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!
FINALLY! I've found the answer to this question. The question in my textbook was as follows : "because earth turns to the east, common sense would seem to say that the body should drift westward. Can the reader think of an explanation?" And it was driving me mad because I couldn't find the answer to that anywhere. Thank you so much for your intuitive explanation.
I started watching your videos after someone linked your channel on reddit... I sincerely don't regret it, your videos are amazing, every next one is getting even better. Thanks for all your effort! Please keep up posting them!
Oh my goodness I was looking for this!! That was so clear and so helpful, thanks so much! The only thing I didn't understand is still why the cos lambda at the very end when demonstrating with the pole? (Why not sin lambda or -cos lambda or anything else?) If you could explain the trigonometry going on there that would be helpful! Thanks!!
Hi Dr Ben Yelverton. I am using your video as reference with a similar situation in relation with the Coriolis force, and have reached the step at around 13:25 with Vx dot. But unlike your video, my 2 at the start of the equation is negative, and from what little I know, I can't use the power rule there, or can I? Would it be t (time) to the power of -2? For context, I'm still at highschool level math, so I don't know enough to solve my problem. Can you help me please? my equation is: Vx dot=-2wFtcosL (F is my force in the x direction instead of gravity in your case, and L is lambda or the latitude)
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.
Wonderful explanation of the mathematics behind the coriolis force and effect. I was just wondering if you could help me clarify my reasoning for demonstrating the parabolic arc formed by an object under the effect of a centrifugal and coriolis force. Say a person sitting on the edge of a rotating circle throws a ball towards the centre of rotation of the circle, the direction of the velocity of the ball would be at an angle since you have your two components one being the tangential velocity of the circles formed by the changing distance of the ball from the centre of the circle and the velocity of the ball being thrown by the person. But because the centrifugal is acting in your non inertial reference frame we can compare it to the trajectory of a ball falling under its own weight due to gravity where the centrifugal force is the ‘gravity’ in this example.
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: ua-cam.com/video/6BFwaS2Ba3Q/v-deo.html
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.
Very interesting video, but there is a catch: I don't agree with one approximation you are making. It is true that ω is small, but it also true that the radius of Earth is much larger than the height of the tower! It means that the centrifugal acceleration isn't negligeable at least at mid latitudes. Let's calculate the centrifugal force in function of the latitude, considering that the height of the tower is negligable compared to the radius of Earth (Re) and taking Earth as a perfect sphere. The distance from the axis of Earth |r|=Re·cos(λ), so the vector r is [0,−Re·sin(λ)·cos(λ),Re·cos²(λ)]. Plugging the vectors r and ω into the expression of the centrifugal acceleration we obtain a(centr)= −ω x (ω x r)= [0,−Re·ω² ·sin(λ)·cos(λ),Re·ω² ·cos²(λ)] Let's plug in some numbers: |ω|=7.292E-5 rad/s, Re≈6.37E6 m and let's consider the case of λ=45°, then a(centr)=[0,−0.0169, 0.0169] m/s². The x component is always 0 and the z component is much smaller than g, so it can be ignored. We are left with a negative y component, though, that accelerates the particle towards the equator. How does it compare to Coriolis acceleration? For a tower 100 m high, vz at the end of the fall is −44.3 m/s, so the x component Coriolis acceleration reaches aₓ(Cor)=−2·ω· vz · cos(λ) =0.00457 m/s², which is actually ~3.7 time smaller than the y component of the centrifugal force! If my calculations are correct, at mid latitudes, a falling particle is more deflected towards the equator by the centrifugal force than it is to the East by the Coriolis force. Notice, however, that the y component of the centrifugal force tends to 0 as we get closer to the equator, while the Corriolis force does not, so at low latitudes the latter prevails and becomes the only source of deflection at the equator.
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.
hello sir, I `ve got a question about work-kinetic-energy theorem. I hope you have time to answer it. Assume a car is moving upon a circular path and increases the speed. the first question is: " can the car increase the speed without increasing the radius of the circular path?" if so, how can we explain it by work-kinetic-energy theorem? The car comes back to the starting point so the displacement and total work is zero while there is a change in kinetic energy and delta K is non-zero. does it have to do with elliptical paths?
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?
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.
ahhhh this came up in section A of the 1B cambridge physics B exam and I just couldn't do it! Oh well, I gave down the coriolis force down and wrote some equations, maybe i wouldn't get completely 0.
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.
What are you talking about? That’s not how objects behave when dropping them from a rotating platform. Ever dropped something when spinning on a merry go round?
awesome video, but there seems to be a little typo at 9:58 whereas cos/sin where switched while writing the z component of the acceleration vector;)
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!
FINALLY! I've found the answer to this question. The question in my textbook was as follows :
"because earth turns to the east, common sense would seem to say that the body should drift westward. Can the reader think of an explanation?"
And it was driving me mad because I couldn't find the answer to that anywhere. Thank you so much for your intuitive explanation.
Yes, it can be pretty frustrating when textbooks do that! Happy to hear that the video helped.
I started watching your videos after someone linked your channel on reddit... I sincerely don't regret it, your videos are amazing, every next one is getting even better. Thanks for all your effort! Please keep up posting them!
Thanks so much for your kind words!
I think this is the best video out there explaining the observation. Thank you so much!
Thanks for your kind words!
Thankyou for the explanation. I was superconfused earlier. This helped
Good to hear! Thanks for watching.
Oh my goodness I was looking for this!! That was so clear and so helpful, thanks so much! The only thing I didn't understand is still why the cos lambda at the very end when demonstrating with the pole? (Why not sin lambda or -cos lambda or anything else?) If you could explain the trigonometry going on there that would be helpful! Thanks!!
in 10:13 shouldn't the Z component be 2vx cos(lamda)-g
Hi Dr Ben Yelverton. I am using your video as reference with a similar situation in relation with the Coriolis force, and have reached the step at around 13:25 with Vx dot. But unlike your video, my 2 at the start of the equation is negative, and from what little I know, I can't use the power rule there, or can I? Would it be t (time) to the power of -2? For context, I'm still at highschool level math, so I don't know enough to solve my problem. Can you help me please? my equation is: Vx dot=-2wFtcosL (F is my force in the x direction instead of gravity in your case, and L is lambda or the latitude)
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.
Wonderful explanation of the mathematics behind the coriolis force and effect. I was just wondering if you could help me clarify my reasoning for demonstrating the parabolic arc formed by an object under the effect of a centrifugal and coriolis force. Say a person sitting on the edge of a rotating circle throws a ball towards the centre of rotation of the circle, the direction of the velocity of the ball would be at an angle since you have your two components one being the tangential velocity of the circles formed by the changing distance of the ball from the centre of the circle and the velocity of the ball being thrown by the person. But because the centrifugal is acting in your non inertial reference frame we can compare it to the trajectory of a ball falling under its own weight due to gravity where the centrifugal force is the ‘gravity’ in this example.
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: ua-cam.com/video/6BFwaS2Ba3Q/v-deo.html
Do you ever find it useful to do rotating frame problems in spherical coordinates?
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.
Great explanation
Great video!
Very interesting video, but there is a catch: I don't agree with one approximation you are making. It is true that ω is small, but it also true that the radius of Earth is much larger than the height of the tower! It means that the centrifugal acceleration isn't negligeable at least at mid latitudes.
Let's calculate the centrifugal force in function of the latitude, considering that the height of the tower is negligable compared to the radius of Earth (Re) and taking Earth as a perfect sphere. The distance from the axis of Earth |r|=Re·cos(λ), so the vector r is [0,−Re·sin(λ)·cos(λ),Re·cos²(λ)]. Plugging the vectors r and ω into the expression of the centrifugal acceleration we obtain a(centr)= −ω x (ω x r)= [0,−Re·ω² ·sin(λ)·cos(λ),Re·ω² ·cos²(λ)]
Let's plug in some numbers: |ω|=7.292E-5 rad/s, Re≈6.37E6 m and let's consider the case of λ=45°, then a(centr)=[0,−0.0169, 0.0169] m/s². The x component is always 0 and the z component is much smaller than g, so it can be ignored. We are left with a negative y component, though, that accelerates the particle towards the equator.
How does it compare to Coriolis acceleration? For a tower 100 m high, vz at the end of the fall is −44.3 m/s, so the x component Coriolis acceleration reaches aₓ(Cor)=−2·ω· vz · cos(λ) =0.00457 m/s², which is actually ~3.7 time smaller than the y component of the centrifugal force! If my calculations are correct, at mid latitudes, a falling particle is more deflected towards the equator by the centrifugal force than it is to the East by the Coriolis force.
Notice, however, that the y component of the centrifugal force tends to 0 as we get closer to the equator, while the Corriolis force does not, so at low latitudes the latter prevails and becomes the only source of deflection at the equator.
Hi ,what would we do if we wanted to include the centrifugal force into the deflection?
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.
hello sir, I `ve got a question about work-kinetic-energy theorem. I hope you have time to answer it. Assume a car is moving upon a circular path and increases the speed. the first question is: " can the car increase the speed without increasing the radius of the circular path?" if so, how can we explain it by work-kinetic-energy theorem? The car comes back to the starting point so the displacement and total work is zero while there is a change in kinetic energy and delta K is non-zero.
does it have to do with elliptical paths?
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.
Amazing sir, thank you
Hello, why is vx a small quantity? Doesn’t the earth a spinning contribute to a very large quantity?
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 really like your work. Keep grinding.. keep uploading good videos. Eventually word will spread.. someday..
Thank you! I appreciate the support.
Oh yeah buddy, what a simplistic explanation.
Wow great stuff 🙏
ahhhh this came up in section A of the 1B cambridge physics B exam and I just couldn't do it!
Oh well, I gave down the coriolis force down and wrote some equations, maybe i wouldn't get completely 0.
These things happen, you'll probably still pick up a couple of marks!
❤❤
In the southern hemisphere the deflection would always be to the west though!
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.
What are you talking about? That’s not how objects behave when dropping them from a rotating platform. Ever dropped something when spinning on a merry go round?