profesor I just wanted to tell you thank you for these videos you have taught me more then the profesors at my university about physics and math (I wish I was joking but its the truth) I am almost 100% sure the only reason I will pass my midterm and finals about phyisics (callculus is a difrent story XD) its because of your videos without them I would have failed miserbly so honestly from the bottom of my heart and I say this for everyone who has a bad profesor thank you in 3 days my finals are coming but how much i know now and how much i knew back then is a lot to say the least thank you stay safe profesor Update: I got a 90 on my finals lesss goo
Yes, I did to keep the problem simple enough. Adding the force of the centripetal acceleration makes it into a very difficult problem beyond this level. (Good catch).
Lovely video, thank you. I'm actually confused by some of the comments and your answers relating to centripetal force. You say that you've ignored it for simplicity. But isn't the centripetal force just effectively the resultant force towards the centre of the circular motion? i.e. here, centripetal F = N - mg cos (theta). And then N - mg cos(theta) = mv(squared)/r The centripetal force isn't 'another force' that can be added or ignored, it is the net effect of the forces that have components towards the centre of the circle, no? A bit like a 'resultant force' isn't another force, but the net effect of other forces. Please correct me if wrong.
You are correct. But because of the centripetal force (the track pushing back on the block towards the center of the circular motion) causing a friction force that is constantly changing, we would need to integrate over the path taken in order to calculate the energy lost by the varying friction force. (That was left out for simplicity)
i am being confused when i integrate the work done against friction 0 degree to 180/360 degree. ( for a half or full loop) the result showing that the work done against friction is zero, but how can that be? my question is , what will be the result if i integrate like this video, 0 degree to 180 degree for a half loop?
Sir , normal force here depends not only on angle but also on velocity , due to centripetal force . And problem here is , I can't just add mv²/R to mgcosθ because "v" at that point depends upon how much potential energy is lost in form of heat which is unknown. Please give me the solution sir . Thank you.
You are right. Newton' s second law projected along the normal to the curve (the radius) says that mv2/R = N - mg cos (angle theta). This is a huge mistake. In my country (Romania) we don' t allow such mistakes, not even to the students.
@@victorburacu9960 yes , here it (mv²/R) has a greater significance. Although if we consider the radius of arc to be very large as compared to the dimensions of the block we can neglect mv²/R as done by sir in this video. But I want solution where radius of arc and dimensions of block are comparable.
@@aniket_Kumar001 There is no closed form solution. We can write the equations of motion along the tangent and along the radius, but we cannot integrate them. Velocity as function of angle theta can be derived only numerically.
@@victorburacu9960 *No dude! I have derived a relation between velocity and theta. and also found out the work done by friction as a function of theta for this case.😌*
Hi. I am wondering if this is correct.? For the Normal force you have said it is equal to mgcos(theta), nhowever since the block is in circular motion, shouldnt you also consider the Normal acceleration of the block. So the Sum of the forces in the Normal direction: N -mgcos(theta) = m(an) therefore. N = mgcos(theta) + m(an) I am stuck on a question like this and cannot work out how to solve it. Your method is solvable, but it doesnt consider the normal acceleration, which will affect the friction force and hence the work done by friction.
You are correct. The centripetal acceleration will add the additional component of the normal force as you indicated. That was ignored in this example (for simplicity). We should do another example with that component included.
The arc length of a section of a circle is calculated with: S = R x theta (where theta is in radians). That is why the circumference of a circle is: C = 2 pi R R x d(theta) is the radius multiplied by a very small section of an angle to give you a very small piece of the arc length of the circle.
Sir, I have some doubts. 1) Wouldn't the integration of cos(theta) be -sin(theta) instead of +sin(theta)? 2) In what situations do we have to take the final KE as 0? For eg. in this video, you considered it but in some of them, you just say that final KE is zero. 3) Is there any way to calculate work done to oppose the friction in a circulate path, (like in this video), if you've been given *mass=1kg, R=5m, Initial angle with tangent (90) and final angle with tangent (127 from initial or 37 from 0), FRICTIONAL COEFFICIENT NOT GIVEN. Options (just in case): a) 80 J b) 40 J c) 20 J d) 60 J Your videos have been of great help!
6 years later, but you confused the integral of cos with the derivative of cos. Integral of cos is sin, because the derivative of sin is cos. Derivative of cos is -sin.
is normal force always equal to mgcos theta in this example? because if an object moves in a circular trajectory there must be centripetal acceleration , which in this case comes out to be zero? is not it true that normal force has to be larger than mgcos(theta) in order for centripetal acceleration?
It all depends on which angle you are referencing. In circular motion there is centripetal force which is always perpendicular to the motion and pointed towards the center of motion. It may be easier then to express that force using vector notation and cylindrical or spherical coordinates.
The way you figured out the components of theta was different than your previous video of Conservation of Energy (9 of 11) sliding in a bowl. In that one the parallel component was mgcos(theta) and in this one it is mgsin(theta). Also the normal force was mgsin(theta) on the other video and this one it is mgcos(theta). Am I missing something? Is there a difference in application between sliding in a bowl and sliding down a quarter circle with the same coefficient of friction? Thank you, love your videos.
How can you neglect the centripetal acceleration.It will change the normal to mgcos(theta)+mv²/R and then will get problem in integration as v is a function and not a constant.
sir u have taken d theta in the anticlockwise direction! but at the same time u have put the limits 90degree to 0 degree in (sin theta) sir the radius 10 is rotating in anticlockwise direction so the limits must be 0 to 90 degree to (sin theta) ..m i right ??if not pls xxplain me how????
+Aman Chaure The only thing that will happen when you reverse the limits on an integration is a change of the sign. If you only care about the magnitude of the result, the order of the limits don't matter.
@@MichelvanBiezen This actually slightly changes the result. I noticed it when I used the DeltaME = Wext formula I get 1.1 gr instead of 0.9 gr like in the video. But once the signs are correct I get 0.9 gr. In my university we do these problems deriving from Wnet = Wconservative + Wexternal = DeltaKE and then Wconservative = -DeltaU, so I can't use your formula even if it comes quite handy.
I know this is already a 2 year old comment, but I am still going to answer it haha. You can actually multiply ds with dtheta/dtheta. And with rearrangement, you'll end up with ds/dtheta x dtheta. Notice ds/dtheta is equal to R. so with substitution, it will be equal to R x dtheta.
Sir if we want to calculate it in a parabola can we take the small elemental displacement to be rdtheta where r is distance of block from the point where we are calculating
sir, the body is undergoing circular motion in that particular segment. won't a centripetal force act on it? also will the motion then be uniform circular motion or non uniform? what would be the direction of friction in such a case?
Ishita Kukreti You are correct, there is a centripetal force. I ignored the centripetal force to simplify the problem. If you want to include the centripetal force, you must add mv^2/R to the mgcos (theta) The direction of friction will be the same.
Solution might be wrong as circular motion of the particle is not considered,, due to circular motion Normal is different than mg cos(theta) ,, please give explanation.
@@MichelvanBiezen but in condition of circular motion we can not take normal reaction as mgcos(theta), normal reaction is greater in that case due to centepetal acceleration , thats why every on every curvillinear motion with friction particle move slowly...
If particle gain a velocity then normal reaction increases and then question become very difficult to solve.. Thats what I understand ,, if I am wrong please let me know..
I'm just slightly confused. You have so many videos (which I seriously like) but they're also scattered into different playlists. For instance, this video should be related to conservation of energy yet it's has its own playlist. But what I'd like to say is, I'd really like if you could recommend me what videos I should watch for preperation of college level mechanical engineering and dynamics. I find more related topics in your phyiscs and mechanics videos than on your mechanical engineering playlist. Maybe because you're already a step ahead in mechanics than what we read in my country. Hope you read this.
If you want to prepare for mechanical engineering, you should watch the physics playlists starting with: PHYSICS 0 GENERAL INTRODUCTION and ending with PHYSICS 19 MECHANICAL WAVES and the playlists on fluid statics and dynamics. (that is about 30 playlists in all). You should also have knowledge of calculus both derivatives and integrals.
Sorry, I'm a little confused. At minute 4, where did the cos(phi) come from? How did you know to use cos and not sin or tan, and what angle is phi in regards to the system?
Dear Prof. Thisproblem is not solved properly. mgcos(theta) is not equal to the normal force. you cannot forget the centripetal force. pleasecheck it. you videos are used by a lot of students. best regards
You brought up a good point. Normally we ignore the centripetal force in order to keep the problem in a simpler format, but we could also add the effect of the centripetal force.
What is ur telegram group I will send u please sir only one question questions is on same pattern but block is projected horizontally on quarter circle with some initial velocity at end of quarter circle find speed of block at end of quarter circle ? Given that coefficient of friction 0.2 mass of block is 0.5 kg initial velocity 2m/s coefficient of restitution e^-π/10 =0.73 rdius of circular path 2 m sir please solve it
profesor I just wanted to tell you thank you for these videos you have taught me more then the profesors at my university about physics and math (I wish I was joking but its the truth) I am almost 100% sure the only reason I will pass my midterm and finals about phyisics (callculus is a difrent story XD) its because of your videos without them I would have failed miserbly so honestly from the bottom of my heart and I say this for everyone who has a bad profesor thank you in 3 days my finals are coming but how much i know now and how much i knew back then is a lot to say the least thank you stay safe profesor
Update: I got a 90 on my finals lesss goo
Thank you for sharing your experience with us and letting us know that our videos have made a difference. We wish you all the best on your finals.
@@MichelvanBiezen thank you professor best of luck 🤞
Aren't you neglecting the contribution to the normal force due to centripetal acceleration,
mv^2/R
Yes, I did to keep the problem simple enough. Adding the force of the centripetal acceleration makes it into a very difficult problem beyond this level. (Good catch).
Lovely video, thank you. I'm actually confused by some of the comments and your answers relating to centripetal force. You say that you've ignored it for simplicity. But isn't the centripetal force just effectively the resultant force towards the centre of the circular motion? i.e. here, centripetal F = N - mg cos (theta). And then N - mg cos(theta) = mv(squared)/r
The centripetal force isn't 'another force' that can be added or ignored, it is the net effect of the forces that have components towards the centre of the circle, no? A bit like a 'resultant force' isn't another force, but the net effect of other forces. Please correct me if wrong.
You are correct. But because of the centripetal force (the track pushing back on the block towards the center of the circular motion) causing a friction force that is constantly changing, we would need to integrate over the path taken in order to calculate the energy lost by the varying friction force. (That was left out for simplicity)
@@MichelvanBiezen Ah, understood. Thank you for clarifying.
@@MichelvanBiezen thank you for explaining
i am being confused when i integrate the work done against friction 0 degree to 180/360 degree. ( for a half or full loop)
the result showing that the work done against friction is zero, but how can that be?
my question is , what will be the result if i integrate like this video, 0 degree to 180 degree for a half loop?
Sir , normal force here depends not only on angle but also on velocity , due to centripetal force . And problem here is , I can't just add mv²/R to mgcosθ because "v" at that point depends upon how much potential energy is lost in form of heat which is unknown.
Please give me the solution sir .
Thank you.
You are right. Newton' s second law projected along the normal to the curve (the radius) says that mv2/R = N - mg cos (angle theta). This is a huge mistake. In my country (Romania) we don' t allow such mistakes, not even to the students.
@@victorburacu9960 yes , here it (mv²/R) has a greater significance. Although if we consider the radius of arc to be very large as compared to the dimensions of the block we can neglect mv²/R as done by sir in this video.
But I want solution where radius of arc and dimensions of block are comparable.
@@aniket_Kumar001 There is no closed form solution. We can write the equations of motion along the tangent and along the radius, but we cannot integrate them. Velocity as function of angle theta can be derived only numerically.
@@victorburacu9960 *No dude! I have derived a relation between velocity and theta. and also found out the work done by friction as a function of theta for this case.😌*
Excellent tutorial. Thanks !
Thanks. Glad you liked it. 🙂
Hi. I am wondering if this is correct.?
For the Normal force you have said it is equal to mgcos(theta), nhowever since the block is in circular motion, shouldnt you also consider the Normal acceleration of the block.
So the Sum of the forces in the Normal direction:
N -mgcos(theta) = m(an) therefore. N = mgcos(theta) + m(an)
I am stuck on a question like this and cannot work out how to solve it. Your method is solvable, but it doesnt consider the normal acceleration, which will affect the friction force and hence the work done by friction.
You are correct. The centripetal acceleration will add the additional component of the normal force as you indicated. That was ignored in this example (for simplicity). We should do another example with that component included.
Please proff michel make an example i have a project and im confused
I didnt understand that why we say Rdtheta ?Does it have any relationship with circular motion like circumference formula?
The arc length of a section of a circle is calculated with: S = R x theta (where theta is in radians). That is why the circumference of a circle is: C = 2 pi R R x d(theta) is the radius multiplied by a very small section of an angle to give you a very small piece of the arc length of the circle.
@@MichelvanBiezen thanks for answering my question I feel lucky about that.
Sir, I have some doubts.
1) Wouldn't the integration of cos(theta) be -sin(theta) instead of +sin(theta)?
2) In what situations do we have to take the final KE as 0? For eg. in this video, you considered it but in some of them, you just say that final KE is zero.
3) Is there any way to calculate work done to oppose the friction in a circulate path, (like in this video), if you've been given *mass=1kg, R=5m, Initial angle with tangent (90) and final angle with tangent (127 from initial or 37 from 0), FRICTIONAL COEFFICIENT NOT GIVEN.
Options (just in case):
a) 80 J
b) 40 J
c) 20 J
d) 60 J
Your videos have been of great help!
6 years later, but you confused the integral of cos with the derivative of cos. Integral of cos is sin, because the derivative of sin is cos. Derivative of cos is -sin.
Great problem....the calc is so good to see! :)
Glad you liked it. 🙂
@@MichelvanBiezen Thanks.....definitely!
is normal force always equal to mgcos theta in this example? because if an object moves in a circular trajectory there must be centripetal acceleration , which in this case comes out to be zero? is not it true that normal force has to be larger than mgcos(theta) in order for centripetal acceleration?
It all depends on which angle you are referencing. In circular motion there is centripetal force which is always perpendicular to the motion and pointed towards the center of motion. It may be easier then to express that force using vector notation and cylindrical or spherical coordinates.
The way you figured out the components of theta was different than your previous video of Conservation of Energy (9 of 11) sliding in a bowl. In that one the parallel component was mgcos(theta) and in this one it is mgsin(theta). Also the normal force was mgsin(theta) on the other video and this one it is mgcos(theta). Am I missing something? Is there a difference in application between sliding in a bowl and sliding down a quarter circle with the same coefficient of friction? Thank you, love your videos.
They are both correct. Note the angle reference.
sorry sir i dont understand about cos theta qual to the angle between angle of motion and angle of friction
How can you neglect the centripetal acceleration.It will change the normal to mgcos(theta)+mv²/R and then will get problem in integration as v is a function and not a constant.
Yes, that would be the more proper way of calculating this. (This is a good approximation)
It is missing the centripetal force and the friction force depends on the velocity of the block...
Yes, that part was ignored. We have examples where it is included.
how ds equal to Rdtheta ?
3.30
The arc length of a section of a circle = (r) (theta) with theta in radians. For a full circle, the circumferance of a circle is (r) (2) (pi)
sir u have taken d theta in the anticlockwise direction!
but at the same time u have put the limits 90degree to 0 degree in (sin theta)
sir the radius 10 is rotating in anticlockwise direction so the limits must be 0 to 90 degree to
(sin theta) ..m i right ??if not pls xxplain me how????
+Aman Chaure
The only thing that will happen when you reverse the limits on an integration is a change of the sign.
If you only care about the magnitude of the result, the order of the limits don't matter.
@@MichelvanBiezen This actually slightly changes the result. I noticed it when I used the DeltaME = Wext formula I get 1.1 gr instead of 0.9 gr like in the video. But once the signs are correct I get 0.9 gr.
In my university we do these problems deriving from Wnet = Wconservative + Wexternal = DeltaKE and then Wconservative = -DeltaU, so I can't use your formula even if it comes quite handy.
Why is ds equal to R x dtheta?
I know this is already a 2 year old comment, but I am still going to answer it haha.
You can actually multiply ds with dtheta/dtheta. And with rearrangement, you'll end up with ds/dtheta x dtheta. Notice ds/dtheta is equal to R. so with substitution, it will be equal to R x dtheta.
Sir if we want to calculate it in a parabola can we take the small elemental displacement to be rdtheta where r is distance of block from the point where we are calculating
sir,
the body is undergoing circular motion in that particular segment. won't a centripetal force act on it?
also will the motion then be uniform circular motion or non uniform?
what would be the direction of friction in such a case?
Ishita Kukreti
You are correct, there is a centripetal force.
I ignored the centripetal force to simplify the problem.
If you want to include the centripetal force, you must add mv^2/R to the mgcos (theta)
The direction of friction will be the same.
Michel van Biezen
Thank you sir
Just one last question.How are we supposed to integrate the v^2/R term wrt θ ?
because we cant multiply theta, can we?
Ishita Kukreti
You need to convert
v = w * r= d(theta) /dt * r
Hello please were you able to do a video where the Fc was added? i cant seem to figure out what you mean. Please help
@@MichelvanBiezen sir,plaese mention this codition in the start of video.....I confused....
Solution might be wrong as circular motion of the particle is not considered,, due to circular motion Normal is different than mg cos(theta) ,, please give explanation.
The circular motion is accounted for by the integral of the work done.
@@MichelvanBiezen but in condition of circular motion we can not take normal reaction as mgcos(theta), normal reaction is greater in that case due to centepetal acceleration , thats why every on every curvillinear motion with friction particle move slowly...
If particle gain a velocity then normal reaction increases and then question become very difficult to solve.. Thats what I understand ,, if I am wrong please let me know..
I'm just slightly confused. You have so many videos (which I seriously like) but they're also scattered into different playlists. For instance, this video should be related to conservation of energy yet it's has its own playlist. But what I'd like to say is, I'd really like if you could recommend me what videos I should watch for preperation of college level mechanical engineering and dynamics. I find more related topics in your phyiscs and mechanics videos than on your mechanical engineering playlist. Maybe because you're already a step ahead in mechanics than what we read in my country. Hope you read this.
If you want to prepare for mechanical engineering, you should watch the physics playlists starting with: PHYSICS 0 GENERAL INTRODUCTION and ending with PHYSICS 19 MECHANICAL WAVES and the playlists on fluid statics and dynamics. (that is about 30 playlists in all). You should also have knowledge of calculus both derivatives and integrals.
Seriously, thank you. All of you guys better watch all his videos as of show of gratitude. He's a master. The true sun!
Does your videos work for a student who studies for both mechanics and dynamics? Sir, what topics on dynamics do you cover?
Sir I'm just wondering, why didn't you mechanical engineering videos? Do you cover everything on those videos?
Thank you prof.
You are welcome.
why is the normal force, mgcostheta? shouldnt it be mgsintheta? PLEASE REPLY SIR
mgcos(theta) is perpendicular to the surface, mgsin(theta) is parallel to the surface
@@MichelvanBiezen please address my concern
@@MichelvanBiezen sir
you are a legend!
Far from it, but we are glad you are enjoying our videos. 🙂
THANK YOU FOR THAT WOTRHY CONTENT REALLY HELPFULL...
BUT KINDLY TELL ME WHY YOU DON'T CONSIDER CENTRIFUGAL FORCE... I AM STUCK IN IT
We will have to make a video on how to include the centripetal force. It is much more complicated.
thank u sir .....
Sorry, I'm a little confused. At minute 4, where did the cos(phi) come from? How did you know to use cos and not sin or tan, and what angle is phi in regards to the system?
Watch a little further, it is supposed to be theta.
Sir I have questions of this type please solve my question
Which question are you referring to?
Dear Prof. Thisproblem is not solved properly. mgcos(theta) is not equal to the normal force. you cannot forget the centripetal force. pleasecheck it. you videos are used by a lot of students. best regards
You brought up a good point. Normally we ignore the centripetal force in order to keep the problem in a simpler format, but we could also add the effect of the centripetal force.
Sir please help me what is ur Gmail id
We respond to questions here when time permits.
What is ur telegram group I will send u please sir only one question questions is on same pattern but block is projected horizontally on quarter circle with some initial velocity at end of quarter circle find speed of block at end of quarter circle ? Given that coefficient of friction 0.2 mass of block is 0.5 kg initial velocity 2m/s coefficient of restitution e^-π/10 =0.73 rdius of circular path 2 m sir please solve it
Why is ds equal to R x dtheta?
The best way to visualize that is to make dtheta equal to 2 x pi (360 degrees), what does ds become?