I once went to a presentation where the speaker's theme was that true success is making a positive impact on the lives of others. You sir, I can tell is truly successful. I cannot thank you enough and you are the hero we all need.
Simply amazing video. I've seen many videos of physics problems on the internet but you're really quick to get to the point, amazing resolution of video, overall congrats man.
I was pulling my hair out over a similar problem involving finding the radius. I completely missed the obvious definition of using the tension given to me to find it. This video opened my eyes to that. Thank you very much.
5 years later and thank you for this. Current teacher is very bad at teaching (shows in how many students dropped the class and the last exam class average of 53%) and this just taught me most of my exam tomorrow.
Thank you. I am glad I was able to give something back with a huge help from my wife who worked very hard on recording, editing, producing and uploading these videos and she made all of the thumbnails.
Thanks for the great video. Just one question. Using vector equations r x F I get -Ri x -T j = RT k (ccw) if mass is left and Ri x -Ti = -RT k (cw) if mass is on the right but equation at the mass : ma j = Tj - mg j stays the same...... If the mass is left, do I just change the sign of T so it is opposite a??
It is arbitrary which direction you assign as positive in each direction. If it were my choice, I would assign the downward direction that the mass falls as positive, and assign the direction the disk rotates as positive. And I would do this, regardless of whether the mass is on the left or the right. To work through the vector equation of torque, if the mass hangs on the left side, then the radius vector points left, and the force of tension on the disk points downward. Using the right hand rule, this means the torque is out of the screen (circle dot), which corresponds to causing a counterclockwise rotation. When the mass hangs on the right side, the radius vector points to the right and the tension force is still downward. The torque would therefore be in to the screen (circle X), which corresponds to causing a clockwise rotation.
Good question. If you think of F = ma as the Net F = ma, then the net force will equal (all the forces aiding the acceleration) - (all the forces opposing the acceleration). In this example mg aids the acceleration (it is in the same direction) and T opposes the acceleration (as seen from the perspective of the mass).
very helpful thank you. Ive got a problem with this same senario but with a bearing frictional torque given in Nm. how would i put that in with workings?
That would be an opposing torque. Therefore you write Torque (aiding the acceleration) - Torque (opposing the acceleration) = (moment of inertia) x ( angular accleration)
Your tutorial vids rock!!! they are the only one I download in 1080p.. help i find more when i need them... Q. when calculating torque for such a system , why dont we use the Fnet that comes from resolving the forces acting on the mass?? why tension and not Fnet?
The force T that affect at the mass of the diskis mg-T , I mean that the force tat affect on the disk is the same that affect on the other mass, we can't ignore the weight of the box while calculating force that make torgue on the disk, is this right?
No. In the case where we have moment of inertia, we need to use the rotational equivalent for F = ma Instead use Torque = I x angular acceleration. (As is shown in the video).
Hi Sir These videos are great. Is there a video that shows how to work out the rotational speed of a flywheel as a result of the energy transferred from the falling mass?
Michel van Biezen Thank you for replying sir. However I'm not sure where to find the video I'm after. My problem only involves one mass and one flywheel like in this video
The moment of inertia of a flywheel is: I = (1/2) mR^2 and rotating KE is (1/2) I w^2. The problem is solved the same was as the others once you know the moment of inertia.
Thank you. I know my inertia is 1.36 Kgm^2 for the pully, 15.54 kgm^2 for my flywheel therefore my inertia for my pully + my flywheel is 16.9 kgm^2. I also know that my PE= 9182.16J. So from this I can work out my speed of the flywheel when the mass (260kg) is dropped (3.6m)?
We have a number of videos that show you how to do that: PHYSICS 12 MOMENT OF INERTIA You can find more of them in the mechanical engineering sections as well.
@@MichelvanBiezen sir i have found these videos.but i want to calculate moment of inertia of bodies which i have mentioned above practically by using fly wheel,so help me regatds this
I have a doubt,, why we are not considering the moment of inertia of the small mass in T=I@.what I am saying is, why" I " is not equal to (MR2/2 +mL2) where L is the length between small mass and large one..? please help me...
OMG .... Your reply was so fast than that I expected. and now I got the idea.... THANK YOU.......THE GOD WILL BLESS YOU FOR THE EFFORTS THAT YOU ARE TAKING....
Hello sir, i have a problem here. if i use the same object for the mass,m and the flywheel,M with radius,R, except there is a second wheel added behind to the flywheel attached with the same shaft/center of rotation with the radius,2R where the string is attached to and its mass is so light that it can be neglected. how does this system react? thanks in advance.
From what I understand, you have two disks fastened together to form the rotating body, and otherwise the same problem. You simply add up the moments of inertia of the various disks, to get the total moment of inertia of the rotor as a whole.
Once you solve for the linear acceleration as shown in this video and others in the playlist, then you'll need the radius since angular acceleration = a / R
If you want to express the acceleration as a vector, then yes. But if you express the acceleration as a magnitude, then no. The magnitude of a vector can never be negative.
It shouldn't matter if you use the either convention for torque, you should get the same answer. However the technique shown in the video makes it less likely to make a mistake (at least in my opinion).
Another INTERESTING fact is this: If the Mass of the Pulley is equal to the Mass of the Weight.. then the resulting Acceleration would equal 2/3 g .. That's the SAME acceleration for the YoYo..!!!!. I wonder how can a Mass unwinding from a Disk have the same acceleration as a YoYo? The YoYo is just a rotating Disk by itself, and yet the Weight and the Mass of the Pulley are TWO components in that system... The MATH doesn't LIE.. but I'm still pondering how those Two Accelerations would be equal.. behold the Unintuitive nature of Physics.. :)
Usually when the mass of the pulley is not give, it implies that they want to consider the pulley "massless" and you don't have to take into account the moment of inertia of the pulley.
In this case we just need to calculate the magnitude. (we only consider the sign on the torque if it is a vector quantity, or we need to add multiple torques)
Torque = the cross product of ( FORCE x PERPENDICULAR DISTANCE FROM THE POINT OF ROTATION TO THE LINE OF ACTION OF THE FORCE) Also tension is a force. We have a playlist on the concept of torque that describes all that in great detail.
When you draw a free body diagram around the mass, and you consider that the acceleration is downward and call that direction positive, the mg is the force aiding the acceleration and the tension T is opposing that acceleration. And then F = ma becomes mg - T = ma
400 watt three phase 60 hz motor at 25 rpm I want it to pull a load capacity of 400kgf.m How many N.M does this motor need to do this task? Drum half diameter 12.5 cm
You will be forever remembered for your services. Students around the world are grateful, thank you so much!
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I once went to a presentation where the speaker's theme was that true success is making a positive impact on the lives of others. You sir, I can tell is truly successful. I cannot thank you enough and you are the hero we all need.
Thanks for sharing!
Simply amazing video. I've seen many videos of physics problems on the internet but you're really quick to get to the point, amazing resolution of video, overall congrats man.
I was pulling my hair out over a similar problem involving finding the radius. I completely missed the obvious definition of using the tension given to me to find it. This video opened my eyes to that. Thank you very much.
Glad it helped 🙂
5 years later and thank you for this. Current teacher is very bad at teaching (shows in how many students dropped the class and the last exam class average of 53%) and this just taught me most of my exam tomorrow.
We are glad our videos are helping.
your have really nourish me in physics my thirst of being physics teacher is now quench ,thanks so much sir we are ever grateful
You are welcome. Glad you found our videos. 🙂
It helps me during my ODL at home, a huge thank for you Sir
Happy to help
Lovely bowtie, nice explanations! This guy knows what he's talking about!
You will remember as a meaningful man❤
Thank you. I am glad I was able to give something back with a huge help from my wife who worked very hard on recording, editing, producing and uploading these videos and she made all of the thumbnails.
bro, you are HIM!!!! ur the GOAT!!!!!!!
Thank you. Glad you like our videos. 🙂
basics are explained in very simple and easy way.... Great teacher
Thanks
Thank you so much kind Sir. Everything you say is so easy to understand.
Glad you find our videos useful.
Thanks for the great video. Just one question. Using vector equations r x F I get -Ri x -T j = RT k (ccw) if mass is left and Ri x -Ti = -RT k (cw) if mass is on the right but equation at the mass : ma j = Tj - mg j stays the same...... If the mass is left, do I just change the sign of T so it is opposite a??
It is arbitrary which direction you assign as positive in each direction. If it were my choice, I would assign the downward direction that the mass falls as positive, and assign the direction the disk rotates as positive. And I would do this, regardless of whether the mass is on the left or the right.
To work through the vector equation of torque, if the mass hangs on the left side, then the radius vector points left, and the force of tension on the disk points downward. Using the right hand rule, this means the torque is out of the screen (circle dot), which corresponds to causing a counterclockwise rotation. When the mass hangs on the right side, the radius vector points to the right and the tension force is still downward. The torque would therefore be in to the screen (circle X), which corresponds to causing a clockwise rotation.
Amazing as always. Waiting for new videos!
2:25 why in other videos they write T-mg and other times mg-T ?
Good question. If you think of F = ma as the Net F = ma, then the net force will equal (all the forces aiding the acceleration) - (all the forces opposing the acceleration). In this example mg aids the acceleration (it is in the same direction) and T opposes the acceleration (as seen from the perspective of the mass).
@@MichelvanBiezen oh that makes sense, thanks a lot sir my final is next week and your videos have always been helpful !
Good luck on your final.
Thanks so much for this! I was racking my brain around a similar problem and this really helped. :)
Thanks from India 😊
You are welcome. Welcome to the channel!
Few seconds watching this...and my mind has just opened to the realities....🔥🔥🔥thank you so much
Glad the videos were helpful. 🙂
Thank you you saved my life
That may be exagerating things a bit, but we are glad that our videos are helping.
may i ask why the mass that hung on the disc is increasing, then the moment of inertia will decrease? This is found from my school experiment
You may have drawn the wrong conclusion from your school experiment. (The inceasing acceleration is not a function of decreased moment of inertia)
Thank for great video! It's very helpful to me
Sir your video are very helpful can you please make some videos on pure rolling and combine rotation and translation motion
very helpful thank you. Ive got a problem with this same senario but with a bearing frictional torque given in Nm. how would i put that in with workings?
That would be an opposing torque. Therefore you write Torque (aiding the acceleration) - Torque (opposing the acceleration) = (moment of inertia) x ( angular accleration)
Thank you so much sir.
Your videos are really helpful and I am extremely grateful to you.
Glad the videos are helpful. 🙂
Your tutorial vids rock!!! they are the only one I download in 1080p.. help i find more when i need them...
Q. when calculating torque for such a system , why dont we use the Fnet that comes from resolving the forces acting on the mass?? why tension and not Fnet?
If the pulley has mass, we cannot use F = ma. We have to use Torque = moment of inertia x angular acceleration
why is it only the Tension "T" that causes the torque, and not the NET force (ie. mg - T)?
If you draw a free body diagram around the wheel, then you can tell that the only force pulling on the pulley is the tension of the string.
You are perfect teacher my respect for you
You are the best l like your way of teaching thanks
The force T that affect at the mass of the diskis mg-T , I mean that the force tat affect on the disk is the same that affect on the other mass, we can't ignore the weight of the box while calculating force that make torgue on the disk, is this right?
I am sorry, but we didn't quite understand the question. But the method used in the video to find the acceleration is the best method.
I'm from India you are amazing sir
Thank you. Welcome to the channel! Indian schools have very high standards. We hope that you'll find these videos helpful. 🙂
What about its angular acceleration? Is it mgR/R^2(m + M/2) or 2mg/RM. Please confirm good people of the Earth. Thanks.
a = angular acceleration x R Therefore angular acceleration = a / R
great teacher Micheal
Thank you
you rock! I'm kinda starting to like physics haha
what about : (Sigma Fy = m* ay) and we can get (ay) from y = 1/2(a)t^2+v0t+y0 ? Is that correct? ((T - mg = -m(ay)))
sir why we didnt take care of big M for F=ma equation, i mean cant we say that mg+Mg-T=(M+m).a ?
No. In the case where we have moment of inertia, we need to use the rotational equivalent for F = ma Instead use Torque = I x angular acceleration. (As is shown in the video).
Hi Sir
These videos are great.
Is there a video that shows how to work out the rotational speed of a flywheel as a result of the energy transferred from the falling mass?
Take a look at this playlist: PHYSICS 8.5 ROTATIONAL KINETIC ENERGY th-cam.com/users/ilectureonlineplaylists?sort=dd&shelf_id=4&view=50
Michel van Biezen Thank you for replying sir. However I'm not sure where to find the video I'm after. My problem only involves one mass and one flywheel like in this video
The moment of inertia of a flywheel is: I = (1/2) mR^2 and rotating KE is (1/2) I w^2. The problem is solved the same was as the others once you know the moment of inertia.
Thank you. I know my inertia is 1.36 Kgm^2 for the pully, 15.54 kgm^2 for my flywheel therefore my inertia for my pully + my flywheel is 16.9 kgm^2. I also know that my PE= 9182.16J. So from this I can work out my speed of the flywheel when the mass (260kg) is dropped (3.6m)?
How would the a(translational) change if m was a disk and connected not at the center but at the surface like M?
Love your videos!
sir how we can measure the moment of inertia of diffferent bodies like hollow ,dics and solid cylinder?
what will be the change in apparatus
We have a number of videos that show you how to do that: PHYSICS 12 MOMENT OF INERTIA You can find more of them in the mechanical engineering sections as well.
@@MichelvanBiezen sir please send me the link where i can find these videos.
@@MichelvanBiezen sir i have found these videos.but i want to calculate moment of inertia of bodies which i have mentioned above practically by using fly wheel,so help me regatds this
Use the examples as a guide and you will be able to calculate the moment of inertia of any object.
thank you for saving me again
Glad our videos are helping. 🙂
I have a doubt,, why we are not considering the moment of inertia of the small mass in T=I@.what I am saying is, why" I " is not equal to (MR2/2 +mL2) where L is the length between small mass and large one..? please help me...
The small mass does not rotate and thus it doesn't have moment of inertia.
OMG .... Your reply was so fast than that I expected. and now I got the idea....
THANK YOU.......THE GOD WILL BLESS YOU FOR THE EFFORTS THAT YOU ARE TAKING....
You sir are a hero
Hello sir, i have a problem here. if i use the same object for the mass,m and the flywheel,M with radius,R, except there is a second wheel added behind to the flywheel attached with the same shaft/center of rotation with the radius,2R where the string is attached to and its mass is so light that it can be neglected. how does this system react? thanks in advance.
From what I understand, you have two disks fastened together to form the rotating body, and otherwise the same problem. You simply add up the moments of inertia of the various disks, to get the total moment of inertia of the rotor as a whole.
how would you solve for the angular acceleration if you are given the value of the moment of inertia?
Once you solve for the linear acceleration as shown in this video and others in the playlist, then you'll need the radius since angular acceleration = a / R
Why is the object's weight not used in the calculation of sum of moments?
We didn't sum any moments (because there is only one rotating object. But the weight of the object was used in the torque equation.
Would the acceleration be negative so TR=-.5MR^2(a/R) since it's moving downward?
If you want to express the acceleration as a vector, then yes. But if you express the acceleration as a magnitude, then no. The magnitude of a vector can never be negative.
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You are welcome.
When i follow the negative clockwise convention for torque i get a different answer
It shouldn't matter if you use the either convention for torque, you should get the same answer. However the technique shown in the video makes it less likely to make a mistake (at least in my opinion).
If the mass accelerates at 9.8 m/s2, then there is no tension and no torque?
That is correct.
Great Explanation
Another INTERESTING fact is this: If the Mass of the Pulley is equal to the Mass of the Weight.. then the resulting Acceleration would equal 2/3 g .. That's the SAME acceleration for the YoYo..!!!!. I wonder how can a Mass unwinding from a Disk have the same acceleration as a YoYo? The YoYo is just a rotating Disk by itself, and yet the Weight and the Mass of the Pulley are TWO components in that system... The MATH doesn't LIE.. but I'm still pondering how those Two Accelerations would be equal.. behold the Unintuitive nature of Physics.. :)
what do I do if the mass of the pulley is not given?
Usually when the mass of the pulley is not give, it implies that they want to consider the pulley "massless" and you don't have to take into account the moment of inertia of the pulley.
@@MichelvanBiezen but I had to find the moment of inertia as well.
What information was given?
@@MichelvanBiezen I had the mass of the block, the time it took for the block to fall a certain height, and radius of the pulley/wheel
I thought it would be Torque = -Ia because it’s spinning clockwise
In this case we just need to calculate the magnitude. (we only consider the sign on the torque if it is a vector quantity, or we need to add multiple torques)
Sir how torque=tension* radius
Torque = the cross product of ( FORCE x PERPENDICULAR DISTANCE FROM THE POINT OF ROTATION TO THE LINE OF ACTION OF THE FORCE) Also tension is a force. We have a playlist on the concept of torque that describes all that in great detail.
I have a question, why is F=ma equivalent to mg-T=ma and not T-mg=ma
When you draw a free body diagram around the mass, and you consider that the acceleration is downward and call that direction positive, the mg is the force aiding the acceleration and the tension T is opposing that acceleration. And then F = ma becomes mg - T = ma
Michel van Biezen that makes a lot of sense! Thank you!
How about if there are two objects hanging with different masses?
We have examples like that as well in the playlist
Thank You, Sir!!1
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00:26 for a second I thought my man said twirk instead to torque
dosen't gravity effect the mass ????
Yes, that is why the force "mg" is drawn on the diagram.
Thank you sir
Thank you
400 watt three phase 60 hz motor
at 25 rpm
I want it to pull a load capacity of
400kgf.m
How many N.M does this motor need to do this task?
Drum half diameter 12.5 cm
You've given the torque you want it to pull, 400 kgf - meters. Simply translate kgf to Newtons, and you get 3920 Newton-meters.
tension is in positive direction and mg is in negative. His force equation is wrong.
The force equation is correct. (relative to the acceleration given in the sketch)
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can make vids for engineering mathematics
+Louis Mwobobia I am working on many topics at the moment, but that is on the list to work on this year.
+Michel van Biezen thanx
woooooooooowwwww this is so so great
Thank you
thank you thank you thank youuuuuu
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Thank you!!
Super sir
great thank you so much
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thankyou sir
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thanks a lot
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Thank you.
Nice
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dam i had this problem on an exam... oh well
rip
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Amazinggggggg
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you rock!
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Thank you very much.. you help me out XD (Y) (Y) (Y)
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that was very helpful and i like your bowtie