Particle sliding off a sphere, using Lagrangian mechanics

Рет қаралды 16,422

Жыл бұрын

Using Lagrangian mechanics and Lagrange multipliers to find the angle at which a particle sliding off a sphere from rest loses contact with the surface. Some other relevant videos are linked below!
Understanding Lagrange multipliers: • Constrained Lagrangian...
Solving the same problem (but with arbitrary initial velocity) using Newtonian mechanics: • Particle sliding down ...
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #dynamics #calculus #calculusofvariations #differentialequations #lagrangian #mechanics #lagrangianmechanics #eulerlagrange #hamiltonsprinciple #action #stationaryaction #leastaction #lagrangemultipliers #constraints #physicsproblems #maths #math #science #education

Пікірлер: 39

@prinduplayzx Жыл бұрын

these are sooo helpful, thank you😁

@marcoghislanzoni 9 ай бұрын

Watching your videos and falling in love (again) with Lagrangian Mechanics. Thank you! :-)

@DrBenYelverton 9 ай бұрын

Glad to hear that, thank you!

@narek323 8 ай бұрын

Thank you. It took me a while to systematically understand the procedure of using Lagrange multipliers.

@DrBenYelverton 8 ай бұрын

Yes, it's an interesting but not straightforward concept!

@narek323 8 ай бұрын

@@DrBenYelverton True. I've realized that the part that I struggle with the most is choosing the constraints, and then from that, finding the number of degrees of freedom. For this problem, I thought of two possible constraints: R-r=0, but also x^2+y^2-R^2=0. But maybe these two constraints are equivalent.

@DrBenYelverton 8 ай бұрын

They're equivalent, just expressed in two different coordinate systems. But choosing polar coordinates here makes things work out much more neatly.

@fredericopires7659 Ай бұрын

Very, very beautiful question!!!!

@DrBenYelverton Ай бұрын

Thanks!

@theibphysicsandmathsgod1000 Жыл бұрын

Could you do a video on the double pendulum pleasee?

@DrBenYelverton Жыл бұрын

As it happens, I already have two! Deriving the equations of motion: kzfaq.info/get/bejne/kJ2Xq8R0nZuuqIk.html Solving them: kzfaq.info/get/bejne/rp6getek1cDWm2Q.html

@essa6722 5 ай бұрын

that's really helpful, but I have a question, why did you assume r=R later? that explains the motion only on the sphere surface, but what if I'm interested in finding r even after falling off the sphere. And thank you:)

@DrBenYelverton 5 ай бұрын

Thanks for watching. You need to apply the constraint r = R in order to find the point where the constraint force becomes zero, but that doesn't mean the constraint actually applies for all time. After the particle loses contact, the constraint no longer applies - mathematically, this is reflected in the fact that λ would need to be negative beyond this point, meaning that the surface of the sphere would have to pull the particle inwards to keep the constraint satisfied. If you want solve for r(t) and θ(t) after the particle loses contact, you should be able to use the same Lagrangian, just without the constraint - it should come out as the usual parabolic trajectory of a projectile.

@bubblegum-iz8zu 5 ай бұрын

Could we ignore/neglect the r dot term from the beginning since it is constrained to the surface of the sphere and doesn’t move in the radial direction?

@DrBenYelverton 5 ай бұрын

I suppose that would work, I just prefer getting the full equations of motion first! Try working through the maths just to make sure it works out in an equivalent way.

@playenz3103 Ай бұрын

What are the additional steps if you are given v0≠0 and some z0 ?

@DrBenYelverton Ай бұрын

The equations of motion would be the same, you'd just end up with different constants of integration when you apply your initial conditions.

@mxminecraft9410 8 ай бұрын

Sir how did you figure out that f(r,thetha)=r-R also what does the 'r' stands for ?

@DrBenYelverton 8 ай бұрын

r is the distance from the centre of the sphere and f is the constraint function which must equal zero. Here the constraint is that the particle remains on the surface, i.e. r = R. This means r - R = 0 and hence f = r - R.

@mxminecraft9410 8 ай бұрын

If we wrote equation for constraint in terms of distance travelled on surface (s) and angle theta ? Sorry I can't find symbol for theta in my keyboard so I would use 'a' for it So the equation could be s-aR = 0 (Here R is radius of cross section of larger body)

@DrBenYelverton 8 ай бұрын

I suppose you could transform your generalised coordinates from (r, θ) to (s, θ) and use this as the new constraint function. Maybe an interesting exercise but I don't think there would be any advantage of doing this in practice!

@mxminecraft9410 8 ай бұрын

@@DrBenYelverton Thanks for clearing my doubt.

@mxminecraft9410 8 ай бұрын

@@DrBenYelverton Sir can you make a video on rotational mechanics using lagrangian and Newtonian methods and comparing whether Newtonian method is more useful or not :) ( Also can you pls take example from JEE advanced to show the results as I am interested in knowing whether JEE advanced problems could be solved using lagrangian or not )

@chriszenker6890 Ай бұрын

Shouldn't the potential energy term be -mgRcos(theta)?

@DrBenYelverton 26 күн бұрын

There are two ways to handle constraints in Lagrangian mechanics. Either you make the constraint implicit in the Lagrangian (i.e. use R instead of r, as you suggested), or use an unconstrained Lagrangian and apply the constraint later via a Lagrange multiplier. In this problem we specifically want to find the point where the Lagrange multiplier becomes zero, so the first method won't work as it doesn't give any information about the constraining force. It will still give the correct equation of motion up until the point where the particle actually loses contact, but assumes that the particle always sticks to the sphere even beyond this point.

@serhatboran3640 Жыл бұрын

👍

@pashaw8380 11 ай бұрын

What's r and R? R is the radius of the circle but r? Need to be specific.

@tclin77 10 ай бұрын

r is simply the function (of time) of radial motion r(t), similar to x(t) in x-y coordinate.

@DrBenYelverton 10 ай бұрын

Indeed, as explained at around 1:45 in the video, r and θ are polar coordinates with their origin at the centre of the sphere.

@padraiggluck2980 Ай бұрын

At ~7:08 shouldn’t there appear s theta double dot?

@DrBenYelverton Ай бұрын

No, what we're doing there is differentiating theta dot squared with respect to theta dot, which just gives two times theta dot. We're not differentiating with respect to time yet, so an extra dot won't appear.

@padraiggluck2980 Ай бұрын

@@DrBenYelverton Thank you, Dr Yelverton.

@mehakkanwal8945 Ай бұрын

r remain same for particle moving on surfaces, why derivatives is jot zero?

@DrBenYelverton Ай бұрын

The time derivatives of r are zero - see 7:50 onwards!

@danilonascimentorj 5 ай бұрын

if you knew that lambda was zero, so why did you decide to use the general lagrange equation?

@DrBenYelverton 5 ай бұрын

I prefer getting the equations of motion in their full form first. I suppose it would work out equivalently in this case if you set λ = 0 from the start, but I find that potentially misleading because then the equations you get from Lagrange's equations are not actually equations of motion, just statements about the coordinates and their derivatives at the particular point where λ = 0.

@noway2831 Жыл бұрын

At 6:40, how do you know d/dt (∂/∂(θ') cosθ) is zero? Applying the identity derived at 9:47, I got an expression involving the first three time-derivateves of θ, sin(θ) and cos(θ). if you're interested I derived this for arbitrary functions psi and theta (LaTeX) \[ \frac{\textnormal{d}}{\textnormal{d}t} \left( \frac{\partial \psi(\theta(t))}{\partial \dot{\theta}(t)} ight) = \frac{1}{\ddot{\theta}(t)} \left( \dot{\theta}(t)\ddot{\psi}(\theta(t)) + \frac{\left[\ddot{\theta}(t) ight]^2 - \dot{\theta}(t)\dddot{\theta}(t)}{\ddot{\theta}(t)} \dot{\psi(\theta(t))} ight) \]

@DrBenYelverton Жыл бұрын

When you differentiate the Lagrangian, the coordinates and their time derivatives need to be treated independently, i.e. the operator ∂/∂θ' by definition means differentiate with respect to θ', keeping θ, r, r' and t constant. So, the cosθ term doesn't contribute to the second term in the equation of motion because it doesn't explicitly depend on θ'. The identity you mention that I derived later on involves the quantity dθ'/dθ; this is different from ∂θ'/∂θ, which would be zero!