Particle sliding down a sphere: when does it lose contact?
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- เผยแพร่เมื่อ 22 ก.พ. 2021
- A particle on top of a smooth sphere is given an initial horizontal velocity. At what point does it lose contact with the surface of the sphere? In this video I solve this classic dynamics problem. Follow-up video (including rotation and moment of inertia) here: • Ball rolling down a ro... .
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.
My website: benyelverton.com
#physics #maths #math #mechanics #dynamics #forces #resolving #particle #smooth #sphere #motion #circularmotion
This is such an excellent explanation! This makes my professor seem like a chimp. Thank you and keep up the great work!
That's got to be some of the best feedback I've had! Thanks for your support.
Nice problem , and interesting cases at the last :)
Thank you, sir!
I always wondered how to calculate that. Great video!
Thanks, I'm glad it was helpful!
loved it! cant wait for the next one. i love just doing this kinda maths
Excellent! I'm planning to upload the next one (with a rolling ball instead of a particle) tomorrow. Thanks for watching.
@@DrBenYelverton from now, in sure I will be! If nobody at home wants to do maths for fun, Ill find someone who does!
Great example and demostration!!..
Thanks!
Subscribing & sharing your channel as a support and hoping that you will bring more life to Mathematics.
Thanks so much, I appreciate your support!
Great video! Really well explained :)
That's great to hear, thanks for watching!
Appreciate for your excellent explaination! Now I can have a good understanding of this type of questions
I'm glad it was helpful!
There are other interesting questions to consider: a) Where does the particle land? (expressed as a horizontal distance from the point where the sphere is touching the "floor"). b) What is the length of this parabola (from the point at which the particle tangentially detaches from the sphere to the "floor")
Indeed, these could be answered by finding the speed at the point where the particle leaves the surface using energy conservation, then resolving the velocity into components using the angle found in the video.
@@DrBenYelverton OK Dr. Yelverton, I think I have the answer to (a): (1.745) x radius
Maybe someone can check my math on that.
Finding the length of the parabola is beyond my calculus knowledge at this point :(
I will try to check that answer when I get a moment! To find the length of the parabola, you can find its cartesian equation (as in th-cam.com/video/7knY1U-ysY4/w-d-xo.html) and then integrate the square root of 1 + (dy/dx)^2 between the initial and final x values. This basically comes from Pythagoras - for more details you can check e.g. wikipedia's article on "arc length". In the particular case of a parabola I think the integral can be done using a trig substitution.
thank you so much
Brilliant❤
Nice video, turns out we did this in class yesterday, good stuff 👍
Excellent - good timing!
that's really interesting specially those last cases
Always interesting to consider limiting cases!
great! I would really loooove to see this done with lagrangian mech as well... cheers!
Thanks for watching. The force-based approach will definitely be easier in this case, given that we're looking for the position where one of the forces becomes zero - but it would be interesting to work through with the Lagrangian method as well. I'll add this to my increasingly long list of video ideas!
@@DrBenYelverton Thank you, and all the best to you prof
@@intemister I've just posted a video on solving the same problem using Lagrangian mechanics - thanks for the suggestion! Here it is: th-cam.com/video/P64-YBtxSAk/w-d-xo.html
Very nice. Can I ask you what tools you are using to do these drawings and write the equations?
Thanks. I use a One by Wacom graphics tablet and the software is Xournal++.
@@DrBenYelverton Thank you very much!
Great video! Dr Y, can you solve this problem using tensor calculus? Given that the motion is restricted to phi = 0 and r = R, I am trying to apply the properties of a particle restricted to move along the surface. I have been able to derive the tensor equation but somehow not been able to apply it properly. I am getting a term V(alpha) V(beta) N B(alpha beta) as the normal component of acceleration being experienced by the particle, where Vs are the contra variant components of velocity , N is the normal vector and B is the curvature tensor. How do we proceed after that? I know that using simple force equations and conservation laws is the easier option here, but I am trying to solve classical problems using tensor calculus. If possible, can you help? Thanks!
It's a good question but it's been such a long time since I used tensor calculus that I'm not sure how to proceed! I studied general relativity around 8 years ago and have not really used this sort of maths since then, so I'm a little rusty at the moment. It's definitely a topic that I want to revisit at some point when I have some more free time, but I'm not sure when that will be.
@@DrBenYelverton Thank you so much!
7:19: I have a question here. the net acceleration of the particle is increasing. How then did we get the equation for uniform acceleration?
Ah, you're thinking of v² = u² + 2as? The equation in the video certainly looks similar, but we can't really compare them directly because v² = u² + 2as applies to motion in one dimension, i.e. when v, u, a and s point along the same line. In our equation v² = u² + 2gR(1-cosθ), v and u are tangential velocities, while R(1-cosθ) is a displacement in the vertical direction, and g is not actually the acceleration of the particle at all.
Okay there is one thing about this I don’t understand. Why are centripetal force and gravity towards the center opposite signs? When I try to solve this I’m inclined to write N-mgcos(theta)-mv^2/r=0 where radially outward is positive. Doesn’t centripetal force always point towards the center of the circle?
The way to deal with circular motion is to find an expression for the net inward radial force and equate that with mv²/r. The idea is that the forces are not balanced (if they were, the object would move in a straight line) and the resultant inward radial force must be exactly mv²/r to keep the object moving in a circle. In this example, there are only two forces acting, the weight and the normal reaction - the centripetal force isn't a separate force, it's just a term we use to refer to the resultant inward radial force. Hopefully thinking of it in that way will help to clear up any confusion with the signs!
@@DrBenYelverton Yeah that actually makes perfect sense. I sort of realized that after being confused for a few hours. Physics I me probably wouldn’t make such a simple mistake. Maybe I would have because we mostly dealt with statics but I haven’t had to draw force diagrams in a while, so I guess I’m out of practice.
There may be a mistake in calculating the radial forces, for cos(theta) would be the adjacent side (mg) divided by the hypotenuse of the triangle which is the radial force, therefore the radial force would be mg divided by cos(theta), NOT their product, please correct me if I’m wrong. Thank you.
The triangle you need to be considering is the one with hypotenuse mg, adjacent side along the radius, and opposite side perpendicular to the radius (so that it goes from part-way along the radius to the tip of the mg arrow).
@@DrBenYelverton Hi! I initially had the same thought but after reading your comment I see the correct triangle. However, how do we know which triangle to use?
how can we calculate the time required to lose contact?
The method would be as follows:
v = ds/dt = Rdθ/dt
--> dt = Rdθ/v
--> t = ∫Rdθ/v
In the video we derived an expression for v as a function of θ (using energy), so in principle you can substitute this into the integrand and do the integral. In practice I don't think this can be done analytically so you'll need to do it numerically (or maybe in terms of special functions).
What would happen if the sphere was accelerating?
You could investigate this by including a fictitious force pointing in the opposite direction to the sphere's acceleration!
Is there a way to do it by only using foces? No energy
I haven't tried this but it should be possible to use Newton's second law to set up an equation of motion in the tangential direction, then integrate to find v as a function of θ, instead of using energy.
@@DrBenYelverton That's what I tried. I coudn't integrate because theta is a function of time and I'm not sure how to do it.
@@jasimmathsandphysics You can write the second time derivative of θ as d/dθ (½(dθ/dt)²) - this follows from the chain rule - then integrate your equation of motion with respect to θ, which gives the same equation as conservation of energy.
@@DrBenYelverton I ended up getting it doing a ds=v dv. Thanks
Hello sir, can you please tell me why this approach is wrong (which gives 45° as answer):
Suppose the angle is X. The force that detaches the object is the force tangent to the sphere namely WsinX and the force that keeps it on the sphere is FcosX. If the detaching force is bigger, than sinX>cosX thus it starts falling at X=45°.
Thanks in advance
The two components of the weight that you mentioned act in different directions, so I don't think it really makes sense to compare them with each other. What really matters is whether the centripetal force can be large enough to maintain circular motion of radius R. Once the particle has reached a high enough speed, an inwards-pulling normal reaction would be required to keep it moving in a circle. This is not physically possible, so it leaves the surface at that point.
@@DrBenYelverton Thanks, sir
Can't see what you're showing at everytime you say "right here". Also I suggest to write whole equation without a minute break between you write "mv²/" and then "R" in denominator. Write down equations and relations in time you name it, not a minute after that, otherwise you load a listener with much information without visualising it, and there's no way for listener to remember every word and equation you just saying. Also it makes video longer, because you anyway say out loud this information again when writing down the equations you named before.
Always looking to improve my videos, so thanks for the feedback. I agree with your first point about it being difficult to see what I'm pointing at - I was considering using cursor highlighting but concluded that that could end up being too distracting. I will have a think about alternative ways of making this clearer...