Second derivation clear all my doubt ...i was confused about the direction of centripetal acceleration due to first derivation as it was mentioned in all the books but you clear in such a way with second derivation and all my doubt flushed away from my mind after that..thank u .thank u thank u .so much sir ..god bless you ❤️❤️❤️and keep growing 👍👍👍👍
Because it vi =vf and these are serving as the two sides of the traingle.two sides equal one side differs that means it it is an isosceles triangle in an isosceles triangle two angles are equal and one angle is different the different angle is delta theta and the same angles are alpha and Alpha
Let dr be delta r and dt be delta t. Regarding the first derivation, as dt -->0, |dr|-->0 because v_f gets closer to v_i, i.e. |dr|=|v_f - v_i| -->0. So why are you saying that |dr|/dt -> v when dt --> 0?
I got confused, there are 3 a's but you kept using acceleration and I did not know to which type of a you were referring to. Why is ar = - ac ? why negative?
Every position on the circle has a different acceleration because the direction is changing. Radial direction points outward from the center of the circle but ac points towards the center of the circle. Those directions are opposite which is why I need a negative sign.
I never understood this. If acceleration is what causes me to move towards something, e.g., gravity pulls me down to the ground, accelerating makes me fall backward into my seat, deaccelerating makes me fly forwards. I have never experienced an inward force on a merry-go-round or UFO amusement park rides, and if you put that spinning device into space, you can make artificial gravity on the external wall, not at the center. The only reason I am touching the center is by means of associatation where I am a part of the rigid body. But I experience tension as a member of this rigid body, not contraction or a force inward. Can you clarity what my issue here is?
9:05 This appears to be a rationalization that is not actually true. If delta t is tending to zero, you cannot also have a velocity, as velocity requires delta t be non-zero. So it would appear in one sense that you have redefined velocity to mean a change in position when the change in time is zero.
11:57 This is also broken, you no longer have a triangle. The delta theta is 0 so the alphas are actually 0 and the 180 is wrong because this is not a triangle but rather is a line, which if you force its existence will report that the angle is 90 degrees. But this means its a line, not a triangle that points towards the center.
So, I went and checked a physical demonstration of this to confirm. Like, I know how calculus works and physics and fictitious forces but I had never felt an inward force. The physics classroom demonstration explained it without saying it in this way, but I don't experience that fictitious force because I am too inert at the scales I am familiar with.
Sit in passenger seat and have a friend go to an empty parking lot and make a hard left turn, you will want to straight but the door will provide a force on you and allow you to go in the circle with the car.
Great question. I agree it's not obvious what to do with the delta t in the first proof (using vectors). At first it seems like it will lead to a division by 0 which is not allowed. As delta v tends toward 0, the numerator also tends to 0. It turns out that the magnitude of delta v is also proportional to delta t, so the delta t's will cancel out in that limit. This can be seen if you watch the 2nd proof where I find an expression for delta v. Hope this helps.
@@PhysicsNinja Thank you! I think I got it now. copied your first proof over and over until I could do it by memory and I think I get it now. I probably watched 8 videos on this and yours was the only one I could stick with.
Remember that the definition of instantaneous velocity is delta position/ delta t when delta t goes to 0. To get the speed we need to find the magnitude of v_instantaneous. So we are just applying the definition.
Think about it like this. If an object is travelling in a straight line, direction doesn't matter. If the object is behind the starting point it has a negative displacement but if it is moving towards the starting point the velocity is positive. We have different directions here but differentiating displacement will still give you velocity. The other thing is the radius vector doesn't actually give radius it gives the position of the object on the path in terms of its across and up value. Differentiating that gives the (change in position)/ (change in time). I don't know if that helped sorry.
@@PhysicsNinja thanks for answering! But I think you misunderstood my point. I totally get the point that the velocities vectors are perpendicular to the radius vectors, which are representing the components of the-change-in-displacement vector. What I don't get is why the angel between the two radiuses have to be equal to the angle between the two-velocity vectors just because the velocity-radius vectors are perpendicular to each other. I don't get the connection between them being perpendicular and having same angle as you mentioned at 4:46
I’ve seen several of the derivation videos on RUclips and this is over and above the best one! Thanks for making it. :)
I've seen many videos with the exact same comment... None of them helped me yet
Second proof was beautiful and much easier to understand! Thanks bro.
indeed
YT must be proud to have Physics Ninja. Amazing lecture sir, thank you very much.
WOW! excellent vid. Physics Ninja, you the best. I liked the calculus derivation showing a_n is always opposite R_vector.
The best best best one ....superb teacher
So nice of you
Second derivation clear all my doubt ...i was confused about the direction of centripetal acceleration due to first derivation as it was mentioned in all the books but you clear in such a way with second derivation and all my doubt flushed away from my mind after that..thank u .thank u thank u
.so much sir ..god bless you ❤️❤️❤️and keep growing 👍👍👍👍
Thanks for the kind words! Good luck with the rest of your class. Best, Physics Ninja
Offcourse sir 🔥❤️👍
I still don't understand why theta angle is the same in both triangles.
Because it vi =vf and these are serving as the two sides of the traingle.two sides equal one side differs that means it it is an isosceles triangle in an isosceles triangle two angles are equal and one angle is different the different angle is delta theta and the same angles are alpha and Alpha
The derivations were both substantive and straightforward. Great video!
The explanation was fantastic
Thanks for two different solution of Centripetal Acceleration.
Most welcome
thank you very much. very nice explanation
you are really ninja of phisic.there arent things in this videos in the youtube
Loved this! Incredible job
This video helped me alot.Thanks bro........
Keep it up Love from India 😊
Let dr be delta r and dt be delta t. Regarding the first derivation, as dt -->0, |dr|-->0 because v_f gets closer to v_i, i.e. |dr|=|v_f - v_i| -->0. So why are you saying that |dr|/dt -> v when dt --> 0?
Excellent
Oh! Wonderful all your proofs. Thank you very much!
I got confused, there are 3 a's but you kept using acceleration and I did not know to which type of a you were referring to. Why is ar = - ac ? why negative?
Every position on the circle has a different acceleration because the direction is changing. Radial direction points outward from the center of the circle but ac points towards the center of the circle. Those directions are opposite which is why I need a negative sign.
why is the angle between the vectors the same???
I love this one
how is the length of v and length of r the same? I calculated the speed of earth rotation and is not the same as the radius
I never understood this. If acceleration is what causes me to move towards something, e.g., gravity pulls me down to the ground, accelerating makes me fall backward into my seat, deaccelerating makes me fly forwards. I have never experienced an inward force on a merry-go-round or UFO amusement park rides, and if you put that spinning device into space, you can make artificial gravity on the external wall, not at the center. The only reason I am touching the center is by means of associatation where I am a part of the rigid body. But I experience tension as a member of this rigid body, not contraction or a force inward.
Can you clarity what my issue here is?
9:05 This appears to be a rationalization that is not actually true. If delta t is tending to zero, you cannot also have a velocity, as velocity requires delta t be non-zero. So it would appear in one sense that you have redefined velocity to mean a change in position when the change in time is zero.
11:57 This is also broken, you no longer have a triangle. The delta theta is 0 so the alphas are actually 0 and the 180 is wrong because this is not a triangle but rather is a line, which if you force its existence will report that the angle is 90 degrees. But this means its a line, not a triangle that points towards the center.
12:44 This is also not true, as a is delta v divided by delta t, which is zero, and so no acceleration at all.
So, I went and checked a physical demonstration of this to confirm. Like, I know how calculus works and physics and fictitious forces but I had never felt an inward force. The physics classroom demonstration explained it without saying it in this way, but I don't experience that fictitious force because I am too inert at the scales I am familiar with.
Sit in passenger seat and have a friend go to an empty parking lot and make a hard left turn, you will want to straight but the door will provide a force on you and allow you to go in the circle with the car.
Thanks so much
8:36 why do you not get a divide by zero error when you take the limit as delta t approaches 0?
Great question. I agree it's not obvious what to do with the delta t in the first proof (using vectors). At first it seems like it will lead to a division by 0 which is not allowed. As delta v tends toward 0, the numerator also tends to 0. It turns out that the magnitude of delta v is also proportional to delta t, so the delta t's will cancel out in that limit. This can be seen if you watch the 2nd proof where I find an expression for delta v. Hope this helps.
@@PhysicsNinja Thank you! I think I got it now. copied your first proof over and over until I could do it by memory and I think I get it now. I probably watched 8 videos on this and yours was the only one I could stick with.
Making my life easier
Why does taking the limit of delta r/delta t = the speed? I was keeping up until that part :(
Remember that the definition of instantaneous velocity is delta position/ delta t when delta t goes to 0. To get the speed we need to find the magnitude of v_instantaneous. So we are just applying the definition.
Sir thank you for your clear teaching but I can't understand the second method
Have you learned calculus yet, if you haven’t, you won’t understand it
Nice.
why would differentiating the radius vector will become the velocity vector? They have different directions
Think about it like this. If an object is travelling in a straight line, direction doesn't matter. If the object is behind the starting point it has a negative displacement but if it is moving towards the starting point the velocity is positive. We have different directions here but differentiating displacement will still give you velocity.
The other thing is the radius vector doesn't actually give radius it gives the position of the object on the path in terms of its across and up value. Differentiating that gives the (change in position)/ (change in time). I don't know if that helped sorry.
Why if they are perpendicular, they are supposed to have the same angle? I didn't get this part :(
Radial vs tangential. They are perpendicular
@@PhysicsNinja thanks for answering! But I think you misunderstood my point. I totally get the point that the velocities vectors are perpendicular to the radius vectors, which are representing the components of the-change-in-displacement vector. What I don't get is why the angel between the two radiuses have to be equal to the angle between the two-velocity vectors just because the velocity-radius vectors are perpendicular to each other. I don't get the connection between them being perpendicular and having same angle as you mentioned at 4:46
THANK YOU
6:22 how you get delta r divided by r equals to delta v divided by v?
They form similar triangles. Vi is perpendicular to ri and vf is perpendicular to rf.
why does this not apply to the surface of the Earth? The alleged counter (gravity) is vertical, centripetal force is horizontal...
can you explain how w^2 r = v^2/r?
V=wr
Then
V²=w²r²
Then
V²/r =w²r
Love from bangladesh
The calculus one was the easier
If you arrange tip to tail, find delta V you will find toward the center , takes only 10 seconds , why all this ???!!!
I hate vectors they make my brain hurt
me too
Me too🙃
Same
Nivilife che
why v1=v2=v
because constant speed
confusing
getting confused is like getting tired/sore when you work out :)
xmtutor does a better job
Only one method= half as good .. LOL