I wish colleges would test teachers based on how well they can teach not based on how much they know about a certain subject because my physics professor is very knowledgable about Physics but is complete garbage when it comes to teaching it So Universities, start hiring teachers who can f**king teach
They wouldn't do that because if they hire professors that could teach (who are always less likely to be capable in research) they wouldn't earn so much money and reputation
@@lingkejiang9244 Ding ding ding! We have a winner! Most of the time, universities are for the money/prestige aspect, and the actual education aspect comes in a very distant second. I don't normally swear, but I feel strongly that it's complete bullshit if you ask me, especially, and I mean ESPECIALLY because they charge you money for your so-called "education" which could be done much better at your local library, and it's not a light amount either.
@@srirampatnaik9164 You learn the same material in physics 1 in college which is equivalent to Ap Physics 1. The only thing that is different about college is that the professor can't teach shit but you are still expected to do good.
@@JimmyCrust THIS. They even told me while I was at a well known college (New England Tech) that most employers don't care about your college - it's more about who you know or your experience.
Many people wonder why radians do not appear when we have radians*meters (rad • m). Here is an attempt at an explanation: Let s denote the length of an arc of a circle whose radius measures r. If the arc subtends an angle measuring β = n°, we can pose a rule of three: 360° _______ 2 • 𝜋 • r n° _______ s Then s = (n° / 360°) • 2 • 𝜋 • r If β = 180° (which means that n = 180, the number of degrees), then s = (180° / 360°) • 2 • 𝜋 • r The units "degrees" cancel out and the result is s = (1 / 2) • 2 • 𝜋 • r s = 𝜋 • r that is, half of the circumference 2 • 𝜋 • r. If the arc subtends an angle measuring β = θ rad, we can pose a rule of three: 2 • 𝜋 rad _______ 2 • 𝜋 • r θ rad _______ s Then s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r If β = 𝜋 rad (which means that θ = 𝜋, the number of radians), then s = (𝜋 rad / 2 • 𝜋 rad) • 2 • 𝜋 • r The units "radians" cancel out and the result is s = (1 / 2) • 2 • 𝜋 • r s = 𝜋 • r that is, half of the circumference 2 • 𝜋 • r. If we take the formula with the angles measured in radians, we can simplify s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r s = θ • r where θ denotes the "number of radians" (it does not have the unit "rad"). θ = β / (1 rad) and θ is a dimensionless variable [rad/rad = 1]. However, many consider θ to denote the measure of the angle and for the example believe that θ = 𝜋 rad and radians*meter results in meters rad • m = m since, according to them, the radian is a dimensionless unit. This solves the problem of units for them and, as it has served them for a long time, they see no need to change it. But the truth is that the solution is simpler, what they have to take into account is the meaning of the variables that appear in the formulas, i.e. θ is just the number of radians without the unit rad. Mathematics and Physics textbooks state that s = θ • r and then θ = s / r It seems that this formula leads to the error of believing that 1 rad = 1 m/m = 1 and that the radian is a dimensionless derived unit as it appears in the International System of Units (SI), when in reality θ = 1 m/m = 1 and knowing θ the angle measures β = 1 rad. In the formula s = θ • r the variable θ is a dimensionless variable, it is a number without units, it is the number of radians. When confusing what θ represents in the formula, some mistakes are made in Physics in the units of certain quantities, such as angular speed. My guess is that actually the angular speed ω is not measured in rad/s but in (rad/rad) / s = 1/s = s^(-1).
at 8:55 you say the lighter object has less rotational inertia and thus has a lower acceleration than the more heavy object, surely it would have a greater acceleration because it hits the ground first???????
I had the same question. Wouldn't the lighter pole have a greater acceleration since the force (gravity) is the same for both poles, but the lighter pole has less inertia?
This kind of reminds me of so I can take fly wheels, and also giant rotating magnets, which essentially are at the center of planets and stars. They have the size and torque necessary in order to generate those huge magnetic fields that extend outward
radians and degrees are considered "dimensionless" An angle is the same whether you are close or far from the center so there is no dimension to it. Therefore you can also just ignore it in the dimensional analysis of your unit calculations.
Cool thanks but does the rotational speed of an object on a turntable increase or stay the same as the object moves further away from the center of the turntable?
Question: For the demo with the mass tied onto the PVC pipe--I understand that the one with the mass on the end has a greater rotational inertia, but at the same time isn't there more torque applied by gravitational force when one has more mass concentrated farther from the axis of rotation? I'm having trouble justifying these results, but I repeated the experiment and got the same thing as you.
I believe how it works is that the gravitational force works equally downward on all points in the object/system of particles, so the torque doesn't really matter because the fact it is equal all over cancels it out - usually torque only causes a rotation if one spot has extra force compared to the others, for example by pushing hard on the edge of a door opposite to the hinges, but all the rest of it has no force applied.
8:45 Silly misunderstanding incoming. Surely the increased weight would increase gravity, meaning it would fall faster? Can someone explain why this isn't the case.
2 objects of different masses when dropped from the same height take the same time to reach the ground. In other words, acceleration due to gravity is independent of mass.
Let us consider a round plate mounted on a bearing. If the mass of the plate is m kg and it has a diameter of d mm. How do I calculate the torque required to bring it in motion?
The formula you would use is torque = linear force * distance (radius) of linear force from center * sin(angle between these two variables), so you would have to figure out the acceleration first as you don't have enough info as is, then use the formula F = ma to get the force applied on the object, then divide d by 2 to get the radius, and then finally use the torque formula described above
OK so I'm trying to figure out how an object can move at the same speed and yet be able to cover bigger distances in the same time, such as what he demonstrated in this video
At minute 1:57 you say that "we have to understand the importance of the radian because that is the base unit when we are measuring angular velocity, for example, or angular acceleration". That is not so, although it is what most of the scientific community believe. The error comes from thinking that in the formula ω = Δθ / Δt θ is in radians. In reality, θ represents the “number of radians” without units. At minute 2:37 you say that "for measuring angular displacement we measure that using the angle θ... we would measure that in radians". If we call β (beta) the measure of the angle, and β = θ rad then θ = β / (1 rad) and with β in radians, θ gives dimensionless [rad/rad = 1]. In the example you pose from minute 3:55 the calculation of ω gives ω = 3.49 rad/s Actually, the unit of ω should be (rad/rad)/s = 1/s = s^(-1). If "it spins at 33.3 revolutios per minute", then β = 33.3 rev β = (33. 3 rev) • [(2𝜋 rad)/(1 rev)] β = 209.23 rad θ = 209.23 t = 1 min t = 60 s ω = 209.23 / (60 s) ω = 3.49 (1/s) ω = 3.49 (rad/rad)/s. Using the formula v = ω • r it is implicit the relation s = θ • r But in that formula θ represents the “number of radians” without units. I will write another comment showing how this formula is obtained and what the variables mean. When you calculate v at minute 6:30 it turns out to be v = ω • r v = (3.49 rad/s) • (.153 m) v = 0.534 m/s and the radians do not appear. The calculation should be v = ω • r v = [3.49 (rad/rad)/s] • (0.153 m) v = [3.49 (1/s)] • (0.153 m) v = 0.534 m/s We all know that v is measured, in this case, in m/s.. What most of the scientific community says is that the radian is a dimensionless unit and is omitted. The latter is not true either, but they rely on the International System of Units (SI) brochure, which states that the radian is a dimensionless derived unit.
The arc length changes as an object is moving further from, say, the center of a turntable, but the angle stays the same no matter how far the object moves from the center.
Centripetal force = mass * centripetal acceleration (Newton's Second Law formula), where centripetal acceleration = linear/tangential velocity times itself (squared) divided by the radius of the circular path you are following.
I think I learned more in this one video than in the ~20 hours we spent on this in class.
I wish colleges would test teachers based on how well they can teach
not based on how much they know about a certain subject
because my physics professor is very knowledgable about Physics
but is complete garbage when it comes to teaching it
So Universities, start hiring teachers who can f**king teach
They wouldn't do that because if they hire professors that could teach (who are always less likely to be capable in research) they wouldn't earn so much money and reputation
@@lingkejiang9244 Ding ding ding! We have a winner! Most of the time, universities are for the money/prestige aspect, and the actual education aspect comes in a very distant second.
I don't normally swear, but I feel strongly that it's complete bullshit if you ask me, especially, and I mean ESPECIALLY because they charge you money for your so-called "education" which could be done much better at your local library, and it's not a light amount either.
Wait? This is a college-level course? I thought AP was high school stuff.
@@srirampatnaik9164 I believe that he is not referencing the video, but rather his general experience/knowledge.
@@srirampatnaik9164 You learn the same material in physics 1 in college which is equivalent to Ap Physics 1. The only thing that is different about college is that the professor can't teach shit but you are still expected to do good.
Enjoying this series very much. Dark Side of the Moon is my favourite album of all time - shocked and delighted to see it.
WHAT A GREAT TEACHER HE IS !!!!!!!!!
You are a life saver. i have my final exam tomorrow and never thought that one day i would understand physics ;) THANK YOU SO MUCH!!!!
No one can explain this topic better than u....... Superb
wow! he's a really good teacher. This video was very helpful.
i just study physics as a hobby. Your animations really makes easier the understanding of the theories.
Great video. Thanks a bunch and keep making videos! You are helping a lot of people whether they say it in the comments or not.
The animation about radians was easy to understand and was a great tool.
I am a senior in high school (Neville) and I found this video to be extremely helpful
This is the best video on angular velocity I've seen. Two big thumbs up to you sir.
Can I have you instead of my current physics prof?
No, not even a chance
Same here
﷽ ﷺ ﷺ
Same here. I sometimes wonder what’s up with most physics professors.
I normally watch your lessons for chem and bio. It's amazing that you can even do physics !!!!
Why do i even go to college when i learn everything from youtube
Colleges give you the connections to internships, research opportunities, fellowships, and jobs, but college makes you learn independently.
College sucks
well most of the jobs people apply for regardless of diploma don't
actually need one
@@JimmyCrust THIS. They even told me while I was at a well known college (New England Tech) that most employers don't care about your college - it's more about who you know or your experience.
NO ONE TOLD ME WHAT A RADIANT IS, THANK YOU SO MUCH
Best chanel for physics lectrues.
this dude taught me more in 10 minutes than my physics teacher has taught me all year
You are an excellent teacher sir. Thank you for your effort, you help me alot.
This is what i was looking for. You have answered every questions i got. thank you so much.
Saved my ass in AP Bio, and you're still saving it now...
Thank you very much for the concise and clear explanation.
Love your videos, although I'm a bit thick, so I have to keep pausing and rewinding to get it :)
You teach a lot better than my current science teacher THANK YOUUUU
it was really helpful , would love to watch more to burn the desire of curiosity .
Many people wonder why radians do not appear when we have radians*meters (rad • m).
Here is an attempt at an explanation:
Let s denote the length of an arc of a circle whose radius measures r.
If the arc subtends an angle measuring β = n°, we can pose a rule of three:
360° _______ 2 • 𝜋 • r
n° _______ s
Then
s = (n° / 360°) • 2 • 𝜋 • r
If β = 180° (which means that n = 180, the number of degrees), then
s = (180° / 360°) • 2 • 𝜋 • r
The units "degrees" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
s = 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r.
If the arc subtends an angle measuring β = θ rad, we can pose a rule of three:
2 • 𝜋 rad _______ 2 • 𝜋 • r
θ rad _______ s
Then
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
If β = 𝜋 rad (which means that θ = 𝜋, the number of radians), then
s = (𝜋 rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
The units "radians" cancel out and the result is
s = (1 / 2) • 2 • 𝜋 • r
s = 𝜋 • r
that is, half of the circumference 2 • 𝜋 • r.
If we take the formula with the angles measured in radians, we can simplify
s = (θ rad / 2 • 𝜋 rad) • 2 • 𝜋 • r
s = θ • r
where θ denotes the "number of radians" (it does not have the unit "rad").
θ = β / (1 rad)
and θ is a dimensionless variable [rad/rad = 1].
However, many consider θ to denote the measure of the angle and for the example believe that
θ = 𝜋 rad
and radians*meter results in meters
rad • m = m
since, according to them, the radian is a dimensionless unit. This solves the problem of units for
them and, as it has served them for a long time, they see no need to change it. But the truth is
that the solution is simpler, what they have to take into account is the meaning of the variables
that appear in the formulas, i.e. θ is just the number of radians without the unit rad.
Mathematics and Physics textbooks state that
s = θ • r
and then
θ = s / r
It seems that this formula leads to the error of believing that
1 rad = 1 m/m = 1
and that the radian is a dimensionless derived unit as it appears in the International System of Units (SI), when in reality
θ = 1 m/m = 1
and knowing θ the angle measures β = 1 rad.
In the formula
s = θ • r
the variable θ is a dimensionless variable, it is a number without units, it is the number of radians.
When confusing what θ represents in the formula, some mistakes are made in Physics in the units of certain quantities, such as angular speed.
My guess is that actually the angular speed ω is not measured in rad/s but in
(rad/rad) / s = 1/s = s^(-1).
at 8:55 you say the lighter object has less rotational inertia and thus has a lower acceleration than the more heavy object, surely it would have a greater acceleration because it hits the ground first???????
I had the same question. Wouldn't the lighter pole have a greater acceleration since the force (gravity) is the same for both poles, but the lighter pole has less inertia?
this will help me in my final exam thank you so much
love the examples
Mr Anderson u r great.Very nice clip. One question ? Why the direction of angular velocity is perpendicular to the plane of ur vinyl record ?
My class liked this.... Thank You :)
Great explanation, thanks for posting the videos.
You are Awesome Mr. Anderson!
Thanks.Explained very well and in detail
Guy, you gats obtain PHD for this physics something, o. Any chance I receive lecture from you?
This kind of reminds me of so I can take fly wheels, and also giant rotating magnets, which essentially are at the center of planets and stars.
They have the size and torque necessary in order to generate those huge magnetic fields that extend outward
thanks a lot for this video sir it made me understand the concept pretty well :)
I respect you. you are doing a very good job keep it up....
Thank you for this kind of tutorial, it would help on my reporting :D
you are the best!
Thanks it helped. Also, thanks for adding subtitles
I also like that you explained it very quickly, it makes it much easier to cram lol
thankyou soo much!!!Amazing video and superb animation!!
Good work Paul
At calculating linear velocity, the unit of angular velocity is rad/s and distance is in m. Multiplying both becomes m/s. where is rad?
radians and degrees are considered "dimensionless" An angle is the same whether you are close or far from the center so there is no dimension to it. Therefore you can also just ignore it in the dimensional analysis of your unit calculations.
Very nice, thanks!. Wonderful simulations and explanations.
Thank you so much you're a lifesaver
Cool thanks but does the rotational speed of an object on a turntable increase or stay the same as the object moves further away from the center of the turntable?
Very very helpful thank you so much
Question: For the demo with the mass tied onto the PVC pipe--I understand that the one with the mass on the end has a greater rotational inertia, but at the same time isn't there more torque applied by gravitational force when one has more mass concentrated farther from the axis of rotation? I'm having trouble justifying these results, but I repeated the experiment and got the same thing as you.
I believe how it works is that the gravitational force works equally downward on all points in the object/system of particles, so the torque doesn't really matter because the fact it is equal all over cancels it out - usually torque only causes a rotation if one spot has extra force compared to the others, for example by pushing hard on the edge of a door opposite to the hinges, but all the rest of it has no force applied.
very good explanation
wow he is a good teacher
2 masses on a spinning table one is heavier than the other, which mass will move farther from the centre?
Always the best
Thankyou very much, I am still working on this.
Helpful. Thank you
1:22 If we apply a force *TO* a mass, we get acceleration. I believe you said "Force * mass = acceleration"
Haha pink floyd, what can stop me from subscribing to you now? 😸
plz give me some tips how can I manage my Science ? I am weak in physics. You are a GREAT TEACHER.
U r doing grt job
8:45 Silly misunderstanding incoming.
Surely the increased weight would increase gravity, meaning it would fall faster? Can someone explain why this isn't the case.
2 objects of different masses when dropped from the same height take the same time to reach the ground. In other words, acceleration due to gravity is independent of mass.
Thank you :)
Let us consider a round plate mounted on a bearing. If the mass of the plate is m kg and it has a diameter of d mm. How do I calculate the torque required to bring it in motion?
The formula you would use is torque = linear force * distance (radius) of linear force from center * sin(angle between these two variables), so you would have to figure out the acceleration first as you don't have enough info as is, then use the formula F = ma to get the force applied on the object, then divide d by 2 to get the radius, and then finally use the torque formula described above
great video
OK so I'm trying to figure out how an object can move at the same speed and yet be able to cover bigger distances in the same time, such as what he demonstrated in this video
best one i have ever seen ..:)
Excellent. Thanks.
It helps me understand.
Thank you sir
this's very good
proof that a professor can make or break your physics experience/grade
my professor doesn't even speak fluent english smh
Thankyou so much for your videos your amazing
I am the unserstand the Chapter ROTATIONAL MOTION Thanks sir
you are very gorgoius sir
Very helpful thank you!!!
amazing video. thank you!!
great video!
It took 10 minutes for me to understand what I couldn't understand in a month
good explanation
You’re so awesome!
Thanks for explanation
it's better when compared to all,up to now
Please how did you get the 3.49 rad for the velocity?
He converted revolution to radian
Thank you
It is amazing
At minute 1:57 you say that "we have to understand the importance of the radian because that is the base unit when we are measuring angular velocity, for example, or angular acceleration". That is not so, although it is what most of the scientific community believe. The error comes from thinking that in the formula
ω = Δθ / Δt
θ is in radians. In reality, θ represents the “number of radians” without units.
At minute 2:37 you say that "for measuring angular displacement we measure that using the angle θ... we would measure that in radians".
If we call β (beta) the measure of the angle, and
β = θ rad
then
θ = β / (1 rad)
and with β in radians, θ gives dimensionless [rad/rad = 1].
In the example you pose from minute 3:55 the calculation of ω gives
ω = 3.49 rad/s
Actually, the unit of ω should be (rad/rad)/s = 1/s = s^(-1).
If "it spins at 33.3 revolutios per minute", then
β = 33.3 rev
β = (33. 3 rev) • [(2𝜋 rad)/(1 rev)]
β = 209.23 rad
θ = 209.23
t = 1 min
t = 60 s
ω = 209.23 / (60 s)
ω = 3.49 (1/s)
ω = 3.49 (rad/rad)/s.
Using the formula
v = ω • r
it is implicit the relation
s = θ • r
But in that formula θ represents the “number of radians” without units.
I will write another comment showing how this formula is obtained and what the variables mean.
When you calculate v at minute 6:30 it turns out to be
v = ω • r
v = (3.49 rad/s) • (.153 m)
v = 0.534 m/s
and the radians do not appear. The calculation should be
v = ω • r
v = [3.49 (rad/rad)/s] • (0.153 m)
v = [3.49 (1/s)] • (0.153 m)
v = 0.534 m/s
We all know that v is measured, in this case, in m/s..
What most of the scientific community says is that the radian is a dimensionless unit and is omitted. The latter is not true either, but they rely on the International System of Units (SI) brochure, which states that the radian is a dimensionless derived unit.
Subscribed.
THANK YOU !!!!!!!!!
Great video...thank u so much....
Super helpful
thx a lot..that was helpful for sure
Why is it that we measure by the angle and not the arc length
The arc length changes as an object is moving further from, say, the center of a turntable, but the angle stays the same no matter how far the object moves from the center.
what is the centripetal force in a rolling disk? the net intermolecolar forces for every "molecule" of disk? thx in advandance
Centripetal force = mass * centripetal acceleration (Newton's Second Law formula), where centripetal acceleration = linear/tangential velocity times itself (squared) divided by the radius of the circular path you are following.
gr8 explanation!!
Very nice.
isnt hollow object have the greater inertia i was waiting for the green one to go down fırst
At 6:51 how'd you go from rad/s * m = m/s, seems it should read = rad*m/s
Excellent 🙏
You da real mvp
good point
Thank you
Thanks man
Cool.
amazing!
damn good explanation