Ball rolling down a rough sphere: when does it lose contact?

Рет қаралды 6,771

Күн бұрын

Previously I considered the simpler case of a particle sliding down a frictionless sphere: • Particle sliding down ... . This time, let's imagine we have a ball of finite size, and that it spins up due to friction as it rolls down. We then have to consider the moment of inertia of the ball.
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 #rotation #sphere #motion #circularmotion #momentofinertia #spinning #spin #rolling

Пікірлер: 25

@shors5841 26 күн бұрын

Unbelievably well-made! Thank you.

@_Be_Still 2 жыл бұрын

Your videos are amazing...logic well explained. More of these. Hope you consider doing videos on Euler angles too

@DrBenYelverton 2 жыл бұрын

Thank you! I'll add Euler angles to my list of video ideas.

@user-vq3ms4ld1f 2 ай бұрын

Thank you sir for this video. I have a question with your video at 2:53. When there is an object on the slope, mgcos(theta)=Normal force.(as far as I can tell) But you writed mgcos(theta)-N is not zero. I think it is because object is moving, that means object has velocity. i understand guesswork, so i want answer for this question. I apologize for my inexperience as I am not an English speaker.

@DrBenYelverton 2 ай бұрын

If an object moves along a circular path, the net force can't be zero because the direction of motion is constantly changing. There must instead be a net force directed towards the centre of the circle, which is called the centripetal force and is in general equal to mv²/r, where v is the speed and r is the radius of the circular path.

@user-vq3ms4ld1f 2 ай бұрын

Thanks for giving me an answer. I got it!

@deltasanderson 7 ай бұрын

You are the best, Dr!

@DrBenYelverton 7 ай бұрын

Thanks for your support!

@supergravity66 Жыл бұрын

Very clear explanation!

@DrBenYelverton Жыл бұрын

Thanks, that's good to hear!

@sureshnair1209 2 жыл бұрын

sir i have a doubt you didnt use coefficieent of friction here and that is also not been used in the expression does that mean this dosent depent of Friction coefficuuents value

@DrBenYelverton 2 жыл бұрын

I made the assumption that the ball rolls without slipping, which gives the constraint u = rω. So as long as there is enough friction to prevent slipping, the angle doesn't depend on the specific value of the coefficient of friction. If slipping were to occur (i.e. because the coefficient of friction is small) then the problem would be more difficult to solve and the expression I derived would no longer apply.

@carlosuzcategui4706 8 ай бұрын

is it u the initial velocity of the ball? if so, assuming is equal to zero as it starts from rest, can we find the actual value of the angle by simplifying the last equation and substituting the moment of inertia value with 2/5?

@DrBenYelverton 8 ай бұрын

Yes, u is the initial velocity. If it's zero then the (R+r) factors cancel and you get cos⁻¹[2/(3+α)]. You can substitute α = 2/5 if it's a solid sphere.

@sergiodevicentesanluis9741 Жыл бұрын

I've got one doubt professor: in 4:49, the angular velocity has two components, one being the rotation of the ball about its center (let it be psi) and the other component the rotation of the ball about the center of the sphere (theta), so wouldn't the constraint be that the linear velocity of the center of mass [(R+r)*theta dot] must be equal to r*psi dot?

@DrBenYelverton Жыл бұрын

If you write the quantity [(R+r)*theta dot] as u (since it's just the linear velocity) and [psi dot] as ω, your constraint becomes u = rω and is therefore equivalent to the one used in the video.

@sergiodevicentesanluis9741 Жыл бұрын

@@DrBenYelverton Isn't omega equal to psi dot (the rotation of the ball about its center) plus theta dot (the rotation of the ball about the center of the spherical surface)? Thanks for the quick reply.

@DrBenYelverton Жыл бұрын

@@sergiodevicentesanluis9741 ω is the angular velocity of the ball about its centre of mass (which is why we can use 1/2 Iω² for the rotational KE), so it's just psi dot.

@nediadarth4999 3 жыл бұрын

nice, very interesting

@DrBenYelverton 3 жыл бұрын

Thanks for watching, glad you enjoyed it!

@hishaamrao8539 Жыл бұрын

When conserving energy, should we not look at the work done by the frictional force?

@DrBenYelverton Жыл бұрын

The friction doesn't actually do any work on the ball because there's no relative motion at the point of contact (i.e. because of the "no slip" condition). In reality there would be some energy losses due to rolling resistance, but this is a whole other effect due mainly to deformation of the ball.

@koulickchakraborty8498 7 ай бұрын

the small ball rolls due to this rolling friction. hence work done by rolling friction is conserative.

@bhupendrakc9609 9 ай бұрын

why is torque due to friction not include?

@DrBenYelverton 9 ай бұрын

The effect of frictional torque is implicitly included in the no-slip condition v = rω. We're using a simple model in which it's assumed that the friction always keeps the ball rotating at just the right frequency to prevent slippage.