zero for each equation. If, for example, there was a moment, M(t) applied at the hinge, then this would appear on the right side of the θ equation. Similarly, if there was a force, F(t) applied at the mass in the radial direction, then this would appear on the right side of the x equation.
@@Freeball99 Got it! Thanks!
in the kinetic energy in the beginning why are you not deriving l (length of the pendulum) with respect to time? the length of the pendulum is l(t) so you should take it into account in the derivation, shouldn't you?
I am including it. This is what contributes to the radial velocity which I've denoted Vr or x_dot.
Hello, your video was very helpful with solving this problem ^^. However I have one question: how do you know that the generalized forces Qi must be equal to zero? Or, at least that's what I understood from the video
Okay, I think I got it. I think it is because the only force that does work (weight force) is conservative. Is this right?
Correct. The generalized forces are the externally applied loads (can be conservative or non-conservative). Typically the internal forces (conservative) are included in the potential energy expression.
I have a doubt! Why is your X an independent coordinate! Isn't that a function of theta as well, coz X= mg cos(theta)/k
While there is definitely coupling in the motion between x and θ, x is not a function of theta. As an example, if this system is at rest (θ=0) and I give the mass a slight nudge in the vertical direction, the mass will simply vibrate up and down like the simple harmonic oscillator.
Why is the heighe not l-lcos(theta) ?
These are, in effect, the same thing. It just differs by a constant, so what you have done is assumed that zero y-location is at the mass while I have assumed that the zero y position is at the hinge. The point is this...we don't really care about the absolute value of the potential energy (and the location of zero potential energy is somewhat arbitrary). Rather, what we care about is the DERIVATIVE of the potential energy and the derivatives are identical (because you simply added a constant, l, which disappears when you differentiate it).
you should not include the minus term on the potential when your positive axis direction is antiparallel with the field of gravity.
Not sure what point of the video you are referring to. Where did I include a negative term incorrectly?
Good...one
We can get the time period
thankyou so much
Shouldn't this problem have just 1 DOF
No. In order to uniquely locate the mass in the 2D plane, we require both x and θ. This is similar to using polar coordinates.
Thank you
Thank you...
Now solve the equations
There you go: th-cam.com/video/mecThHHV4eg/w-d-xo.html
One arm? Come one, you could get it even worse than that. How about three arms? ;)
this was delicious
Coulc you please spend a little bit of your time and answer my question please?