Well, that certainly explained it in the most simple terms. I watched other videos which strain through multiple substitutions and painful steps to get there. You did it simply and easily. Well done! I now have another resource for learning!
Huge thanks for the explanation, I tried using Khan Academy where he compared two circles, one with position vectors, the other with velocity vectors. I didn't exactly get through my head what he meant, so I think this more clearly explained the proof :) Well done!
This is a great job of being explicit about how the angles are "similar triangles". Every other derivation just makes the claim they are "similar triangles" which is NOT at all clear because they are not 90 degree triangles.
N.Padmanabhan N.Padmanabhan angle between tangents is equal to the angle between radius....and velocity is along the tangents so angle is similar to that of 2 radius vectors
Why like most if these geometrical proofs of this did u skip the proof of the equality of the angles? It is the key to the whole thing where can I find more about this?
I have found that many people skip over the geometry, they often get the magnitude BUT forget to really find the direction that the force is acting. You can also get around having to use anything about similar triangles either, just by looking at the angles and using the fact that all interior angles sum to 180 degrees in a circle.
+Natalee Martin cuz the velocity on the tangent line of the circle, so the velocity is always vertical to the radius. if shifting v1 to the letf, and when v1 touching the initial point of v2, is pretty easy to find both the angle are equivalent : )
Thanks, this is very nicely done. I tried to do it 'in my own words' and in a series of visual steps so I can teach it. (This is a great way to learn.) I ran into some bumps. OK, so I thought: take a look at a different video, maybe it's got something different.... I looked at three including the Kahn Academy (great idea lousy execution always!), they all completely miss key elements. I really like the way you've slipped in delta v over delta t. Another video attempted this by just shoving it in... so I came back to watch your's again.
+Gavin Li This demonstration assumes constant speed, and thus uniform circular motion. The speed of the object at any time must be constant in order to have uniform circular motion.
+Cale Hover thx for the video and the explaining tho : ) i understand how "|V1|" ="|V2|", just cant find the relationship between them and V, where delta X= V(delta t). and delta V can also represented by V*delta(t). however, since delta X may not equal to delta V, and so V can have different value. and theta between both angle are always the same. with the same curvature, delta X and delta V are both related to the same delta t. over all, V is some constant, but it may not always equal to V1. anyway, im not the smartest student, I was trying to figure out how it works. and when i first watched ur video, i was getting more confused . but your video actually is great, and helping a lot of ppl. : ) and now i can proof it with calculus, and thank you again. : )))))
+Cale Hover actually, you are right. the idea is right, but still needs calculus's knowledge to understand how |V|=|V1|. delta X= curvature times (delta t), which is the displacement onto the velocity. and the V= r*d theta/ d t which has same speed with V1. and @#$%$#@!@#%^. it really is very confusing there @.@, but i gotcha now...
v represents the speed of the object, which is constant. v2 and v1 represent the velocity vector of the object at two different times. Since the speed is constant, the length (magnitude) of v2 and v1 will be the same.
@@thailandfutsal5508 this is the special case where the velocity is tangential to the circular path only. There is no centripetal component to the velocity.
If the object is moving in a circular path at steady speed, the acceleration is constant. It's not speeding up or slowing down and the direction of the velocity vector is changing at a steady rate due to the fact that it is on a circular path.
You are correct, I should have said the *magnitude* of the acceleration is constant. The direction of the acceleration changes uniformly, always pointing toward the center of the circular path.
Well, that certainly explained it in the most simple terms. I watched other videos which strain through multiple substitutions and painful steps to get there. You did it simply and easily. Well done! I now have another resource for learning!
our physics teacher just told us that a=v•v/r and nothing more. So this helps a lot.
Huge thanks for the explanation, I tried using Khan Academy where he compared two circles, one with position vectors, the other with velocity vectors. I didn't exactly get through my head what he meant, so I think this more clearly explained the proof :) Well done!
This is a great job of being explicit about how the angles are "similar triangles". Every other derivation just makes the claim they are "similar triangles" which is NOT at all clear because they are not 90 degree triangles.
Same I was so confused about that
I don't understand how theta in both the triangles are equal. can u explain me
N.Padmanabhan N.Padmanabhan angle between tangents is equal to the angle between radius....and velocity is along the tangents so angle is similar to that of 2 radius vectors
Great explanation, much better than a High School Physics text.
Glad you think so!
Why like most if these geometrical proofs of this did u skip the proof of the equality of the angles? It is the key to the whole thing where can I find more about this?
I tried to show by inspection that this is true, showing their congruency when you rotate the angles on top of one another.
Thank you, I have understood. This is very simple derivation for everyone.
Excellent clear proof, thanks a lot. I had got stuck on my textbook's proof on the isosceles triangle's equal ratios part. :-)
Great breakdown man!
I have found that many people skip over the geometry, they often get the magnitude BUT forget to really find the direction that the force is acting. You can also get around having to use anything about similar triangles either, just by looking at the angles and using the fact that all interior angles sum to 180 degrees in a circle.
Just a question for a science practical I am writing on centripetal force... just want to know how you can deduce that those angles are equivalent?
+Natalee Martin cuz the velocity on the tangent line of the circle, so the velocity is always vertical to the radius. if shifting v1 to the letf, and when v1 touching the initial point of v2, is pretty easy to find both the angle are equivalent : )
GREAT VIDEO YOU DESERVE A SUBSCRIBE.
Thank you! I hope it was helpful.
Thanks, this is very nicely done. I tried to do it 'in my own words' and in a series of visual steps so I can teach it. (This is a great way to learn.) I ran into some bumps. OK, so I thought: take a look at a different video, maybe it's got something different....
I looked at three including the Kahn Academy (great idea lousy execution always!), they all completely miss key elements. I really like the way you've slipped in delta v over delta t. Another video attempted this by just shoving it in... so I came back to watch your's again.
Great explanation, thanks!
You are simply the smartest man I know good sir
You're too kind
|V1| and |V2| are not necessarily equal to V. this proof makes not sense unless can proof they are equivalence.
+Gavin Li This demonstration assumes constant speed, and thus uniform circular motion. The speed of the object at any time must be constant in order to have uniform circular motion.
+Cale Hover thx for the video and the explaining tho : ) i understand how "|V1|" ="|V2|", just cant find the relationship between them and V, where delta X= V(delta t). and delta V can also represented by V*delta(t). however, since delta X may not equal to delta V, and so V can have different value. and theta between both angle are always the same. with the same curvature, delta X and delta V are both related to the same delta t. over all, V is some constant, but it may not always equal to V1. anyway, im not the smartest student, I was trying to figure out how it works. and when i first watched ur video, i was getting more confused . but your video actually is great, and helping a lot of ppl. : ) and now i can proof it with calculus, and thank you again. : )))))
+Cale Hover actually, you are right. the idea is right, but still needs calculus's knowledge to understand how |V|=|V1|. delta X= curvature times (delta t), which is the displacement onto the velocity. and the V= r*d theta/ d t which has same speed with V1. and @#$%$#@!@#%^. it really is very confusing there @.@, but i gotcha now...
Why v2 and v1 is equal to v
v represents the speed of the object, which is constant. v2 and v1 represent the velocity vector of the object at two different times. Since the speed is constant, the length (magnitude) of v2 and v1 will be the same.
@@HoverCraftPhysChem does the object has the centripetal velocity
@@thailandfutsal5508 this is the special case where the velocity is tangential to the circular path only. There is no centripetal component to the velocity.
at 3:34 how can we use that formula the acceleration is not constant can we apply that formula the also we need calculus here don't we
If the object is moving in a circular path at steady speed, the acceleration is constant. It's not speeding up or slowing down and the direction of the velocity vector is changing at a steady rate due to the fact that it is on a circular path.
@@HoverCraftPhysChem How is the acceleration constant if the direction is changing last time i knew it was supposed to be a vector,right?
You are correct, I should have said the *magnitude* of the acceleration is constant. The direction of the acceleration changes uniformly, always pointing toward the center of the circular path.
Thanks man!
I could kiss ye; beautifully explained, thanks
Thanks...great job..👍
That was great sir
Thank you so much
THANK. YOU!!
Thanks
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