Thanks for feedback. I am trying out the whiteboard on several of the vids on Quantum Mechanics. The advantages from my perspective are that I can get more info on the board at the same time (eg the force diagrams while I am doing the equations, and with larger writing my writing is likely to be marginally more readable! But I'll take stock once I've had more feedback.
I first show that you dont actually need a frictional force to go round a banked turn. But there will of course usually be a frictional force. The direction will depend on whether you are travelling faster or slower than the velocity calculated for the non-friction option. If faster, then friction acts inwards to stop you flying off the top fo the bank. If slower, then friction acts outwards to stop you sliding down the bank. Friction doesnt balance the centripetal force. It contributes to it.
Delighted to hear that all students need to learn this material. In England, A levels are exams taken by students aged 17-18 just before they go to university. Students usually choose 3 or 4 A level subjects.
You need to learn all this in India too .well if preparing for competitive examinations just before going to university .I SUGGEST YOU TO CHECK THE LEVEL OF PHYSICS IN JEE ADVANCED IN INDIA .
Ignoring friction there are only 2 forces. mg acts down. N acts perpendicular to the surface and in the up direction. You resolve N into vertical and horizontal components. The vertical component is N cos theta. That must equal mg.
Not ignoring as such. The first calculation shows that if the surface were frictionless (sheet ice) then we could still travel around at velocity v. Faster, and you fly off. Slower and you slide down. But if the road had a frictional force, then that force would allow v to increase or decrease by a certain amount before you shoot off the top or slide down the bank.
oh and i would like to add wile there is little to no friction there are side forces on the aircraft from the fuselage and more so the vertical stabilize and feeding in opposite rudder (ie bank to the left one would add right rudder) at high theta can help to hold that bank angle also the amount of elevator has an effect on all this too and if you REALLY want to confuse people the rudder input induces a rolling moment opposite the direction of input
This was extremely helpful and encouraging to me! I am so happy I finally understand the basic physics of banked turns and I look forward to using my new skills on my upcoming physics 1 test!
sorry for nagging, but the wing does not redirect airflow downwards for lift, it is the pressure difference that causes the upward force on a wing. pressure difference comes from airflow travelling faster via top of the wing compared to the bottom of the wing.
your description of lift is more then good enough for 99% of things most of the lift is from the AoA any way very little comes from the pressure change :D btw have you ever looked in to the physics of rotary wing fight all kinds of stuff going on with a heli helis are flying physics lessons lol would love to see you do some videos on flight in general
Suppose the speed is zero then the car will not slid down unless the mgsin(theta) greater than fsmax. This means that the min speed is zero. So you should assume that mgsin(theta)is less than fsmax.
it may look the same but it isnt. it leads to wiping out an replacing stuff while prgressing on and that in my opinion makes it less comprehendable, espacially if u r just looking for a particular formula and not watching the whole video. in addition usually the blackboard gets quite crowded after some time and so stuff gets put left or right on free space, pen and paper on the other hand provides a linear, easy to jump to "timeline" of things in proper order and completly written out formulas.
If there was sheet ice on a banked curve, the vehicle would not move tangential to the circle. There need to be friction to cause forward motion. The banked curve supply only a centripetal force.
The forward motion would've been caused by the car's existing speed, that it developed in a part of the road where there wasn't sheet ice. The vehicle would move tangent to the circle, if all of a sudden it encountered sheet ice while attempting to travel around an unbanked curve. The banked curve doesn't only supply the centripetal force, it also supplies the vertical component of the normal force that keeps it from vertically accelerating.
Hey, I've got a question about the last part, where you were talking about the 3 components that the pilot can control in order to turn. With the equation, you stated that 'v' can be decreased in order to reduce 'r'. In other words, the pilot can decrease his speed in order to turn tighter. This makes sense in the equation, but isn't it practically incorrect because by decreasing his speed, the pilot also decreases his lift component (due to less air passing over the wings)? Or would this be represented by 'g' no longer being the correct constant (as he wouldn't have enough lift to stay in the air)? Either way, I still don't see how that equation alone can account for change in turn. I think it's the same one that you used for the car, so maybe it doesn't totally work because the 'lift' component for the car isn't affected by its velocity. Is that right? Sorry if that sounds confusing at all. Thanks HEAPS for posting these videos; I've learned a lot from them, and was using this for revision as I'm hoping to start a music/science conjoint degree this year :)
As my annotation a little earlier points out, this is a simple explanation and things in practice are inevitably more complicated. But for A Level, I suspect this is all you'll need.
Can I ask why you quote the velocity V as V squared and not as SQR(R x g x tan theta)? You don't seem to take the square roots at the end of each derivation of velocity v in order to get the final velocity.
If the car is travelling on a flat surface, it can move along a curved path because of friction. And so, where does the centripetal force apply in this scenario?
If you resolve the forces perpendicular to the plane ignoring the components of N, surely N should equal mg cos theta? But also I understand why it equals mg/cos theta when resolving vertically using the components of N. Confused!
Is there a frictional force acting between the tyres and the banked road? If so, the frictional force is acting upwards to the banked road, right? And if we resolve the frictional force into horizontal component, wouldn't it balance the centripetal force which is acting onto the left side of the horizontal whilst the frictional force is acting to the right side of the horizontal. I've been going through this for 2 hours and I'm still confused. Hope that you can help me out :)
sir why centripetal acceleration doesn't act along the wedge i.e without the component of friction. the centre is radially inside the circle not horizontal so centripetal force should act radially towards the centre?
Does A level cover analytical mechanics? In the US AP Physics C which is calculus based physics only covers calculus based physics with distance as the integral and instantaneous velocity or acceleration as a derivative as well as applying calculus to electricity and magnetism and many more things, but it does not cover Lagrangian nor Hamiltonian mechanics.
You are fantastic! But I wanted to know this: that when the car's v=0 and if bank is small enough, will the car still slide inwards? If so what would the answer be in the minimum velocity equation?
In my comment earlier, I should correct the last sentence to"So you should to assume that mgsin(theta) greater than mu mgcos(theta) or tan(theta) greater than mu" Also if tan(theta) less than mu the car will not slide down even if it at rest.
When the mass at 21:20 slides down the ramp, surely there must be a vertical resultant force as well as the horizontal centripetal force? I don't understand when you say vertical forces balance, surely they can't if the mass is gradually travelling down a ramp?
That's right. The forces won't balance if the vehicle slides. I am talking about stability where the vehicle drives round the ramp without either shooting off the top or sliding down.
The normal force will be AS LARGE AS NECESSARY to prevent the car from sinking through the asphalt. It is a constraint force. Its purpose is to repel two objects apart once they are in contact, and stop them from passing through each other.
wait, this is impossible. Ncostheta can not be equal to mg. the normal force is already smaller than mg, and the cosine of theta times that smaller force is even smaller. it is legitimately impossible for them to equal each other. i don't understand.
Not Legato I think I can maybe help? My understanding of this topic is quite broken but I believe that the normal force is actually greater than the force of its weight, this is because the vertical component of the normal force MUST be equal to the weight so that the car stays at the same vertical height, but then the horizontal component of the normal force is the reason that the car is pulled towards the centre of the circle (the centripetal force)
Could someone help me with an issue which I believe could be a simple answer, but still confuses me. Why is the vertical Cosine theta and why is the horizontal sine theta? What am I looking for in those two right-angled triangles that tells me to use them trig functions? Thank you
on a flat, horizontal surface, the direction (vector) of gravity on a mass is downwards, or think of it as inwards, towards the center of the earth. When there's no incline, there's no horizontal component, or..as you could say, the horizontal component is zero. One natural way of viewing such a system would be on the unit circle; at what angle is the sin negative and the cosine zero? The sin(3π/2) = -1 and the cos(3π/2) = 0, it also just so happens that the angle sum trig identities tell us that cos(3π/2+θ) = sinθ and sin(3π/2+θ) = -cosθ from this we can find that ||mg||*cos(3π/2+θ)*i + ||mg||*sin(3π/2+θ)*j = ||mg||*sinθ*i - ||mg||*cosθ*j notice what happens, on both sides of the equality sign, when you set the angle θ = zero; ||mg||*cos(3π/2+0)*i + ||mg||*sin(3π/2+0)*j = ||mg||*sin(0)*i - ||mg||*cos(0)*j ||mg||*cos(3π/2)*i + ||mg||*sin(3π/2)*j = ||mg||*(sin0)*i - ||mg||*(cos0)*j ||mg||*0*i + ||mg||*(-1)*j = ||mg||*0*i - ||mg||*(1)*j :. ||mg||*(-1)*j = -||mg||*j hope that gives you enough to mess around with it a bit! :)
+lolzomgz1337 You use the formulae in this video which balances the force which would cause the car to slide down the slope and the force which would cause it to go over the top.
im confused. the way i would do it is tilting the plane theta(@) degrees then i would say mg.cos@ = N and mg.sin@ = mv2/r the reason why i would do that is because i thought the centripetal force had to come on an angle @ not simply to the right. if you are oscillating at an angle wouldnt make sense that the force would also be at that angle?
(In the friction version) If you set the forces in x and y to both equal 0, set them equal to each other, and solve for v, why doesn't it work out the same way? Mathematically it seems correct, but I always get the wrong answer. HALP!
Forces shouldn't add up to zero, if an object is accelerating. They should add up to m*a. In the vertical direction, the forces should add up to zero, because all acceleration is horizontal in this example. In the horizontal direction, forces should add up to mass*acceleration, which in this case is centripetal acceleration, equal to v^2/r. Vertical force addition gives us: N*cos(theta) - F*sin(theta) - m*g = 0 N*cos(theta) is the vertical component of the normal force. Cosine applies here, because when theta = 0, the normal force is vertical, and the trig term would equal 1. F*sin(theta) is the traction force holdiing the car on the road, and the vertical component of it. Sine applies here, because there would be zero vertical component of this force when theta = 0, and the trig term should also be zero. Horizontal force addition gives us: N*sin(theta) + F*cos(theta) = m*v^2/r The opposite trig terms apply in this case. When theta = 0, normal force doesn't act horizontally at all, and traction is the only force enabling the car to steer. Two equations, two unknowns (F & N). Now it's just algebra to isolate one of the unknowns, and combine it with the other equation.
He used the word curvature, rather than radius. Curvature is by definition equal to the reciprocal of the radius of curvature. A tight curve with a small radius has a lot of curvature. A large circle has a large radius and very little curvature.
Sir, why can't we let mg cos theta = N rather than N cos theta = mg ?? this appears to be a contradiction and I'm confused. this occurs at 11 minutes into the video
+Ali Moazzam I see your point. I think you have missed that mg cos theta does not equal N. We say that N cos theta = mg in the vertical direction because there is no vertical movement. therefore the forces are equal. But if you resolve mg along the N direction the forces are not equal because a part of the N force is providing the centripetal acceleration.
+DrPhysicsA But if N cos theta = mg then mg /cos theta = N which means N is greater than mg. And as theta increases N remains constant while mg shrivels to practically zero? On the other hand when we measure static friction using a slope we always use Fn (or N) = mg cos theta meaning mg is always greater than Fn (or N) which makes more sense than N greater than mg. So I have the same problem as Ali Moazzam.
+Ali Moazzam I agree. If N cos theta = mg then mg /cos theta = N which means N is greater than mg. And as theta increases N remains constant while mg shrivels to practically zero? On the other hand when we measure static friction using a slope we always use Fn (or N) = mg cos theta meaning mg is always greater than Fn (or N) which makes more sense than N greater than mg. So I have the same problem as you are having. There has to be an explanation but it is eluding me at this time. I don't like N being greater than mg. It should always be the reverse.
+Ali Moazzam Like you I was initially puzzled myself but I think I know the answer. When measuring Us with object (mass) on slope it should be noted that the mass is at rest so the Fn is derived from mg = mg cos theta i.e Fn is LESS than mg. However when finding V on Banked turns the object (mass) is not at rest but has a velocity = V. Now the Fn is not derived from just mg. Here Fn = V^2/r sin theta and is greater than mg. So the two Fns are quite DIFFERENT. One when mass on slope is at rest and the other where it has a velocity = V. It has nothing to do with different coordinate systems or axii as is often claimed.
Please explain why the theta is the same Between the bank of the ground and normal force and the normal force y component. I know it's a geometry law I'm missing.
How do you come up with the angles as you did at 10:02 . I am not getting it , why you put the theta there??? It's really confusing for me .... please help!!!
DrPhysicsA Yes , I know that the sum of all the the angles of a triangle is 180 . The thing is that which angle should I take theta and which one 90-theta?
Hi there. I noticed that there maybe an error on explanation of the banked turn of the aircraft in the video. Some where when you said that the air was forcing down. Airplane wings develop lift by lowering the pressure on top of the wing. The air pressure pushes the wing up. In a bank turn there's a pressure differential between wings causing the turn. Reference: internet explaining of thing.
Hi I don't understand why Ncos = mg , because i got used to N = mgcos where is mistake ? (i mean when we are making cases about work and pushing the objects up the incline we are saying that N=mgcos) Best KJ
DrPhysicsA it is not on this video , symply when we are deriving Ff (uN) we are saying that Normal force is the component of mg perpendicular to the inclain and i was trying to substitute the N by that and i got confused
N isn't always equal to the same function of m, g, and theta. It depends on the specifics of the problem. N is a constraint force that will be as large as necessary to prevent an object from penetrating the surface.
v^2=g*r*tan(θ).. you have the velocity and the radius and g=9.81, plug it in and find the angle. If I'm not wrong, and if what I understood from the video is correct, that question should be a frictionless banked turn.
Indeed it's N=mgcostheta vertically woth respect to inclined plane. But taking forces vertically in y-axis irrespecitive of the inclined plane, we see that Ncostheta = mg.
but if that were true, the y component of the mg with respect to the slope would have to equal mg/cos(theta) to balance out the normal force(they must be balanced or else the object on the slope would rise off the slope). This cant be true because a component of the gravitational force can't be greater than the gravitational force itself. mg
We must remember that this is a banked turn not just an object on an inclined plane. So the object is moving and the normal force is therefore a combination of the gravitational force and the centripetal force. Consider for example a motorcyclist who rides round the so called wall of death at a funfair. The walls are almost vertical and the motorcyclist rides round at sufficient speed that he doesn't fall off. The gravitational component of the normal in this case is almost zero. So what normal force is acting on the wall. It is the component of the centripetal force.
DrPhysicsA But how is this possible in a frictionless banked curve scenario where the centripetal force is due to the normal force, which is also due to gravity. If gravity is the force responsible for everything, it doesn't make sense for the normal force component to be greater than gravities component because the centripetal force is caused by Fn due to gravity. I could understand where the extra leverage would be coming from if there was an external force such as friction or tension acting upon the object towards the center.
N = mg costheta is only true if there is no acceleration. The normal force is determined by the method described in the video when there is acceleration.
Basically, we are saying that the horizontal component of the normal force is providing the centripetal force. since the mass term m appears on both sides of the equation we can cancel it.
In this case it isn't because the object is on an inclined plane. There is a normal force which is perpendicular to the plane but it does not equal the weight of the block.
DrPhysicsA Am I right in saying that it's because, in the initial situation (where you state N = mg), there's no incline, so θ = 0. But cos 0 = 1. Therefore, 'N.cos θ = mg' is actually always applicable, as cos 0 = 1, so N.cos θ = N, when θ = 0. Is that correct?
Alex Khouri If θ = 0, then the ground is flat and so the normal force will be perpendicular to the surface (as usual) but in this case it will be wholly in the vertical direction. So N =mg as you say.
You did a mistake by resolving forces at 9:50 N = mgcos(a) u said Ncos(a) =mg which is not correct ur mistake is where u have made the right angle and u put the right angle on the wrong spot and got the wrong triangle to resolve forces. Pls correct that mistake as others will believe what u are saying
Please delete this comment,People will believe what you are saying. Jk you're not clear on the concepts.Both ways are correct. I know this is 7 months old but leaving this here for other people.
Can I ask why you quote the velocity V as V squared and not as SQR(R x g x tan theta)? You don't seem to take the square roots at the end of each derivation of velocity v in order to get the final velocity.
