"The sum of the forces in the radial direction is equal to mv^2/r". This ultimate simplification of this concept was exactly what I needed. Thank you so much.
thank you so much from the bottom of my heart on behalf of anyone viewed your valuable videos. I really appreciate all your effort. God bless you and I wish you all bestز
+Libyan Libyan Thanks Libyan Libyan (love that username). It gives me great pleasure to lend a physics hand. Glad you're finding these useful. Cheers, Dr. A
Still helping today, thank you so much! Sometimes teachers don't quite explain why the topics we are learning are truly valuable to real-world problems so I sincerely appreciate your dedication to making sure we/your students understand. You're awesome, thank you again!
So now my question is this: if the rollercoaster relied entirely on gravity and mass for speed; how high would the ramp leading up to the loop-the-loop have to be?
sir .... how do the forces act when the roller coaster in between the peak and but has covered more than 1/4th of the track ?..... great lecture by the way
Weightlessness without puking on spaceships. 🙌PROFESSOR ANDERSON FOR PRESIDENNNT!!!! 🙌 and why the hell does your class seem so quiet?!? I was shooting out answers that whole time!
Light *4* Love, I think my class needed coffee. And I do not accept your nomination for president. Who would want that job? Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
I'm thoroughly enjoying your lectures, but am unclear on one thing: the argument for the normal force at the bottom is that the seat pushes up on you as you push down on it, but then why is the argument that the normal force pushes down on you when you are NOT pushing up at it at the top?
Jossy, Great question. Think about riding a roller coaster. When you go around the loop very slow, you could fall out at the top (non-ideal scenario). If you go much faster, the chair is still pushing on you at the top. Which way? Down. Because it is trying to get you to move in that direction. If the chair is trying to accelerate you faster than gravity can, it pushes on you (hence you also push back on it). You become "weightless" when this normal force goes to zero. Which means the chair would be trying to accelerate you at g, but gravity is already doing this, so the chair doesn't have to push at all. Hope this helps, Cheers, Dr. A
Excellent! So for a moment, the the normal force vector pushes me down while it decreases to zero as I approach this "weightless" state (at the top) and lose contact with the seat which puts me at free-fall correct? Thanks Professor Anderson. (The comments section = your office :))
All of this depends on the speed of the rollercoaster, but if the roller coaster is at just the right speed to achieve weightlessness, then yes you are correct. Cheers, Dr. A
@@JossinJax Roller coasters are designed to avoid that weightlessness condition, so that the occupants always stay in their seats. When you are at the top of the loop in an upside-down roller coaster, your downward acceleration is designed to be greater than g, so that you still feel a force of your seat pushing on you. In a right-side-up roller coaster, your apparent weight decreases, but they design the speed and curvature so that your acceleration to follow the hill is less than g, and that you still have a force from the seat. The restraint harnesses would carry tension if the normal force went negative, but this is a condition that is avoided by design. The restraint harness is there in case of an emergency when the ride malfunctions, and doesn't get to its proper speed for getting through the loop. This is why upside-down loops on water slides are no longer allowed. If a rider tries to slow down by bracing against the side walls, the rider may not make it to the top of the loop at a fast enough speed, and will get injured after losing contact with the slide.
The coaster is accelerating downward at the top of the track, in order to follow the curvature of the track. As a result, the net constraint force from the track on the roller coaster is downward, rather than upward like it is for a roller coaster at rest in the boarding zone. This means as an occupant of the roller coaster, you feel your apparent weight toward the floor of the roller coaster, rather than as tension in your restraint harness. The intent is that your restraint harness never needs to carry tension, unless there is an emergency because the ride malfunctions. Your seat is always supposed to push on you, rather than depending on your restraint harness to hold you downward.
4 года назад
what is the velocity going from the bottom of the loop to the top then? how much loss is there?
circle drawing skills OP
"The sum of the forces in the radial direction is equal to mv^2/r". This ultimate simplification of this concept was exactly what I needed. Thank you so much.
thank you so much from the bottom of my heart on behalf of anyone viewed your valuable videos. I really appreciate all your effort. God bless you and I wish you all bestز
+Libyan Libyan
Thanks Libyan Libyan (love that username). It gives me great pleasure to lend a physics hand. Glad you're finding these useful.
Cheers, Dr. A
Masterclass, professor, loved this!
Ahh. Finally I understood this concept after seeing this video. Thanks for the beautiful explanation and illustration.
Wonderful lecturer that makes physics fun and understandable !!!
@MA, Thank you for the excellent presentation, I now finally understand the maths explaining weightlessness,..regards,..P
Excellent. Now go ride some roller coasters!
Cheers,
Dr. A
LC APPLIED MATHS 2024 😂
👇🏼
Still helping today, thank you so much!
Sometimes teachers don't quite explain why the topics we are learning are truly valuable to real-world problems so I sincerely appreciate your dedication to making sure we/your students understand. You're awesome, thank you again!
38 minutes ago interesting, are you perhaps doing rollercoasters for a certain project as well? 👀
@@chaosforever🤫
So the centripetal force is the same during all the motion?
So now my question is this: if the rollercoaster relied entirely on gravity and mass for speed; how high would the ramp leading up to the loop-the-loop have to be?
Great question. Work it out then see:
ruclips.net/video/D63ru7ygpQc/видео.html
Cheers,
Dr. A
sir .... how do the forces act when the roller coaster in between the peak and but has covered more than 1/4th of the track ?..... great lecture by the way
Abhinav,
Great question. I'll try to make a video about this soon.
Cheers,
Dr. A
HI sir, what this video created? Link? Thank you :)
@@astock95 did u find
Watching your extremely understandable lecture videos help remind me of why I'm going into STEM and need to persevere ...
Outstanding concept delivery ,
AMAZING VIDEO, PERFECTLY EXPLAINED thank youuuuu
Thank you for helping me understand my FE practice problem
Thank you so much prof, you are singlehandedly saving me in freshman engineering physics. Best wishes!
this makes so much sense thanks
YOU EXPLAIN IT SO WELL
THANKS!
Cheers,
Dr. A
This dude / professor is good... very clear. Thank you for sharing.
This video is so so helpful... My prof has explained this problem in class over 4 times and I haven't understood it till now! Thank you!!
Weightlessness without puking on spaceships.
🙌PROFESSOR ANDERSON FOR PRESIDENNNT!!!! 🙌
and why the hell does your class seem so quiet?!? I was shooting out answers that whole time!
Light *4* Love,
I think my class needed coffee. And I do not accept your nomination for president. Who would want that job?
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
amazing sir.. you arr a genius
I'm thoroughly enjoying your lectures, but am unclear on one thing: the argument for the normal force at the bottom is that the seat pushes up on you as you push down on it, but then why is the argument that the normal force pushes down on you when you are NOT pushing up at it at the top?
Jossy,
Great question.
Think about riding a roller coaster. When you go around the loop very slow, you could fall out at the top (non-ideal scenario). If you go much faster, the chair is still pushing on you at the top. Which way? Down. Because it is trying to get you to move in that direction. If the chair is trying to accelerate you faster than gravity can, it pushes on you (hence you also push back on it). You become "weightless" when this normal force goes to zero. Which means the chair would be trying to accelerate you at g, but gravity is already doing this, so the chair doesn't have to push at all.
Hope this helps,
Cheers,
Dr. A
Excellent! So for a moment, the the normal force vector pushes me down while it decreases to zero as I approach this "weightless" state (at the top) and lose contact with the seat which puts me at free-fall correct? Thanks Professor Anderson. (The comments section = your office :))
All of this depends on the speed of the rollercoaster, but if the roller coaster is at just the right speed to achieve weightlessness, then yes you are correct.
Cheers,
Dr. A
@@JossinJax Roller coasters are designed to avoid that weightlessness condition, so that the occupants always stay in their seats. When you are at the top of the loop in an upside-down roller coaster, your downward acceleration is designed to be greater than g, so that you still feel a force of your seat pushing on you. In a right-side-up roller coaster, your apparent weight decreases, but they design the speed and curvature so that your acceleration to follow the hill is less than g, and that you still have a force from the seat.
The restraint harnesses would carry tension if the normal force went negative, but this is a condition that is avoided by design. The restraint harness is there in case of an emergency when the ride malfunctions, and doesn't get to its proper speed for getting through the loop.
This is why upside-down loops on water slides are no longer allowed. If a rider tries to slow down by bracing against the side walls, the rider may not make it to the top of the loop at a fast enough speed, and will get injured after losing contact with the slide.
If all the forces, at the top, are downward, what keeps the coaster on the track? What's the math for that?
The coaster is accelerating downward at the top of the track, in order to follow the curvature of the track. As a result, the net constraint force from the track on the roller coaster is downward, rather than upward like it is for a roller coaster at rest in the boarding zone.
This means as an occupant of the roller coaster, you feel your apparent weight toward the floor of the roller coaster, rather than as tension in your restraint harness. The intent is that your restraint harness never needs to carry tension, unless there is an emergency because the ride malfunctions. Your seat is always supposed to push on you, rather than depending on your restraint harness to hold you downward.
what is the velocity going from the bottom of the loop to the top then? how much loss is there?
Use the Conservation of Mechanical Energy for that. Should work fine, if you don't consider friction. :)
Is he writing backwards?
Nope. Secret here: www.learning.glass
Cheers,
Dr. A
@@yoprofmatt I see, that's pretty cool. Thanks for the video btw
Thanks a lot
Most welcome.
Cheers,
Dr. A
Thank you very much
Thanks sir