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When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it. WHAT THEN IS THE FORMULA TO KNOW THE TIME WHEN IT PRACTICALLY STOPPED?
Good question Roderik! Yes the time period can be calculated but it requires calculations around drag force due to the air. Essentially air is assumed as a fluid in which the pendulum is swinging. In this case we assume vacuum in which the pendulum swings
@@TheScienceCube I know how the exponential dies and can figure out the time in an RLC series or parallel circuit through some formulas but I cannot seem to figure out what the analogies are in a 'MECHANICAL engineering' mentality. I have looked at the tables of the properties of air, and now trying to figure this simple phenomena to 'SIMPLY' come out with an approximate formula for the 'stop time function' with all the constraints of this 'simple pendulum system'.
Well it depends on the sign you agree on for +Y direction and -Y direction. The most commonly used convention is all forces or say vectors in downward direction will be taken as negative and upward as positive. We see tension is upward and therefore positive and gravity is downward therefore negative. So you can use the signs you have suggested but then you should be consistent in all your calculations.
@@TheScienceCube but the answer changes as both the equations have different meaning. also i have seen that most of the time people take the components of mg only not T for these type of question. for eg. when we derive equations for tension in string in vertical circle we take components of mg only.
@@sbti6051 The two equations are the same for small angle approximation. The equation you get is T - mg Cosθ = may (ay = acceleration in y direction) but if ay is considered near zero and Cosθ = 1, you get T = mg. The equation for XX would become -mgSinθ = max and you get ax = -gSinθ = -gx/l. So you see you get the same result. It is just that the reference frame or the cartesin coordinate has changed
The vector that is alongside the angle would carry Cos and the other Sin. I would however encourage you to go through basic trigonometry again to understand why this is so
Thanks for posting the video. Question, how do you determine what the maximum force is of the mass on the string with an initial velocity (as if someone pushed the mass not just let it drop)?
Hi! How are you? In brief - At highest points: Energy is gravitational potential (U = mgh). At lowest point: Energy is kinetic (K = 0.5 * mv^2). Energy Conversion: Swinging pendulum converts between kinetic and potential energy. Total Energy: Remains constant (E = K + U) in the absence of non-conservative forces like air resistance. I would also suggest you watch this video that would be helpful. th-cam.com/video/7hRlok5Vmog/w-d-xo.html
Then ax is an x component of centripetal acceleration??? Even then for small change in angle partial should move in constant velocity and not accelerating, since velocity difference between point a to point b for small change in angle should be neglible
Good question Victor! Normally we would have assumed no change in velocity if we were talking in terms of differential element like d theta but the change in angle here is measurable enough to change the velocity as well
When you say that forces in y direction are approximately equal to zero because of angle,can’t you say that they are equal to zero because the body isn’t moving up or down?
Good question Caronline. Well if you see the video again (particularly from 3:36) what you will see is that the movement does happen in YY direction as well, but,if the angel is small, the movement in YY direction is so small that there is very little acceleration and we assume it to be zero. This YY movement becomes higher if the angel increases. So you cannot say that there is no movement in the YY direction. I suggest you see the video again and listen carefully. You will get it then :). Write back if it is still not clear
Thank you for responding to me.I get it,it’s quite clear.I have an other question.If the pendulum was in a car and the car was moving forward,can we say that the alteration in y direction is 0,because tha car and the pendulum have the same alteteration?
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@@jhanolaer8286 Ok...if you know the value of T from the formula we derived that is 2 pi root l by g, you can find the force at any angle by taking sine of the angle and multiplying with T
Good question! Well, you can but you don't need to. Basically we are looking for some equations that can eventually help us find the time period T of the pendulum. What we see is that if you can write 2 equations - one in XX direction and the other in YY direction, then we can have our equations to find T. mg being in YY direction becomes part of the equation. Well, if for some reason, mg was not in YY direction, we would have resolved it to find the YY component and the XX component and used it in the equation
Thank you sir.... And one question out of this topic ...can we resolve mass in Motion in Horizontal circle :conical pendulum ?? If so why is it not resolved while finding Time period ...
@@kingzer3676 We do not resolve mg here also for the same reason as I gave earlier. I'd encourage you to make the free body diagram of a pendulum moving in a conical formation. You'll see that again the force mg is acting vertically only and therefore does not need resolution
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How can t=mgcos theta when the vertical component of mg is sin theta and horizontal is cos theta ? How u said t cos theta as vertical when sin theta is for vertical components? Sir please reply
@Ayesha if you resolve T into its component then the one that is along side angle theta will be the component that is vertical that is TCos theta and this should equal the other force in the opposite direction that is mg. Let me know if I answered your question.
it's simple, assume @ = theta, then it's sin ( 90 - @ ) = cos @ and cos ( 90 - @ ) = sin @. since you don't have the theta that you need you just work with the theta and minus it from 90.
@@TheScienceCube I didn't say you were not clear, but you haven't answered the Question Raised by "@Ayesha Safeer". he asked why are you representing vertical component as T Cos@ instead of T Sin@ as we normally do, the answer is -> is 90-@ changes Sin to Cos and cos to sin.
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Wareesha - When we talk about forces, there is nothing like upward or downward force. We have a convention around force acting in a certain direction aligned with more acceptable axis like X,Y and Z (both +ve and -ve directions). Coming to Tension as a force, you should think of Tension as a pulling force. So imagine yourself pulling a heavy box with a rope along the X axis to the right then Tension in the string is the pulling force that is pulling the box to the right. You could have never thought of the rope pushing the box! However if the rope is inclined at an angle to the ground, you can imagine that not all the force you are putting is going in pulling the box in X direction. Some of it is actually trying to pull it vertically and so goes wasted. Hence we can and need to resolve forces, including tension for working out problems
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@@mikaelafrias7786 oh yes! of course. That's what we derived. However, lower the angle, more accurate this equation would be, which also we discussed in the derivation
Great to know Raj! Please share this channel with 2 of your friends on whatsapp, that will help me grow this channel and enable me to make more videos for students like you :)
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Hi! Please also join my Telegram Channel t.me/TheScienceCube_Community for PDF lesson summaries and other physics solutions. You can always leave if you wish to, but trust me, you won't want to miss this! 😉
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@@TheScienceCube artificially intelligent ka calculators bana lo phir Sarey desh main supply karo muft main Sabhi tarah ki economy apkey naam Chalney lagegi
@@TheScienceCube Eh... I don't understand what you mean. Yes, it's different than what I'm "used to", but I was kinda aiming for an objective statement of unclarity. I mean, I think this video is awesome and think it would be miles better if it was spoken more clearly. That's all.
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thank you so much it helped a lot!!
@ÖZGEŞAHİN-j3x so glad to know :) Where are you from?
@@TheScienceCube Turkey
@ÖZGEŞAHİN-j3x 🎅
@@ÖZGEŞAHİN-j3x Great! Do check out my website that has useful resources. Its all free www.thesciencecube.com/
I have so many confusion about this small topic, now it is clear to me. Thank You sir.
Great to hear Mahananda! Are you preparing for a competitive exam?
@@TheScienceCube Yes for Jee Advanced.
@@mahananda_maiti Best wishes Mahananda! :)
Hi! i have just started adding some serious students to a WhatsApp group i have created. The max students the group can takes is 256. If that interests you, please join.
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Thank you so much, you're the only one I found who explained why to put ay = 0.
:) Where are you from Aymen?
@@TheScienceCube Algeria. why did you ask hh
@@aymenzidane7130 I have students from 112 countries. Always curious to know who is from where? Are you in school still or preparing for some exam?
@@TheScienceCube I am at university. Please, can I get your Facebook account so that I can ask if I need something?
@@aymenzidane7130 I am ot on Facebook. Feel free to ask here
thanks for clearing the concept
@@surajmishra3334 very happy to know! Are you preparing for an exam?
Bro.. Your mathmetical proof is very logical.. Thanks a lot bro
Thanks Manazir! Where are you from?
Thank you so much this was really clear xxx
@@Lily-tp9ui glad you liked it :)
When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it. WHAT THEN IS THE FORMULA TO KNOW THE TIME WHEN IT PRACTICALLY STOPPED?
Good question Roderik! Yes the time period can be calculated but it requires calculations around drag force due to the air. Essentially air is assumed as a fluid in which the pendulum is swinging. In this case we assume vacuum in which the pendulum swings
@@TheScienceCube I know how the exponential dies and can figure out the time in an RLC series or parallel circuit through some formulas but I cannot seem to figure out what the analogies are in a 'MECHANICAL engineering' mentality. I have looked at the tables of the properties of air, and now trying to figure this simple phenomena to 'SIMPLY' come out with an approximate formula for the 'stop time function' with all the constraints of this 'simple pendulum system'.
YOU ARE THE GREATEST
:) Thanks so much Luis! Where are you from?
So clear, thank you!
Where are you from Ishan?
@@TheScienceCube UK but I teach in China.
@@ishanr8697 that is so cool! You teach Physics?
Why do you write 'm · a_y = T · cosθ - mg'?
Can't it be 'm · a_y = mg - T · cosθ' in the case when ball is moving down?
Well it depends on the sign you agree on for +Y direction and -Y direction. The most commonly used convention is all forces or say vectors in downward direction will be taken as negative and upward as positive. We see tension is upward and therefore positive and gravity is downward therefore negative. So you can use the signs you have suggested but then you should be consistent in all your calculations.
@@TheScienceCube 👍
why you took Tcos theta = mg ?you could also take mgcos theta = T when we take components of mg.
which one is correct
Yes you are right. That could also have been done but the Cartesian coordinate would be tilted by theta, making the calculation more difficult
@@TheScienceCube but the answer changes as both the equations have different meaning. also i have seen that most of the time people take the components of mg only not T for these type of question. for eg. when we derive equations for tension in string in vertical circle we take components of mg only.
@@sbti6051 The two equations are the same for small angle approximation. The equation you get is T - mg Cosθ = may (ay = acceleration in y direction) but if ay is considered near zero and Cosθ = 1, you get T = mg. The equation for XX would become -mgSinθ = max and you get ax = -gSinθ = -gx/l. So you see you get the same result. It is just that the reference frame or the cartesin coordinate has changed
I struggle with how you know where and when and how to use sin and tan and cos?
How do you know that Tcosø is used and TsinØ?
The vector that is alongside the angle would carry Cos and the other Sin. I would however encourage you to go through basic trigonometry again to understand why this is so
But we use torque for angular shms how we r able to find time period by force analysis ???
Thanks for posting the video. Question, how do you determine what the maximum force is of the mass on the string with an initial velocity (as if someone pushed the mass not just let it drop)?
You mean instead of letting go the ball, it is given some push?
Sir can you tell about the energy in pendulum at a given instant
Hi! How are you? In brief - At highest points: Energy is gravitational potential (U = mgh).
At lowest point: Energy is kinetic (K = 0.5 * mv^2).
Energy Conversion: Swinging pendulum converts between kinetic and potential energy.
Total Energy: Remains constant (E = K + U) in the absence of non-conservative forces like air resistance. I would also suggest you watch this video that would be helpful. th-cam.com/video/7hRlok5Vmog/w-d-xo.html
Thanks for explaining this so well. My professor made this overly complicated.
:) Thanks! Where are you from?
Then ax is an x component of centripetal acceleration???
Even then for small change in angle partial should move in constant velocity and not accelerating, since velocity difference between point a to point b for small change in angle should be neglible
Good question Victor! Normally we would have assumed no change in velocity if we were talking in terms of differential element like d theta but the change in angle here is measurable enough to change the velocity as well
@@TheScienceCube thank you sir
@@victormaxwellpeters9771 Sure! Where are you from Victor?
@@TheScienceCube Sir, I am from mangalore which is in karnataka
@@victormaxwellpeters9771 Great! Please do share this channel with your friends :)
When you say that forces in y direction are approximately equal to zero because of angle,can’t you say that they are equal to zero because the body isn’t moving up or down?
Good question Caronline. Well if you see the video again (particularly from 3:36) what you will see is that the movement does happen in YY direction as well, but,if the angel is small, the movement in YY direction is so small that there is very little acceleration and we assume it to be zero. This YY movement becomes higher if the angel increases. So you cannot say that there is no movement in the YY direction. I suggest you see the video again and listen carefully. You will get it then :). Write back if it is still not clear
Thank you for responding to me.I get it,it’s quite clear.I have an other question.If the pendulum was in a car and the car was moving forward,can we say that the alteration in y direction is 0,because tha car and the pendulum have the same alteteration?
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Sir plse correct me
mgcos theta =T
Tcos theta = mg
Both can be done or not??
Plse answer asap
how to get the perpendicular force?
Which force are you referring to? T Cos theta?
@@TheScienceCube T sin theta..
@@jhanolaer8286 Ok...if you know the value of T from the formula we derived that is 2 pi root l by g, you can find the force at any angle by taking sine of the angle and multiplying with T
Why was not mg resolved into its components??
Good question! Well, you can but you don't need to. Basically we are looking for some equations that can eventually help us find the time period T of the pendulum. What we see is that if you can write 2 equations - one in XX direction and the other in YY direction, then we can have our equations to find T. mg being in YY direction becomes part of the equation. Well, if for some reason, mg was not in YY direction, we would have resolved it to find the YY component and the XX component and used it in the equation
Thank you sir....
And one question out of this topic ...can we resolve mass in Motion in Horizontal circle :conical pendulum ?? If so why is it not resolved while finding Time period ...
@@kingzer3676 We do not resolve mg here also for the same reason as I gave earlier. I'd encourage you to make the free body diagram of a pendulum moving in a conical formation. You'll see that again the force mg is acting vertically only and therefore does not need resolution
...also do check out this new video on "How to prepare for an exam", particularly the 11th tip th-cam.com/video/TIa3v4UaVRc/w-d-xo.html
Let me know there which tip you liked the most...I'm doing a survey :)
How can t=mgcos theta when the vertical component of mg is sin theta and horizontal is cos theta ? How u said t cos theta as vertical when sin theta is for vertical components? Sir please reply
@Ayesha if you resolve T into its component then the one that is along side angle theta will be the component that is vertical that is TCos theta and this should equal the other force in the opposite direction that is mg. Let me know if I answered your question.
@@TheScienceCube no, you haven't.
it's simple, assume @ = theta, then it's sin ( 90 - @ ) = cos @ and cos ( 90 - @ ) = sin @. since you don't have the theta that you need you just work with the theta and minus it from 90.
@@sky-xk5be Which part is not clear. Please specify
@@TheScienceCube I didn't say you were not clear, but you haven't answered the Question Raised by "@Ayesha Safeer". he asked why are you representing vertical component as T Cos@ instead of T Sin@ as we normally do, the answer is -> is 90-@ changes Sin to Cos and cos to sin.
Hi Sir, I Have Joined The WhatsApp Group. I Had Some Doubts For The Simple Pendulum, Could You Explain?
Sure tell me
I don't understand really.. Y channel like urs is so underrated.
Thank-you! In that case, please do share with your friends :)
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why does the tesion have a verical component? isn't it suppose to be the upward force
Wareesha - When we talk about forces, there is nothing like upward or downward force. We have a convention around force acting in a certain direction aligned with more acceptable axis like X,Y and Z (both +ve and -ve directions). Coming to Tension as a force, you should think of Tension as a pulling force. So imagine yourself pulling a heavy box with a rope along the X axis to the right then Tension in the string is the pulling force that is pulling the box to the right. You could have never thought of the rope pushing the box! However if the rope is inclined at an angle to the ground, you can imagine that not all the force you are putting is going in pulling the box in X direction. Some of it is actually trying to pull it vertically and so goes wasted. Hence we can and need to resolve forces, including tension for working out problems
Where does the w^2x come from???
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Hello, but do i still use the T equation if i have a given angle?
Hi! What is T equation?
@@TheScienceCube the period equation
@@mikaelafrias7786 oh yes! of course. That's what we derived. However, lower the angle, more accurate this equation would be, which also we discussed in the derivation
@@TheScienceCube But my experiment requires to test angles from 45 to 5 degrees so that's why.
@@mikaelafrias7786 this would give incorrect results for large angles like 45 degree
Dont we need to take torques like we use comp shm
Can you elaborate on your question? I am not sure I understood
thank you so much it helped me a lot
Great to know Raj! Please share this channel with 2 of your friends on whatsapp, that will help me grow this channel and enable me to make more videos for students like you :)
Raj, i have just started adding some serious students to a WhatsApp group i have created. If that interests you. Please join the link.
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thankyouuuuuuuuuuuuuuuu😭
So glad to know you liked it!! Are you appearing for some exam?
Hi! Please also join my Telegram Channel t.me/TheScienceCube_Community for PDF lesson summaries and other physics solutions. You can always leave if you wish to, but trust me, you won't want to miss this! 😉
thanks a lot sir
:) any particular topics you want me to make tutorials on? solved problems or more lessons?
Thank u sir...😄
Thanks Parvathy! Where are you from?
@@TheScienceCube India(Kerala) 😇
@@parvathykrishna3898 Great to know! Do share some videos with your friends and relatives if you feel it could help! Best wishes and good luck :)
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Thank you
Thank you 🙏💕
Cool!
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Very good 🙏🙏🙏🙏🙏
:) Where are you from Surendra?
Thanks sir
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If you solve it with torque the differential equation pops up immediately.
Thats right Aleph :)
why do we subtract mg from sincos theta and why do we not have a "-" sign in ay
Which equation are you talking about?
Thnx sir
U cant cancel m=-m
Only m is being cancelled here. You see, the minus sign is retained
Thanks brother
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Thanks
Hi! where are you from Flash?
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I’m here cause I’m adding weight to my tennis racket
:) What does that mean!!
So you have watch walter lewin video then tought us
Ok well done
Yes Walter Lewis is a great teacher! Are you preparing for some exam Mandeep?
@@TheScienceCube no
I am not preparing for any competition exams or else but i want to know practical application hence i watch his video
@@MandeepSingh-ge3fu Very cool! Appreciate your interest
@@MandeepSingh-ge3fu You may like to share this video with someone who might be preparing for an exam - th-cam.com/video/TIa3v4UaVRc/w-d-xo.html
Just take mgsin@ and mgcos@
It will be easy
@=theta
Where? Please explain
And I am sorry that I can't explain
But he writen it very clearly
th-cam.com/video/2Qfr_nspZo8/w-d-xo.htmlsi=Ei8Ufi7Cn29YkyA9
Watch that video
Tum mujhe funding do
Main tumko research doonga😃😃😃😃
:) what was that?
@@TheScienceCube artificially intelligent ka calculators bana lo phir Sarey desh main supply karo muft main Sabhi tarah ki economy apkey naam Chalney lagegi
Physics is interesting
Yes! Where do you study Aakanksha?
@@TheScienceCube I study in Loyola higher secondary school kunkuri in Chhattisgarh
@@aakankshaambastha8092 OK, are you preparing for some exam?
@@TheScienceCube no right now not preparing
@@aakankshaambastha8092 best wishes! And please share this channel with students who might find it useful 🙂
it's hard to understand this accent
Oh! Where are you from Arisoda?
@@TheScienceCube Europe, not english-native or anything, but still, maybe if you accentuate the different words, it's maybe more easily digestible
@@arisoda I am sure some people might be finding the accent different from what they are used to. What class are you in?
@@TheScienceCube Eh... I don't understand what you mean. Yes, it's different than what I'm "used to", but I was kinda aiming for an objective statement of unclarity. I mean, I think this video is awesome and think it would be miles better if it was spoken more clearly. That's all.
@@arisoda Ok got it! thanks for the feedback! Which country are you from?