You have great skills to explain such physical problem in most comprehensive way. I need to know, why in this problem, rotation of mass m was not considered for deriving EoM.
Not sure if I completely understand the question, but hopefully this covers it... As a result of the problem being geometrically nonlinear, this effectively rotates the coordinate system with respect to the global coordinates. We see this in the video as the vectors r1 and r2 are rotated with respect to the e1 and e2 vectors. As a result the spring forces are not always along the e1 or e2 axis. That said, the mass will not rotate unless the spring forces produce a moment, which cannot happen if the spring are attached to the center of mass. On the other hand, if the springs were welded to the outer surface of the mass, then the mass will rotate as a result.
@@Freeball99 thanks a lot for your reply. I am going through all of your videos these days. You have cleared so many concepts. I like your way of explaining concepts with detailed mathematics. In connection with the above query, I was actually referring to the rotation (or revolution) of mass m about the axis passing through the fixed point of the spring(s). If the stiffness of two springs are not equal then I think it is likely that mass will also have rotation about the fixed point of spring. It may have been ignored due to very small rotation. Please correct me if I have understood incorrectly. Regards
Hi, I have a few questions: -Shouldn’t the kinetic energy be (m/2)(v1+v2)^2? There is only one mass, mass m. - The total energy of the system is constant, but for this problem, the energy of each position of m is not unique. As a result, shouldn’t the equations be derived differently? Stating only the energy tells us very little about this system’s motion. Maybe you could make another video? - in your second method, the units don’t match. You have force on one side, and energy (newton metres) on the other.
Kinetic energy is (m/2)(v1^2 + v2^2). You are squaring it in the vector sense, so you're really taking to dot product of the vector and itself. This has the effect of squaring each orthogonal component. Yes, the total energy is constant, but the contribution from each component keeps shifting. According to the Lagrangian formulation, the changes in each energy contribution is all we need (plus an initial condition) to determine the state of the system at any time. Not sure what you are referring to here. F = kx?? This has units of force on both sides.
Yes, in problems such as this one which are geometrically nonlinear, the problem becomes linear (and far less interesting) for very small displacements. In this case, the vibrations in the x-direction and the y-direction are decoupled.
@@sayanjitb It means that the geometry of the problem is changing as the system deforms. In other words, the coordinate system shifts as a result of large deformations. Initially, the two spring forces act in the vertical and horizontal directions respectively. However, for large deformations (like at point P as shown in the video), the springs no longer act in the vertical and horizontal directions, but instead at some other angles. In this context, the coordinate system of the springs has change (rotated). It is this rotation of the coordinate system (due to large deformations - or more precisely, due to moderate to large rotations). This produces a nonlinear behavior.
Interesting video, Would like to see more videos on non linear systems
You have great skills to explain such physical problem in most comprehensive way. I need to know, why in this problem, rotation of mass m was not considered for deriving EoM.
The mass is not rotating as all. It might be helpful to think of it as a point-mass.
@@Freeball99 But, will there be any rotation of mass with respect to fixed point of spring as the springs will get inclined by angle theta?
Not sure if I completely understand the question, but hopefully this covers it...
As a result of the problem being geometrically nonlinear, this effectively rotates the coordinate system with respect to the global coordinates. We see this in the video as the vectors r1 and r2 are rotated with respect to the e1 and e2 vectors. As a result the spring forces are not always along the e1 or e2 axis.
That said, the mass will not rotate unless the spring forces produce a moment, which cannot happen if the spring are attached to the center of mass. On the other hand, if the springs were welded to the outer surface of the mass, then the mass will rotate as a result.
@@Freeball99 thanks a lot for your reply. I am going through all of your videos these days. You have cleared so many concepts. I like your way of explaining concepts with detailed mathematics.
In connection with the above query, I was actually referring to the rotation (or revolution) of mass m about the axis passing through the fixed point of the spring(s). If the stiffness of two springs are not equal then I think it is likely that mass will also have rotation about the fixed point of spring. It may have been ignored due to very small rotation. Please correct me if I have understood incorrectly. Regards
Very interesting
Hi,
I have a few questions:
-Shouldn’t the kinetic energy be
(m/2)(v1+v2)^2? There is only one mass, mass m.
- The total energy of the system is constant, but for this problem, the energy of each position of m is not unique.
As a result, shouldn’t the equations
be derived differently? Stating only
the energy tells us very little about
this system’s motion.
Maybe you could make another
video?
- in your second method, the units don’t match. You have force on one side, and energy (newton metres) on the other.
Kinetic energy is (m/2)(v1^2 + v2^2). You are squaring it in the vector sense, so you're really taking to dot product of the vector and itself. This has the effect of squaring each orthogonal component.
Yes, the total energy is constant, but the contribution from each component keeps shifting. According to the Lagrangian formulation, the changes in each energy contribution is all we need (plus an initial condition) to determine the state of the system at any time.
Not sure what you are referring to here. F = kx?? This has units of force on both sides.
Okay, unit vectors are unitless.
So that means that an oscillator becomes nonlinear just if you change it's initial position?
Yes, in problems such as this one which are geometrically nonlinear, the problem becomes linear (and far less interesting) for very small displacements. In this case, the vibrations in the x-direction and the y-direction are decoupled.
@@Freeball99 what is the meaning of geometrically non linear here?
@@sayanjitb It means that the geometry of the problem is changing as the system deforms. In other words, the coordinate system shifts as a result of large deformations.
Initially, the two spring forces act in the vertical and horizontal directions respectively. However, for large deformations (like at point P as shown in the video), the springs no longer act in the vertical and horizontal directions, but instead at some other angles. In this context, the coordinate system of the springs has change (rotated). It is this rotation of the coordinate system (due to large deformations - or more precisely, due to moderate to large rotations). This produces a nonlinear behavior.
Interesante...