Physics 13.1 Moment of Inertia Application (10 of 11) Acceleration=? When Pulley Has Mass
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- čas přidán 17. 03. 2016
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In this video I will find the acceleration, a=?, of an object hanging from a atwood machine.
Next video in this series can be seen at:
• Physics 13.1 Moment o...
Bro this guy is a legend. Makes it so clear and easy compared to my lecturer...
I know this was made 7 years ago but damn thank you so much.
Here i am pulling my hair out over this then he does it so easily.
Man this question is easy
@@srisai2744personally I was pulling out hair on a test about 10 minutes ago doing this so I would say we have different opinions on that
great video, totally missed a question like this on my last exam.
You'll get it next time. 🙂
In the case of ideal pulleys, we considered tension at either side of the pulley to be same. Thats what i was getting wrong, it was quite confusing that the tension seemed to change but then i saw this video. Thanks for the help
In the case of the "ideal" pulley where the disk of the pulley doesn't have mass, the tension on both sides is the same. But in a real pulley we have to take into consideration the moment of inertia of the pulley and the tensions on both sides are not equal.
once again.... if you look at at the result for problem 5 of 11.. and look at m1 in this problem... if you let m1 = 0 .. you end up with the same result as in problem 5 .. Amazing... these problems are so self consistent.. thanks michel!! :
Again, thank you so much for dragging my household through AP Physics.
Happy to help! 🙂
Thanks for the help! I have been struggling with tension for a while now.
Happy to help!
What will happen if the string is attached to one end of the pulley and the mass block is released.... Will a variable touque act on the pulley?
Excellent Sir. Regards
u da real mvp
Omgg thank you... This is what i was looking for... I am preparing for JEE ADV exam and i had difficulty with pulleys with mass but now i got it
You came to the right channel to help you prepare for the JEE ADV. We have thousands of physics videos as well as math vicdeos that will help you prepare.
bro did you noticed the catch in the formula of acceleration?
here the formula is a=(M₂-M₁) g/ (M₁+M₂+ *½M* ), where did this *½M* came from? This ½M can be written as (I/R²) which may be seen as Rotational Mass (we know I=MR² so M=I/R² , let's call this M as rotational mass ) .
So overall formula comes to be , a= Net Translational FORCE / (Net Translational Mass of blocks + Rotational Mass of Pulley). = M₂-M₁/(M₁+M₂+I/R²) = M₂-M₁/M₁+M₂+½M.
Great work love from INDIA i am preparing for NEET thnx sirr
Welcome to the channel. Alll the best on your studies and your NEET exam.
Why are the tensions different if they are connected? Shouldn't they be the same? Thank you for your videos by the way, they are very helpful.
Not if the pulley has mass and therefore a moment of inertia.
This is amazing. Question: how do you incorporate friction from the pulley (given in torque) in this method? Tried looking for a video of yours that has this but no luck.
We have some videos like that in this playlist: MECHANICAL ENGINEERING 11 FRICTION starting with Mechanical Engineering: Ch 11: Friction (31 of 47) (Flat) Belt Friction: Deriving the Equation Is that what you are looking for?
@@MichelvanBiezen thanks! I'll check them out and let you know. The question is basically the same as in this video, but the pulley has friction to it
@@MichelvanBiezen Hi again...I'm afraid not actually, checked them out but they are more advanced material, I'm still just in physics 1...I really like how you manipulated the equation and got the answer without needing to calculate the moment of inertia..I'll work on understanding the other method for solving it...thanks so much for your videos, sincerely
Yes, I think you are correct that this is a different type of friction. With pulleys it depends on how they express the friction. One way is to state: It takes x Nm of torque to overcome the friction of the pulley and then you have to subtract that from the torque that aids the acceleration.
Actually, static friction is already part of this problem. It is not the tension in the ropes that accelerates the pulley, it is the static friction between the rope and the pulley causing torque on the pulley. The rope is NOT attached to the pulley (but laying on top of it) so T1 and T2 are just the tensions within the rope and not the force between rope and pulley. The static friction causes the rope to have grip or traction on the pulley, taking the pulley along with it. Then why do the calculations give the correct answer without taking friction into account? Actually, the equation of motion in the video is not the equation of motion for just the pulley. It is the equation of motion for the system consisting of pulley + piece of rope in contact with the pulley. In that case you only need to worry about the external forces working on the system and you dont need to consider the forces between the different parts of your system, like the friction between rope and pulley and vice versa. The only external forces on the pulley + rope that cause a torque are the tension T1 and T2. Therefore the calculations are correct, but it is a little more subtile when you look closer at the problem.
❤Thank you sir, for the great explanation
You are welcome. Glad it ws helpful. 🙂
Hello, great video! I just had a question. I learned that by convention, torque is positive if the disk is spinning counterclockwise. Wouldn't we have to take the negative into consideration? Also, are you using the magnitude of of T1 and T2?
It is better not use the "convention" of positive or negative torque. (We use that in other situations, but not here). Here we only care if the torque aids the acceleration or opposed the acceleration and the net torque will cause the acceleration we are trying to calculate.
such a nice and cool video! Thank you!
Glad you liked it!
🔥🔥🔥🔥🔥🔥🔥🔥amazing ....and mind blowing
Thank you. Glad the videos are helpful. 🙂🙂
I'm not completely clear why the tension isn't the same along the length of the rope aka why T1 is not T2. Is there a video done by you that I can watch that addresses this? Thank you!
If the pulley has mass (and therefore moment of inertia) it requires tension to provide the torque to give it angular acceleration and thus T2 must be greater than T1. If the pulley does not have mass then the tension will be the same on both sides.
Not good
If the tension is uniform through the rope, there would be no rotational motion on the pulley, as the only torques are created by the tensions.
If the whole system being accelerated upwards with acceleration "a" then should we use "g+a" instead of "g"? (I mean, if the system is in the elevator) thanks for your reply in advance
We have a video exactly like that. Take a look.
I couldn't find that video. Can you please write the link?
I found out that you can get the same answer by setting acceleration equal to the mg of the right mass minus mg of the second mass, and dividing that quantity by the sum of masses of each block and half of the mass of the pully. Is that appropriate to do? Is it cheating?
Its cheating youre going to be arrested
really well made!
Thank you. Glad you liked it.
amazing video, i learned so much
Glad you found our videos 🙂
Thank you so much!!!
You are welcome.
very clear explainatuin helps alot
Glad it helped!
ur the best michel
Thank you. Glad you like the videos. 🙂
Thank you so much !!❤
You are welcome. Glad you found our videos. 🙂
How do you do it with two different radius, one for each mass.
Then the torque on one side will be (m1g - m1a) R1 and the torque on the other side (m2g + m2a) R2 . And you will have two moments of inertia so the right side of the equation will be: (I1)(alpha1) + (I2)(alpha2)
As there are masses attached to the pulley shoudnt the moment of inertia change?
No, the attached masses cause torques to act on the pulley, but they do not change the moment of inertia of the pulley. Moment of inertial only depends on the shape and the mass distribution.
@@MichelvanBiezen oh thankzz :DD.
T1= ma+mg , But why is that when you substitute it to the equation, you changed it to ma-mg?
No, it wasn't changed to ma-mg.
I was wondering the same thing, you didn't start doing this until example 9
Guys, basic algebra. Yes, T1=m1a +m1g. But when you subtract T1 in the second equation, both + terms become negative. -(ma + mg) = - ma- mg.
T2 -- T1
( m2g - m2a) -- ( m1g + m1a)
m2g - m2a -- m1g -- m1a
Understand?
Thank you sooo much sirrr
You are welcome. Glad you found our videos. 🙂
if the question only ask for tension in both string only, do we have to calculate acceleration first as well ?
Yes. You will not be able to determine the tension without calculating the acceleration.
alright, thank you!
how do you find the moment of inertia of the pulley ?
Assuming the pulley is a solid disk, I = (1/2) MR^2
@@MichelvanBiezen but what if we were told to calculate the moment of inertia of the pulley and then compare it to that of the solid disk (1/2MR^2)
can you please help, i was given the same question but there is no torque?
i tried solving it but i dont get the correct answer
Does the pulley have mass (and moment of inertia) in your question? If not use the techniques shown in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
no, okay thank you
okay i just saw the example, in my question they are saying the accelerations are different and tensions are the same
The magnitude of the accelerations must be the same if the masses are connected to the same string. (directions are different)
+Michel van Biezen
their tensions are the same but the diameters of the pulleys where they are connected are different
i watched this video before my midterm and my midterm had an exact problem like this haha
Glad you watched the video! 🙂
Doesn't the gravity cause the Torque too? I'm so confused. I thought that when we do use the Torque Equation T=F x R, we would say (mg-Tension)*R
It depends on which direction the object is accelerating. For the block accelerating upward, T = mg + ma for the block accelerating downward T = mg - ma
@@MichelvanBiezen my question is about different. You are saying that the tensions supplies them torque, but I’m asking: doesn’t the gravity and tension isn’t the string supply the torque
@@MichelvanBiezen you drew a tension force down on both strings? Both isn’t the tension going up?
relative to the pulley, the tension acts downward for both strings.
The equations as presented in the video are correct.
How do we get the radius if the acceleration was given?
R gets cancelled out from both sides
@MichelvanBiezen
Normally you would work out the problem in exactly the same way, but once you have the equation, you solve for R. In this case you can notice that R cancels out which means that the radius can be of any size.
Can acceleration be a negative value?
If you express it as a vector, yes. If you express it as a magnitude, no.
Michel van Biezen tq sir
@@MichelvanBiezen I'm trying to get a negative acceleration just to convince myself that setting any sign convention you want will yield the same result. For e.g. I'm using sign convention up as +ve for acceleration and forces, and counter-clockwise as +ve (Opposite to how the system would behave in your diagram) however I can't reach a negative answer.
Is the right handside of equation τ = Iα the difference in torque values (Can it be stated as a vector summation)? I've tried (with my sign convention CCW as +ve):
-T2+T1 = 1/2*m(pulley)*a (cancelled r's)
(right side of the equation resisting pulley motion in +ve direction) but it doesn't yield a negative answer.
Thank you for these videos!
It's ok answered my own question.. You have to define coordinate system for each masses separately
www.youphysics.education/rigid-body/rotational-motion/rotational-motion-problems/rotational-motion-problem-2/
czcams.com/video/FlHKTvUjD6g/video.html
Nice
Thanks
why is T1 not equal to T2?
If they were equal the pulley wouldn’t rotate as the net torque would be 0
@@JohnDoe-hf1zw i understand now, thanks
T1 is not equalt to T2, because the pulley has mass (therefor it has a moment of inertia) and only the difference in tension can cause it to accelerate.
@@MichelvanBiezen yea makes sense, thanks. phys 1 was light today thx for the help