Particle sliding down a sphere: when does it lose contact?

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.

#physics #maths #math #mechanics #dynamics #forces #resolving #particle #smooth #sphere #motion #circularmotion

[apexdisease]

This is such an excellent explanation! This makes my professor seem like a chimp. Thank you and keep up the great work!

[williamjuniorolivaburgos]

I always wondered how to calculate that. Great video!

[mxminecraft]

Nice problem, and interesting cases at the last :)

[charliemaynard]

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")

[bibsp]

loved it! cant wait for the next one. i love just doing this kinda maths

[suryavardhansinghshekhawat]

that's really interesting specially those last cases

[SandeepSingh-hcno]

Subscribing & sharing your channel as a support and hoping that you will bring more life to Mathematics.

[fridayyc]

Appreciate for your excellent explaination! Now I can have a good understanding of this type of questions

[cambriolage]

great! I would really loooove to see this done with lagrangian mech as well... cheers!

[DaTopDJ]

Nice video, turns out we did this in class yesterday, good stuff 👍

[surry]

Very nice. Can I ask you what tools you are using to do these drawings and write the equations?

[bubblegum-izzu]

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.

[shlokdave]

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!

[introvertedavgeek]

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?

[jacobharris]

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?

[Johnny-twpr]

What would happen if the sphere was accelerating?

[nazmusweb]

how can we calculate the time required to lose contact?

[IamBATMAN]

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

[jasimmathsandphysics]

Is there a way to do it by only using foces? No energy

[scanwork]

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.