I have a question about the model of Coulomb friction you used in this video. When the velocity of the block is zero, the signum function gives the value of friction to also be zero but is this actually true? For example, in the final resting position, if we assume the block rests in a non-equilibrium position, its velocity is zero but there is still some friction force acting on it to counteract the small amount of spring force. I assume the reason you excluded this nuance is because in its normal cycle, when the block reaches its farthest point, the spring force is large enough to overcome the static friction. So does this mean, in a more generalized context for complex systems, we need to take the value of signum function as [-1,1] when velocity is zero? Using this model introduces non-smooth behavior which makes the solution a lot more involved. Maybe you could consider solving a system that has this general Coulomb friction in a future video?
please solve this Mcqs Thank you. 1) All vibrating bodies have following Degree of Freedom: a)1 b)2 c)3 d)4 . 2) The frequency of vibrations with increase of damping in the case of free vibrations with coulomb damping will A.Remain mine B.Increase C.Decrease D.May increase or decrease depending upon the damping coefficient
1. All vibrating masses, in general, can have 6 degrees of freedom (3 translations and 3 rotations). Since 6 is not an option (which it ought to be) I am going to have to go with 3 - perhaps they are asking about translations only. 2. In the case of Coulomb damping, this does not affect the frequency of vibration, it only decreases the amplitude of each successive oscillation. The mass will continue to vibrate at its natural frequency regardless of the amount of Coulomb damping applied.
Wonderful videos and terribly instructive! Maybe I'm misunderstanding the scenario here though. The mass is at some equilibrium position, is released, and then displaces to the right. It traverses to the right a distance x0 until the instant where its velocity is 0, and then begins its movement left. It then displaces to the left a distance x0 to the equilibrium position, then a distance x1 until the spring is entirely compressed and the velocity is again 0. Finally, the mass travels, again, a distance x1 to the equilibrium position and then another distance, x2, such that x2 < x0. If that sounds about right, I guess I'm not sure how you are able to pick those exact times for your boundary conditions? The units of pi/wn work out to seconds, but I'm just not sure how you arrived at that.
Sorry for the delayed response, but I somehow missed this until now.. What you have stated above is exactly correct. I think the piece you are missing is this.... the frequency of oscillations, f, is related to the angular frequency, ωn, by f = ωn / 2π. Also, f is related to the period of vibration by f = 1/T. So the period of vibration can be written as T = 2π / ωn. In this particular problem, we are looking at half-cycles. Consequently, the start of the 2nd half-cycle begins at a time equal to half the period or t = π / ωn and so we use this as the initial time for the 2nd piece.
@@Freeball99 Does this mean for other cases in SDOF with damping, if we want to know each amplitude, have to be solved half-cycle by half-cycle? Is there any other way to write the EOM? And if in each half-cycle the EOM are greatly influenced by initial condition, is this mean the solution is not in closed form? Thank you.
@@andrealiu8650 This is only for the case of Coulomb damping and if we want to get some analytical solutions for it. Also, we can solve this numerically. For other kinds of damping we can find closed-form solutions.
@@Freeball99 For this mass-spring system then...I've seen your videos in this channel in recent days, and hope to write a python code to solve it. Well, if to solve this numerically, does it mean to use matrix form to represent the EOM and apply forward Euler step to solve 2nd ODE?
When finding an "equivalent" viscous damper we do this on an energy basis. In other words, we can replace the coulomb damping with a viscous damper such that both dissipate the same amount of energy with each cycle/oscillation. The equivalent viscous damping constant can be shown to be C = 4F/πωX (where F is the magnitude of the frictional force). In order to find the damping ratio, simply divide by the critical damping so, the damping ratio ζ = C / 2*SQRT(m*k) = 4F / (πωX * 2*SQRT(m*k))
In both cases it is assumed that the mass is displaced to the right of the equilibrium position. This is why the spring force is to the left. The difference is that the velocity is negative in the second case.
@@Freeball99 So when the mass is displaced to right of equilibrium position and moving left, isn't acceleration is actually pointing left, and the m x'' also pointing left?
@@andrealiu8650 To be clear, in BOTH cases the acceleration is to the left, because the net force is to the left. I have simply applied Newton's 2nd Law without making any assumptions as to the direction of the acceleration.
Dear sir, I had a question. So, when the block starts moving from left most position where spring is in compressed condition then shouldn't the direction of spring force (-Kx) change and just become in same direction as motion of the block? But you took in these both cases direction of spring force in the left direction, shouldn't we add another signum function into it?
In both cases that I considered, the mass was assumed to be displaced to the right of the equilibrium position. So the spring force is to the left in both cases.
In BOTH cases the acceleration is to the left, because the net force is to the left - this is not determined by the direction of the velocity, but rather by the direction of the net force.
What I have have done is apply Newton's 2nd Law without making any assumptions as to the direction of the acceleration. But to be clear, in BOTH cases the acceleration is actually to the left, because the net force is to the left.
This is an awesome channel. You are awesome!! Thank you!!
Great! Shorter than other videos and better explained
I was Waiting for this... thank you
thank you , you are a life saver
very well explained....
I have a question about the model of Coulomb friction you used in this video. When the velocity of the block is zero, the signum function gives the value of friction to also be zero but is this actually true? For example, in the final resting position, if we assume the block rests in a non-equilibrium position, its velocity is zero but there is still some friction force acting on it to counteract the small amount of spring force. I assume the reason you excluded this nuance is because in its normal cycle, when the block reaches its farthest point, the spring force is large enough to overcome the static friction. So does this mean, in a more generalized context for complex systems, we need to take the value of signum function as [-1,1] when velocity is zero? Using this model introduces non-smooth behavior which makes the solution a lot more involved. Maybe you could consider solving a system that has this general Coulomb friction in a future video?
please solve this Mcqs Thank you.
1) All vibrating bodies have following Degree of Freedom:
a)1
b)2
c)3
d)4
.
2) The frequency of vibrations with increase of damping in the case of free vibrations with coulomb damping will
A.Remain mine
B.Increase
C.Decrease
D.May increase or decrease depending upon the damping coefficient
1. All vibrating masses, in general, can have 6 degrees of freedom (3 translations and 3 rotations). Since 6 is not an option (which it ought to be) I am going to have to go with 3 - perhaps they are asking about translations only.
2. In the case of Coulomb damping, this does not affect the frequency of vibration, it only decreases the amplitude of each successive oscillation. The mass will continue to vibrate at its natural frequency regardless of the amount of Coulomb damping applied.
@@Freeball99 Thank you so much for my clearification
it's very awesome. I wish there will be a video too on the translational and rotational cases of undamped systems. Thank You :D
I have some videos for this already. Is there something specific you are thinking of? ruclips.net/video/nZw8Z1z8vTI/видео.html
Thanks
WRight to the point
awesome, your understanding is better than prof.....
Thank you for the wonderful Display.
this equations can be derived by Lagrange Approach, What will change in Lagrange function when friction is considered in problem ?
You can just include the frictional force as a generalized force and add it to the right-hand side of the equation.
Wonderful videos and terribly instructive!
Maybe I'm misunderstanding the scenario here though. The mass is at some equilibrium position, is released, and then displaces to the right. It traverses to the right a distance x0 until the instant where its velocity is 0, and then begins its movement left. It then displaces to the left a distance x0 to the equilibrium position, then a distance x1 until the spring is entirely compressed and the velocity is again 0. Finally, the mass travels, again, a distance x1 to the equilibrium position and then another distance, x2, such that x2 < x0.
If that sounds about right, I guess I'm not sure how you are able to pick those exact times for your boundary conditions? The units of pi/wn work out to seconds, but I'm just not sure how you arrived at that.
Sorry for the delayed response, but I somehow missed this until now..
What you have stated above is exactly correct. I think the piece you are missing is this.... the frequency of oscillations, f, is related to the angular frequency, ωn, by f = ωn / 2π. Also, f is related to the period of vibration by f = 1/T. So the period of vibration can be written as T = 2π / ωn. In this particular problem, we are looking at half-cycles. Consequently, the start of the 2nd half-cycle begins at a time equal to half the period or t = π / ωn and so we use this as the initial time for the 2nd piece.
@@Freeball99 Does this mean for other cases in SDOF with damping, if we want to know each amplitude, have to be solved half-cycle by half-cycle? Is there any other way to write the EOM? And if in each half-cycle the EOM are greatly influenced by initial condition, is this mean the solution is not in closed form? Thank you.
@@andrealiu8650 This is only for the case of Coulomb damping and if we want to get some analytical solutions for it. Also, we can solve this numerically. For other kinds of damping we can find closed-form solutions.
@@Freeball99 For this mass-spring system then...I've seen your videos in this channel in recent days, and hope to write a python code to solve it. Well, if to solve this numerically, does it mean to use matrix form to represent the EOM and apply forward Euler step to solve 2nd ODE?
What is the value of damping ratio in case of coulomb damling ?
When finding an "equivalent" viscous damper we do this on an energy basis. In other words, we can replace the coulomb damping with a viscous damper such that both dissipate the same amount of energy with each cycle/oscillation. The equivalent viscous damping constant can be shown to be C = 4F/πωX (where F is the magnitude of the frictional force). In order to find the damping ratio, simply divide by the critical damping so, the damping ratio ζ = C / 2*SQRT(m*k) = 4F / (πωX * 2*SQRT(m*k))
Thanks! I just wonder why this is not included in Hibbeler's book
Me tooo
I wish someone would just post an animation of the two (three) types of damping systems.
Why the spring force is appointed to the left in the 2nd case?
In both cases it is assumed that the mass is displaced to the right of the equilibrium position. This is why the spring force is to the left. The difference is that the velocity is negative in the second case.
@@Freeball99 So when the mass is displaced to right of equilibrium position and moving left, isn't acceleration is actually pointing left, and the m x'' also pointing left?
@@andrealiu8650 To be clear, in BOTH cases the acceleration is to the left, because the net force is to the left. I have simply applied Newton's 2nd Law without making any assumptions as to the direction of the acceleration.
@@Freeball99 Thank you~
Dear sir, I had a question. So, when the block starts moving from left most position where spring is in compressed condition then shouldn't the direction of spring force (-Kx) change and just become in same direction as motion of the block? But you took in these both cases direction of spring force in the left direction, shouldn't we add another signum function into it?
In both cases that I considered, the mass was assumed to be displaced to the right of the equilibrium position. So the spring force is to the left in both cases.
Sir y u have shown inertial force to the left even block Is moving to the left
In BOTH cases the acceleration is to the left, because the net force is to the left - this is not determined by the direction of the velocity, but rather by the direction of the net force.
maybe i have a stupid question why when the block is moving to left the (m*x^..) not have minus sign ??
What I have have done is apply Newton's 2nd Law without making any assumptions as to the direction of the acceleration. But to be clear, in BOTH cases the acceleration is actually to the left, because the net force is to the left.
Best video
5 *****
can we get download link of your note used for the derivations please
There you go: www.dropbox.com/s/by3x8sucafovylg/RUclips_Videos_2020-09-11_21.59.52.pdf?dl=0
@@Freeball99 thank you very much