Video není dostupné.
Omlouváme se.
Physics 11.1 Rigid Body Rotation (1 of 10) Basics
Vložit
- čas přidán 7. 08. 2024
- Visit ilectureonline.com for more math and science lectures!
In this video I will explain the translational, rotational, and combined motion of rigid body rotation.
When I graduate and make some more extra savings, it would be my honor to donate to you and your awesome videos! Please keep it up.
did the dream come true?
Every video I need to review for my test, this guy has. All contain definitions, examples, and are short and concise.
Same! I have a final covering rotational motion, angular momentum, and torque, and Michel has it all bundled into one playlist. It's so perfect! He also explains these things that I couldn't understand for weeks (when asking TA's/other videos) in just a few minutes!
This comes in so handy as a revision for my exam(1st year undergraduate) which is tomorrow and I've been preparing for a month really intesively, however never enough of clarifications of concepts or making them simpler, and who does it better than you! Thank you!
I love you, man. You're the best physics teacher on CZcams :)
Researching rigid body dynamics for my own personal use, and even as a highschool student you do a great job explaining this :)
Glad you found our videos! 🙂
Thank You So much !!!!
Best online course teacher
Your lectures help so much, keep up the work prof
Amazing to-the-point lectures!
Not all heroes wear caps
*capes
Leen Minr thank you
true that
Well.. he's not wearing a cap either Lol
some wear bow ties
This account is incredibly helpful. You're an awesome teacher compared to my physics prof, thanks for putting up these videos!
Thank you for making this concept simple to learn!
What a great great simple explanation :-)Goooo Michel :)
Sir, I have a problem that seems to be pretty simple but I cannot get the right answer. I hope you can take a look and give me some hints about it, I think I have trouble calculating the couple C.
The problem is: "A homogeneous cylinder of mass m and radius R is at rest on a horizontal plane when a couple C is applied clockwise. Determine the magnitude of the coefficient of friction between the wheel and the plane so that rolling will occur."
The answer is: coefficient of friction >= (2C/3)/(mgR)
Your videos are always awesome. You teach us what school is supposed to teach...Thank you very much, sir.
Thank you So much for miking life easier
This video was my thousandth like on CZcams.
Best Physics fundamentals explained.
sir i have a fixed-joint hydraulic cylinder pushing a circular bender with a 2 ft radius. my initial force is 1.5 tons to start the bender rotation. my quetion is that what is my final force apllied when i the circular bender turn at 90 deg.? should i used the resultant force?
magnificent !
oh thank god you did these questions
Thanks a lot :)
Thank you sir...
Awesome, I'm studying exactly this topic in physics right now.
Laurelindo me too
sir could you please tell the average velocity of all the points of the rim
is it 0 or 2*v
Hello there,
First. thank you for this video!
You said that the tremendous acceleration of a particle tire when it's at the bottom hitting the road is part of the reason why there is a lot of heating effect taking place by the rotation of the tire.
I have 3 questions about that:
1. How can I calculate the exact amount of heat generated from this acceleration?
2. What are the other reasons causing the tire's heating? How can I calculate them?
3. How much (+/-) of the total heating is caused by this effect (heating caused by acceleration)?
Thank you!
The heating comes from the stress created during the rapid acceleration and deceleration as well as the friction it experiences on the road and the deformation of the tire as it makes contact with the road. These are difficult to calculate and cannot be measured directly.
@@MichelvanBiezen , this is incorrect. _All_ of the heating comes from friction and compression due to contact with the road. To see why, simply consider a spinning tyre not in contact with a road, and with no translational motion. Using your own numbers, the top of the tyre would then be moving at 60 mph and the bottom at -60mph, a difference of 120 mph. The acceleration _is exactly the same_ as for the tyre in contact with the road.
thanks sir...
Amazing explanation professor. Thank you for you videos.
You are very welcome
thanks very much
Can you do a vector analysis of the rotating globe while it is "translating" through space around the sun at 66,000 mph? Would there be a momentary instance where the velocity would be zero like on a rotating tire going down the road? Will we have "tremendous accelerations" and then reduced velocities repeating over and over again just like the spinning wheel "translating" down the road? Will you do it or as one CZcams channel called Kryptonite Physics has pointed out "There is a reason why in Astrophysics they do not use rotating reference frames that are moving is because it all falls apart if they do".
It will not be like the motion of a tire on a moving car. That is very different. The speeds of stars moving through the galaxy, are small compared to the size of the galaxy.
@@MichelvanBiezen Okay, but will you do it. I would like to know what the speed at the equator will be while the earth is rotating at 1,030 mph at the equator while "translating" through space around the sun at 66,000 mph. ( or whatever speed it's supposed to be going). Maybe you will get some interesting results.
@@jimcapp5077 Visit GreaterSapien's channel where he discusses Kryptonite Physics' videos and he is wise to the flat-earth stuff.
You should write a book!! I wish you were my mentor
Amazing lectures!
Glad you like them!
Yashash,
Wouldn't the average velocity of each portion of the tire be the velocity of the car?
Isn't the right wording actually " average of the velocities" instead of "average velocity" bcz average velocity means something else; it means total displacement/total time ig. Byw are you a professor? at which institute?
Professor, what if there is deformation? The normal force is shifted more towards the "front" leading edge but does this mean its acting as 'rolling friction' because it now has a horizontal component thats opposing the motion (eg rotating clockwise and the normal force has a component to the right thus, tangential v is reduced)? Thank you!
If there is deformation then you have to include the rolling friction. That is covered in this playlist: MECHANICAL ENGINEERING 11 FRICTION
Thankyou so much
Super sir
awesome
the best =')
Thank you, but still dont understand bottom of tire not moving,,but it is not stay at same contact point with earth and changed point of contact??. not moving mean not changing place
It means that wherever the tire makes contact with the road, there is no relative movement between the surface of the road and the surface of the tire. (no slipping or sliding)
What would happen if the shape of the wheel is different i.e a triangular or square wheel.
In what respect? What aspect of the rotational motion are you thinking about?
thanks alot sir.
Most welcome 🙂
Sir please provide a link of video of restitution of moving body
I would have to go look for it myself. But I believe it is in the playlists on conservation of linear momentum.
Great!!!打扮的很像柯南有木有哈哈~
thank you
You're welcome
You said that the tremendous acceleration of a tyre particle when it's at the bottom hitting the road is part of the reason why there is a lot of heating effect taking place by the rotation of the tire.
The inertial frame of reference for any particle of the tyre is the centre of the wheel, which is moving at constant velocity, the acceleration felt by a particle at the edge of the tyre is due to its' rotation.
The heating of the tyre at the contact patch with the road is due to the deformation of the tyre and sliding friction with the road surface, as not all of the contact patch is stationary wrt. the road.
The tremendous acceleration you mention is wrt. a stationary observer by the side of the road, a different inertial frame of reference.
"The inertial frame of reference for any particle of the tyre is the centre of the wheel, which is moving at constant velocity, the acceleration felt by a particle at the edge of the tyre is due to its' rotation."
That is WRONG.
- Only the center point of the wheel is moving with constant velocity.
- Every other point is experiencing acceleration and deceleration.
Think about an airplane turning.
- A F16C can make a turn of 14 degrees/s at Mach 0.8 (15,000 feet)
- This results is a centrifugal force of 6.78G
- Obviously the plane cannot be viewed as a inertial reference frame.
@@decadent. WRONG.
What are you talking about?
@@stevegreen8262
Your statement is incorrect.
- The whole point being made here is that both motions must be considered.
- When you combine linear velocity with rotational motion . ONLY the center point of the rotation has CONSTANT VELOCITY.
- So every other point must experience acceleration/deceleration..
- The laws of physics do not suddenly change because you chose to view something as being in a box.
- At the risk of having this comment deleted ..... [ highly likely - although Michel van Biezen I would respect you more if you leave it :) ]
- When I did my degree , I believed what I was told regarding relativity . When I started working on real projects I realized that it is B.S.
- Relativity is maintained for political and ideological reasons. It is wrong as all engineers/pilots/snipers know.
- Read the comments on the video on my channel especially with the debate I had with the PhD professor for more :)
/
Plane flying on a rotating earth at the equator.
- 1040 mph + 650 mph EAST
- 1040 mph - 650 mph WEST
- Now , if we accept the idea that planes are being “pulled down” by gravity as they fly level.
- Then, there must also be a corresponding centrifugal force acting opposite.
/
Standard passenger flight (35,000 feet , 650 mph) EQUATOR
/
Speed (mph) = 390 (west)
Height (Miles)= 6.62
Gravitational acceleration (m/s^2) = 9.78091
Centrifugal acceleration (m/s^2) = 0.00476309
Remaining acceleration (m/s^2) = 9.77614
/
Speed (mph) = 1690 (east)
Height (Miles)= 6.62
Gravitational acceleration (m/s^2) = 9.78091
Centrifugal acceleration (m/s^2) = 0.0894402
Remaining acceleration (m/s^2) = 9.69147
/
Modern mems accelerometers are extremely accurate and would EASLY be able to show this difference (9.77 m/s^2 vs 9.69 m/s^2)
Why does an institution like a university not carry out this simple experiment ?
/
Here is WHY ….
- What about a plane making a turn ?
- Based on the F16C manual.
- In a level turn (15,000 feet) , at Mach 0.8 (272.232 m/s) , 7 Gs, and 14 deg/sec, the F-16 can turn 180 degrees in about 13 secs.
Google “F16 turn performance 15000 feet”
/
Based on the formulas on this page ( converted to more useful units)
GOOGLE “Wiki Standard_rate_turn”
/
Radius (meters) = 57.29579594 * Velocity (m/s) / Turn Rate (degrees/s)
/
Centrifugal Acceleration (m/s) = Velocity (m/s) ^ 2 / Radius (meters)
/
At 272.232 m/s :
/
Turning Radius (slipage) = 1114.12 meters
Centrifugal Acceleration = 6.78 G
/
This is ASSUMING a stationary earth ( of course ) .
/
Now, Factor in the 1040 mph the airplane would have it if took off from a rotating earth ( flying EAST )
/
Add on 1040 mph (464.9216) = 737.1536
/
Turn radius must be the same !!!!
/
Turning Radius (slipage) = 1114.12 meters meters
Centrifugal Acceleration = 49.7 G
/
GAME OVER fuzzy haired idiot 'Einstein" which means ONE STONE in German :)
I was amazed to see a very basic misunderstanding on display at 3:00 in this video. Consider a spinning tyre that is _not_ in contact with a road, and has _no_ translational motion (indeed, the exact situation shown in your first diagram). Sticking to the same numbers as the video, suppose the top of the rim moves to the right at 60 mph. The bottom of the rim will then be moving left, i.e. at -60mph, a difference of 120 mph. Acceleration is the rate of change of velocity. So this is _exactly the same acceleration_ as for the car in the video which is in translational motion. And this, of course, is what we'd intuitively expect: a fly clinging to the rotating rim would feel exactly the same centrifugal acceleration whether or not it was in translational motion. It follows, then, that the heating of the tyre has nothing to do with the acceleration as the video claims. (Why would it? ... there's no physical principle that says acceleration causes heating). _All_ of the heating comes from friction and compression due to contact with the road.
The acceleration causes internal stresses that add to the heating.
@@MichelvanBiezen , the acceleration is exactly the same whether or not the wheel is in translational motion. Again, look at your own examples: a change in velocity by 120 mph between the bottom and top of the rim whether you look at your first or last diagram. That's the _same_ acceleration. You say in the video that the tyre rim is undergoing a "massive" acceleration when in contact with the road, as if to imply it was a greater acceleration than the case _without_ translational motion.
And yes, without contact with the road, and that contact pushing the car forward, there wouldn't be any internal stresses
@@MichelvanBiezen , yes, there is stress due to the deformation of the tyre. Just to check, though, you're happy that the _acceleration_ on the wheel rim is the same whether or not the wheel has any translational motion. In other words, it would be the same for a car on a treadmill that was going nowhere. Or, to put it more precisely -- the accelerations can be calculated for the frame of reference in which the wheel axis is stationary without regard for translational motion. (Which is a simple statement of Galilean invariance).
Rigid body rotation and translation prove that people on a globe earth rotating and translating(orbiting the sun) would experience acceleration and deceleration. I don't feel the earth beneath me accelerating or decelerating which disproves the globe model. What do you think Michel?
The accelerations and decelerations are so small that you cannot "feel" them. The rotational speed on the Earth's surface is much smaller than the speed of the Earth orbiting the Sun.
@@MichelvanBiezen Okay but I suggest if you want make a video showing the calculations for the accelerations that would be felt by a man standing on equator of the globe surface.
Allegedly the radius is 3959 miles and orbit around sun is 67,000mph.
The accelerations are so small they cannot be "felt".
@@MichelvanBiezen Okay I get you but you are just saying that so that's why I suggest if you would like to present the calculations as to how much acceleration would be felt by a man and then people can experiment by accelerating at that force and test whether they feel it.
@@MichelvanBiezen I'll be open and up front. I'm not here to argue either. Just looking for truth. I'm a 30+ yr M.E. and a "FE'er".
If the ground were accelerating under us from 12pm to 12am and then decelerating from 12am to 12pm there would be observable evidence. Keep in mind that a delta V of 2000 mph spread over a 12hr time frame would not be linear. We'd see highest transitional forces from approximately 6pm to 12am and 6am to 12pm.
I see zero evidence of ground acceleration in large bodies of water (tides), ballistics, aircraft, gyroscopes, clouds, ect. All observable phenomena reflects a stationary ground.
I also have an expensive and very accurate scale that measures down to 0.01g (+/- 0.002 tolerance). I see no weight changes at those purported times of heliocentrism / rigid body rotation w/translation. An object simply weighs the same 24/7.
Now... if one subscribes to the theory of big bang, where we have other celestial translations going on, the effects get even greater.
I hope you can reply with a detailed answer (or video) and not just a summary statement. There are MANY of us FE'ers out here who observe real world phenomena that do not match what we've all been taught. The common man now has ability to acquire high quality equipment like telescopes, gyroscopes, accelerometers, lasers, scales, ect.. and we are looking at sciences for ourselves. We are seeing big discrepancies in the academic teachings and reality. I've gone to absurd lengths to study this and I simply see no evidence the ground is moving. Nor is it spherical in shape. (bonneville salt flats with a high zoom lens was my best observation)
Shalom to you Mr. Biezen and I truly hope you take some time to consider, go look and measure real world phenomena for yourself, and make a report or video. ❤
What if the rate of rotation is not equal to rate of translational motion?
What is the question? (Any question starting with "what if" must have a second part to the statement)
@@MichelvanBiezen
He is asking about wheel spin/slip
I don't know what happen to me, and I study every day, why the bottom is not moving? I can not understand,can't comprehend.
The most simple explanation of that is when the bottom would be moving, the wheel would had slip. Imagine you want to accelerate the car on the frozen lake, then yes, the bottom would move, but the car wouldnt go anywhere either. Because of the slip
WHAT??? The point of the tyre in contact with the road does certainly NOT stand still, and then accelerates. You are dead wrong, mate. The tyre (and all of it's `points`) have a constant, angular velocity. The `standing still` only happens to the point of the ROAD where the point of the TYRE passes. (Else the tyre would be slipping/spinning) - So the tyre does NOT (not) experience a change in velocity, and hence no acceleration, other than the constant roundabout-like acceleration around the wheel hub. What have you been smoking? I want some...
In the discussion, you need a point of reference. The point of the tire making contact with the road does not move relative to the road.
assignment bring me here
Guys, a terraplanist has used this vídeo for "proove" Earth don't rotate and orbit around of sun.
That should be an interesting "proof". 🙂
@@MichelvanBiezen that guy doesn't understand physics, he just use graphics in this video, is really funny.