Physics 13 Application of Moment of Inertia and Angular Acceleration (5 of 5)
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- Опубликовано: 16 окт 2024
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In this first of the two part series I will show you how to find the angular acceleration of a pulley system where the pulley has a mass of 2kg.
Next video in this series can be seen at:
• Physics Ch. 13 Moment ...
If just my professor was only a fraction as good as you I'd be so happy
Only when the pulley has no mass (no moment of inertia) and no friction is the tension in the string the same on both sides.
+Michel van Biezen If there is not a friction between string and pulley then string must slip over the pulley. That means pulley must not rotate instead of that string is slipping over the pulley. It is not clear how tangent force of the string which has a tension acts on pulley?. Again it is clear if tangent is acting on pulley then friction force must act between pulley and string. Other wise how tangent force acted
Gareth M
The tension in the string acts in both directions.
Thus when you draw a free body diagram, the direction of the tension will be relative to the body in consideration.
Relative to the pulley the tension acts downward.
Relative to the two masses, the tension acts upwards.
Your explanations are extraordinarily clear. I feel every word is chosen accurately, but apart from the terminology the explanations are clear AND adequatly concise. Moreover, the vast range of problems (in difficulty and in topics) makes this probably the best channel for physics on youtube.
Thank you for your kind words.
There are two ways to do this problem. One way is to call the direction of the acceleration the positive directions. Then I call all the forces helping (or pushing) the acceleration positive forces and all the forces opposing the acceleration negative forces. The other way is the draw free body diagrams and up is always positive and down is always negative. This method is more difficult and you will have to solve multiple equations and multiple unknowns.
This free-body diagram also serves as a handy reference for the volume of this video's left and right audio channels.
Joking aside, thank you a lot. It was very clear and concise.
In the case of a problem like this where you have to take into account the moment of inertia of the disk, you have to assume a positive and a negative direction for the torque. Traditionally the counter clockwise direction is positive and the counterclockwise direction is negative.
I have my finals in 2 weeks and this mans a genuine life saver!
Glad you found our videos! 🙂
thank you so much! i wish my teachers were as clear as you
That was the greatest lecture I have ever seen
Thank you for linking the previous and next video! Very convenient for me to learn step by step!
You are welcome. We started doing that on our newer videos.
Great job Professor! You have made the world a better place.
We need professors like this guy asap
THANK YOU FOR DOING THIS YOU HAVE NO IDEA HOW BIG OF A HELP THIS WAS THANK YOU THANK YOU THANK YOUUUU
thislittlemiggy
Glad to be of help. Thank you for the comment.
Much love prof! The algebra always gets me and I knew you would come through.
Thank you so much! This lecture was amazing! I'm taking a college level physics class and this helped me so much!
It couldn't be explained any better. Thank you so much
Glad you liked it! 🙂
Glad it helped.
thank you for helping me in my dynamics exam
It's our pleasure
You make it look so easy and clear :) thank you
Idk why but the sound is too low on the right side... This only has happened on this video until now. Just warning...
Im loving your videos, they are helping me a lot!!! Thx
Yes, our older videos were recorded in mono sound (not stereo) before we figured out what we were doing.
How do we know which tension is larger and which one is smaller? thank u for ur invaluable work! I hope u will response my question! am waiting!
It is easy to check. Once you have calculated the acceleration, for the larger mass accelerating downward: T = (10)(9.8) - (10)(a) and for the smaller mass accelerating upward: T = (5)(9.8) + (5)(a)
Very clear instructions. Great video!
Thank you, hopefully this is enough to get me a passing grade 🙏🏼
You can do it!
I did this problem by using the energy conversations, first found the speed then got the accelaration.
Yes indeed, that is a good method.
Thank You so much! This cleared a lot of misconceptions that I had
Tanq so much sir... Exact video i was looking for .... Upload more...
We have uploaded many videos on this topics.
Thank you so much for all the content you put out! 🙏🙏🙏
Thanks, I have been searching RUclips for this exact problem. Are there situations in which there would be a center friction on the circle that would act against the flow of motion? If so, any videos on this?
Why are the Tension forces T1 & T2 in the same direction as m1g and m2g?
Because we only look from the pulley's point of view. Had we instead taken a look at the forces acting on m1, the direction of the tension T1 would be upward.
The professor used a pulley as a system. So the tension in pulley's point of view must point down.
because tension is pulling force
thank you so much!!! this was so helpful
Glad you found our videos and found them helpful! 🙂
Thank to you i’m finally get it
you're a pretty great teacher !
Thanks a lot, really helpful. Your channel is amazing. Subscribed
Good tutorial! Thanks for sharing your knowledge! :)
Great Explanation.👍
Thank you. Glad you found our videos. 🙂
Hi Michael, if we were to do this same problem but ignore the pulleys mass, can we say that T1=T2? If so, why? And if not, why?
Yes, if the pulley does not have mass and there is no friction, then T1 = T2
dear teacher
my name is mesut from Turkey
ı have a question for you. ı want to spin 10 kw electric engine by flywhell and other low power electric motor. my flywhells weight is 30 kg and radius 30 cm. how many power will be use other electeic motor?
Excellent, as always. Thanks
Does taking the pulley's mass into account differs the tension of T1 and T2? I always thought of it as the same.
Anyways, excellent video. Especially with how rigorous the explanation was.
The tension will be the same only if we don't take the mass of the pulley into account. The pulley will have a moment of inertia which will cause the tension on both sides to be different.
@@MichelvanBiezen awesome,thanks
I have a problem about the friction. Because we know that the rope doesn't slip of the pulley because of friction so I have a problem wether or not we need to add fr to our torque or not.
The friction between the rope and the pulley allows the rope to apply a torque (which would not be possible without friction), but the friction does not add torque.
Thank you sir your videos are great.
I got a problem related to this topic. two objects of mass 2kg and 1kg are hanging from a frictionless pulley. What will be the acceleration of the centre of mass?
Does the pulley have moment of inertia?
no sir.
You said this one is a bit tricky. If I use the energy equation to get the final velocity, isn't angular acceleration equal to v^2/R and a = alpha × R, still cool
Those two equations are always correct.
Please make a video on degree of freedom of various atomic gases.
You may find your answer here: PHYSICS 32 KINETIC THEORY OF A GAS
Are you subtracting the torques from each other because they move in opposite directions?
That is correct.
Sir, why is I = 1/2(mR^2)? shouldn't it be I = mR^2? sorry for the clarification
The moment of inertia is mR^2 if all the mass is located at distance R from the point of rotation, but for a solid disk the moment of inertia is (1/2) mR^2
Free online Physics courses for CBSE n ISC school's.... Join Now...ruclips.net/video/wskkuE6XR3I/видео.html
Thank you so much for making your videos!!!
If you try to find the angular acceleration(alpha ) from the net torque equation=I*(alpha) you would not find it equal to the value you would find if you try (alpha)=a/r
why does T1=m1g - m1a, when both forces are in the same direction, shouldn't they be added? and shouldn't the tension forces be opposing the m1g and m2g?
No, since the blocks move as a system. The block on the left moving down is the same motion as the block on the right moving up since they are connected. That is why the technique in the video is used.
@@MichelvanBiezen i still dont get it
@Mert Kaplan The Tension in that equation is from the perspective of Box 1, not the pulley. You can tell because the masses are not marked "P".
So for m1, m1a = m1g - T1, solve T1, T1 = m1g - m1a
Why isn't T equal to the force of gravity? Isn't the tangential force being applied onto the pulley just the weight of each object? Why do you have to subtract ma?
+Aaa Aaa
The tension T is only equal to mg when the objects are motionless (or when they are moving at a constant velocity).
Let's pause for a moment and think about it. When an object is hanging from a string and not moving, the tension T must be equal to the weight of the object (mg).
Now imagine that you pull harder on the string in order to accelerate the object upwards. The additional force must be equal to:
F = ma. Therefore the total tension T must now be the sum of the two forces: mg + ma
Can you use the energy method to solve for angular acceleration or is this the only way?
You can use Newton's 2nd law. See the videos in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
Awesome, thanks!
You sir are a life saver ! :')
Say the heavier block was sliding down a ramp with friction, how would that affect the acceleration?
+pepsi1241
There are lots of examples in this playlist:
PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
why are you able to separate t1 and t2 into 2 diff force equations? Don't they rely on each other in some way?
T1 and T2 will be different because the pulley has mass and therefore a moment of inertia.
Hats off, you are amazing. thank you!
Thank you, Sir.
You are very welcome
Do you have any videos over the rotational inertia of multiple disks attached to an object in free fall? Also, if the first disk is situated so that the string connecting the first disk and the second disk is at a 45 degree angle above horizontal, how does this effect the moment of inertia of the second disk?
www.kwantlen.ca/science/physics/faculty/mcoombes/P1101_Solutions/RotDynamics/P08_00006.gif
^^^This image is a similar representation of my question.
Thomas. Not yet. It takes a while to cover every possible topic in physics. I am currently going back and adding additional examples on specific topics. (currently working on conservation of momentum and impulse).
Michel van Biezen Understood. Thanks for the videos they help a lot.
your audio is always stuck in the left headphone, but the videos are very helpful so I deal with it
Yes our older videos were not recorded in stereo. (Part of our learning process)
Can we use the conservation of energy concept to solve?? Please reply sir I have a test.. Thank you
To find the acceleration of the system, this is the best method. To find the position and the velocity of the masses you can use the conservation of energy. There are many examples like that in the playlist. PHYSICS 8 WORK, ENERGY, AND POWER ruclips.net/p/PLX2gX-ftPVXWtv0wrHIHTCUrZHUkVDJiZ
Thank you sir... :)
thank you professor
Welcome!
sir, you are a life saver :D
what you have written on white, is not visible clearly.
Very helpful, thanks!
Can anyone help me with the understanding of the tensions direction? T1 and T2 both point towards the mass, which I, for the life of me, can not understand. The way I see it, if mass1 was not attached to the rope it would accelerate faster, therefore, tension should be in the up direction, working against gravity. And the same for mass2, if tension if pushing down on the mass, then how would the mass accelerate against gravity??
Thanks.
It is better to ignore the internal forces of the system (the tensions) and only concentrate on the external forces acting on the system and causing the whole system to accelerate. Since m1 is larger m1g will be aiding the acceleration (acting in the same direction as the acceleration of the system) and m2g will be opposing the acceleration (acting in the opposite direction as the acceleration of the system) and a can be found by: a = (m1g - m2g) / (m1 + m2)
I was looking into that too for acceleration ( a = (m1g - m2g) / (m1 + m2) ) (Atwood's Machine 'n stuff) While using that equation, the acceleration comes out to 3.27 m/s^2. which is 7% larger than the value calculated in the video.
Which result is closer to true value of the systems acceleration?
Also, thanks for the FAST reply and video to help students like myself understand physics!
Since the pulley in this example has mass, you must also include the moment of inertia and the energy required to get the pulley spinning, thus the actual acceleration will be less.
hi, sir if I make clockwise as positive torque=- I a /R? Is torque negative?
sorry i mean anticlockwise as positive
It is better here to only use the magnitude of the torque and not worry about the direction of the torque. We only need the magnitude to find the acceleration.
Sir this means that friction is present? Because without friction why would pulley rotate
The friction (or lack of friction) referenced here is the friction on the bearings of the pulley.
if i cut the string will the distance of the two bodies be the same as they fall?
+Gugu Nyambe
That depends on the initial conditions when you cut the strings.
1) What are the positions?
2) What are their velocities?
Both of those conditions affect how long it takes for each block to hit the ground when the string is cut.
Thank you very much sir
Most welcome
Why is T1 and T2 pointing downwards?
I'm not sure how you get the force equations using them both pointing in the same direction.
Since the pulley causes the direction of the tension to change it is better to think about it like this: F net = F aiding the acceleration - F opposing the acceleration = (total mass) x ( a)
If I understand this correctly, wouldn't that mean that T1 = ma - mg ---> (ma = T + mg => T = ma - mg)?
Both tension (T1) and mg are pointing in the downward direction, which is the direction that the acceleration is pointing, so the Forces are positive (aiding acceleration). By getting T1 = mg - ma, it seems that your original equation was ma = mg - T.
You need to add the subscripts to your masses and T in the equations before I can answer your questions. Also T1 = T2
I suggest you watch several examples in the playlist, including those with the inclined plane and a pulley to see how to work these types of problems.
Sorry, from above: m[1]a = T[1] + m[1]g => T[1] = m[1]a - m[1]g
Both T[1] and m[1]g are pointing in the downward direction, so that's why I set up my above equation like I did.
Since you got T[1] = m[1]g - m[1] it seems that your original equation was m[1]a = m[1]g - T[1]
And I'm currently on the Torque playlist :)
Ur the best
Many thanks, great explanation!!
My problem is almost the same but it adds a constant torque of .35Nm due to friction. can you help me find the acceleration with this added factor?
+Bryant Flores
Where you given the radius and moment of inertia of the pulley? If yes,
you'll have to use the principles seen in the videos of this playlist:
PHYSICS 13 MOMENT OF INERTIA APPLICATIONS
If not, the you can find the force by taking the torque and dividing it by the radius of the pulley. Then add the force to F net = m total * a
sir u calculated m.I. of the wheel as half mr^2 but it looks like a ring of m.I. mr^2 so whether it is a disc or a ring
+Rahul Tiwari
consider the pulley to be a solid disk.
Sir u told that when the pulley is massless and frictionless then the tension on both sides of the string is same ? But how can the mass of a pulley affect the tension and not the acceleration? inspire of the fact that it is a single string connecting both the masses m1 m2
if we're looking for tensions do we first need to find the acceleration?
Yes, unless it is a statics problem
If it was a smooth pulley, and tensions on the both side are equal, will the net torque become zero?? Pls answer
Torque is defines as F x perpendicular distance. However if the pulley is smooth then there is no way to apply a force on the pulley and therefore you cannot apply a torque.
So if pulley is smooth , the pulley will not rotate, the rope just slips on pulley with some acceleration ..im I right sir?
We are making as assumption that the pulley is massless and frictionless, which of course is impossible. (Like we assume no wind resistance, which is also impossible). All pulleys have some mass and some friction. Thus in real life, there is a small amount of torque and a small amount of friction with all pulleys. That said, if there was a perfectly smooth pulley, then the pulley would not rotate, but then if the pulley had no mass, and minute amount of friction would make it rotate.
Thanks.master very good
Glad you enjoyed it.
YOU'RE A WIZARD, BIEZEN
Glad you found our videos. Welcome to the channel!
Why isn't the tension equal to only the weight force?
Only if the objects are not accelerating. If the objects are accelerated you have to add the force required to accelerate the object.
@@MichelvanBiezen Thank you so much
you are amazing!!!!!!
Awesome now I understand this!
great one
Bless your soul
Thanks!
Welcome!
could energy work for this problem, if not why?
I just get confused on what method to use
It depends on what is being asked. If they want to know what the acceleration is, then the method shown works best. But if they want to know the velocity when the object reaches a particular height, then energy conservation works best.
@@MichelvanBiezen thank you for replying!
for the tension 1 why do we subtract m1g - m1a?
+Luis Alva
To show yourself that this is correct, draw a free body diagram and use F net = m1 a
how did +m2a become -m2a?
Amazing
Glad you liked it. 🙂
why is the moment of inertia equal to 1/2mr^2 .. shouldnt it be equal to mr^2!!?
The moment of inertia of a solid disk is (1/2) MR^2
Wait why is the moment of impulse 1/2mr^2?
My textbook says its Σmr^2.
Which one is correct?
+rurutchi
The moment of inertia depends on the shape of the rotating object. For a solid disk the moment of inertia is (1/2) mR^2
For a hollow disk it is mR^2 Take a look at the playlist: PHYSICS 12 MOMENT OF INERTIA
Fundamentally, moment of inertia is Σm*r^2. Or in its continuous form, ∫r^2 dm.
When you account for the distribution of mass, you often will get a formula for moment of inertia equal to X*M*R^2, where X is a fraction that you get after carrying out this integral. For a thin uniform hoop, X=1. For a solid disk or cylinder around its circular axis, X=1/2. For a sphere, X=2/5. For a thin spherical shell, X=2/3.
thank you 😼
Good video, very detailed explanation! I figured that there is an easier way to solve this, for those interested. Consider a wheel and a frictionless axle. There is a rope that is tied to the wheel, and you drew a black line on the rope. When you pull the rope with force F, how fast does the black line accelerate? It accelerates at F / [1/2 mass of of the wheel]. From this, we can say that the resistance of the wheel to linear acceleration is half that of the resistance of a rectangle to a linear acceleration. After that, solve this like any dynamics problem. The net force is g(m1 - m2). The inertia of the system is (0.5mass_of_wheel + m1 + m2)
Thank you
Thank you!
Could you just use the formula a= [(m1-m2)*g]/ (m1+m2+(.5)(mpulley) ??? Like all the time?
+TRICKYY11500 Only if the pulley is a solid disk with moment of inertia = I = (1/2) m R^2. The equation can be different with a different moment of inertia.
Thank you sir :)
why is T1 = m1g-m1a..acceleration, tension, and mg are in the same direction; using F=ma I end up with t1=m1a-m1g
Adelle,
There are two methods you can use to find the tension in the string. The first method is to use free body diagrams. When using that method one must take into account the positive and negative directions and simply add them like vectors and use F - ma. The second method (which is simpler) sets the tension equal to the weight of the object suspended from it +/- ma. Use + when the mass is being accelerated upwards for the additional force required to accelerate it upward (F = ma) beyond the force required to hold it against gravity (mg). Subtract ma when the object is accelerating downward because less force is needed than mg because it is accelerating downward. In the limit when the acceleration downward reaches g (a = g) then the tension goes to zero.
Michel van Biezen is it okay to use the freebody diagram method?
Samuel Akorede
Definitely.
It is typically harder and more students tend to make errors with that method.
However every student should know how to use the free body diagram method.
I believe that I have a few videos devoted to that method as well.
what is frequency here?
There is no frequency in this type of problem.
lifesaver, thank you sir
Thank you so much!!!
I'm trying to figure out how to create a pulley that will help me pull up and pull down a bed. Learn all this is gonna help me what I need
Look in this playlist for multi-pulley systems: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS