Thanks! Good point, I hadn't noticed that. Fortunately it didn't affect the final answer since we ended up ignoring that term anyway. Pinning this comment so that others can be made aware of the error!

Yes, it can be pretty frustrating when textbooks do that! Happy to hear that the video helped.

Thanks so much for your kind words!

Thanks for your kind words!

Good to hear! Thanks for watching.

Thank you! I appreciate the support.

Thanks! For the case of a ball in a rotating frame, I don't think we can really make an analogy with free fall under gravity. The reason is that the fictitious forces in the rotating frame case depend strongly on both the position and velocity of the ball, whereas for the free fall case the only force acting (weight) is constant to a good approximation. This means the two objects actually follow differently shaped trajectories - the ball in the rotating frame traces out part of a spiral, while the ball falling under its own weight moves along a parabola. If you haven't seen it, I have another video on exactly this topic (well, almost - I consider a ball thrown outwards rather than inwards, but the same kind of argument can be made in either case) that I hope might help to understand the shape of the trajectory: kzfaq.info/get/bejne/bKh2qsSEl6jFY4U.html

As long as F doesn't vary with time, you can integrate with respect to t using the power rule as in the video, to get -wFt²cosL. It's still t², not a negative power because the power rule always involves increasing the power of t by 1, regardless of whether the coefficient is negative. Another note - your right hand side has dimensions of force instead of acceleration, so you should probably divide it by the mass of the particle to make the equation dimensionally consistent.

These things happen, you'll probably still pick up a couple of marks!

I don't think I've ever actually done this, but can imagine scenarios where it might be useful. For example, in this video local Cartesian coordinates were sufficient because our falling particle doesn't move over a distance scale where the curvature of the Earth is important, but if you were trying to launch a projectile over a very long distance then this would no longer be the case and spherical coordinates might be more appropriate.

The centrifugal force is -2mω×(ω×r). We know ω, and r would be the position vector relative to the centre of the Earth which you can work out in terms of x, y and z. Doing the cross products will give you another vector that contributes to the acceleration. You'd need to be careful about which terms you ignore; since the centrifugal force is proportional to ω², the other second-order terms we neglected in the video should probably be kept as well for consistency. You'll end up with a system of three coupled differential equations and I'm not sure whether they can be solved analytically! I haven't worked through the details myself but hopefully that gives you a starting point.

The car can increase its speed without increasing the radius, as long as the frictional force can become large enough to provide the required centripetal force - this will depend on the road surface and on the tyres. The work done is not just force × displacement, but rather the integral of force · displacement, so zero displacement overall doesn't necessarily mean zero work done. The accelerating force will be acting in the direction of motion of the car, so the dot product (force · displacement) is positive for each infinitesimal displacement of the car and the work done is positive overall.

@@DrBenYelverton I really appreciate you. thanks a lot. In this problem friction was centripetal force? In addition, I`ve got few more questions; I know mathematically that due to the perpendicular angle between centripetal force and displacement the work done by this force is zero like the weight for a satellite or string tension for a rotating body. but how come it changes the direction of the body without transfering energy? it is not so easy to grasp by attitude. thank you.

Yes, friction would be acting as a centripetal force in that example. The centripetal force doesn't change the speed of an object if it doesn't have a component in the direction of motion. If this doesn't seem intuitive, maybe think of a vector diagram showing the initial velocity v, the change in velocity δv due to the centripetal force acting over some short time δt, and the final velocity v + δv. You'll see that the final velocity is pointing in a different direction to the initial velocity, and that as we let δt become infinitesimal, the magnitude of the final velocity tends towards that of the initial velocity.

​@DrBenYelverton thanks a lot. Yeah the vector diagram is good👍 .due to Teaching to high school students, I was searching for a more intuitive and not mathematical explanation on it. Zero work of centripetal force means that no energy is needed to keep the body in circular path?

That's right - the energy of an object moving at a constant speed in a circle is constant.

vₓ comes purely from the Coriolis acceleration, which itself is small because it's proportional to ω. If you're thinking that there should be an extra contribution of rω, where r is the perpendicular distance to the Earth's axis of rotation, remember that the frame (x, y, z) is co-rotating with the Earth and therefore already moving with this velocity.

I’m pretty sure the deflection should always be to the east, both on mathematical and physical grounds. The derivation in the video doesn’t assume anything about the hemisphere, and the answer comes out proportional to cosλ, which is positive regardless of the sign of λ. I also find using the right-hand rule that the Coriolis force vector -2m(ω × v) always points to the east. The physical argument presented at the end also applies regardless of the hemisphere, because all points on the Earth are rotating eastward, and the particle will always start with an ‘excess’ of tangential velocity in that direction. My guess is that you’re thinking of the other classic example of the Coriolis effect, which is a particle moving along the Earth’s surface (e.g. a projectile or a train). In that case, the direction of the deflection does depend on the hemisphere (it’s proportional to sinλ, which has a different sign in each hemisphere).

@@DrBenYelverton Thank you for your detailed explanation. Indeed I was thinking about the Coriolis forces, which are antiparallel on opposite sides of the equator. I also assumed the lambda angle was the polar angle of spherical coordinates and taking it past the equator in the southern direction would place it in the 2nd quadrant-where the cosine is negative. I appreciate your series covering aspects of classical mechanics.