The friction force is static friction. You cannot use the f=uN formula until the break away moment. Having said so, you will get the same result without using f=uN formula.
sir,your effort for the video is mind blowing i am a big fan of your channel and if you dont mind i am in love with your purple shirt and the black bow sir thank you so much for this videos
He should do an example of two cylinders, and leave the R dependency in there so we can see how the acceleration is affected by the radius of the cylinder.
hmm.. after knowing this, I can asume that the acceleration can be written as a = gsin theta / (k+1) whereas k = the coefficient from I. Thank you very much
Based on final formula, if we increase friction(mu) , the acceleration will not change. Seems weird for me, i expect if we increase mu , acceleration will decrease.
@@MichelvanBiezen Thank you dear Professor for reply. I greatly appreciate this. Assuming the theta and mass is constant, in the formula on right top of the board(beneath the f=ma), when i decrease the "mu" on the left side of equation, the "a" on the right side should increase.
Mr. Michel, I defined clockwise to be a positive torque direction; on this basis I made force of static friction to be positive and the vertical component of the Force of gravity to be negative, while I seem to be having sign discrepancy in my problem. What should I do?
We often get hung up on negative signs. This can be remedied by simply caluculating the magnitude (which is always positive). The sign becomes important when we work with vector quantities.
What if the ball was filled with some volume of fluid, say water? would that change the centre of mass of the ball and cause a different acceleration down the ramp?
Just a small question, lets say we were given the radius and mass of the ball and there is some kinetic friction force present on the table, then, what is the acceleration of the same round object?
The direction of the friction force is in the opposite direction of motion of the object without friction. (Note that if there was no friction, the wheel would slide and not rotate). To make the wheel rotate in the correct direction, the friction must be directed upwards.
Sir, I think friction force should act along the direction of motion not opposite because the point of contact is going backward due clockwise angular velocity
The direction in the video is correct. Think about it this way: If there was no friction the wheel would slide and not rotate. With friction, what direction does the friction need to be in order for the wheel to rotate in a clockwise direction?
Sir,then it means if wheel was made to move up the inclined plane then friction would assist the torque and its motion would be along the direction of motion!! Am i right ??
Hello Professor Beizen, Will Fnet equation change if the ball is rolling up? If the ball is rolling up, would the friction be directed downwards? Will both mgsin(theta) and friction point downwards? Will the equation become mgsin(theta) + F = ma if ball is rolling up?
With a box, all of the potential energy is converted to translational kinetic energy and heat (due to friction). With a hollow ball (or any ball), the potential energy is converted to both translational AND rotational kinetic energy as well as heat (due to friction).
Sir. Could you please answer me? What happens if frictional force(kmgCosA) equals to mgSinA? Net force will be zero but there will be still net torque due to frictional force. What will happen?
This is just an approximation. If you want to learn more about rolling friction, take a look at the videos in this playlist: MECHANICAL ENGINEERING 11 FRICTION
@@MichelvanBiezen Thank you for your reply. But i did not understand it. What are we approximating? Could you please give some more explanation? Meanwhile I have watched all your videos related to rolling motion. I am looking forward for your answer. Thank you so much.
Just like we ignore wind resistance in kinematics, we ignore rolling friction and just approximate it with the coefficient of friction, which is not the whole friction effect. (see the mechanical engineering videos on rolling friction).
Take a look at: PHYSICS 11 ROTATIONAL MOTION ruclips.net/p/PLX2gX-ftPVXXp-ceOBqew_PiJ-nQnuwOl or PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 ruclips.net/p/PLX2gX-ftPVXUV3FuKaomSwPLowv4qoOth
How is F=ma valid here when it doesn't account for the moment of inertia? You could have two spheres with the same radius and mass but different moment of inertia but they will both have different linear accelerations even though according to F=ma they should have the same acceleration. If F=ma is still valid as you claim then why not just use a=F/m to find the acceleration? Clearly you can't which suggests to me that it's not valid here. I'm just more confused now!
I agree there is a certain amount of "mystery" regarding this problem. But the problem is correct as shown in the video, and F=ma still works here as well. But in this case the force is doing 2 things: 1) it accelerates the object down the incline AND 2) it causes to object to rotate, both which require energy, and thus work, and thus the force is used to do both.
@@MichelvanBiezen thanks for your response. I started thinking about this problem recently when asking myself if I could prove that it doesn't matter if a force acts through an objects centre of mass or not. I figured that if I found the net force acting on the ball (the component of gravity pulling down the incline minus the force of friction pushing up the incline), multiplied by the distance (the hypotenuse of the triangle) and added it to the torque times the angular distance the ball rolls through as it goes down the slope then it should be equal to mgh which it is, and this supports what you said about gravity doing all of the work - both linear and rotationally speaking. Also the force of friction pushing up the slope multiplied by the distance also gives the amount of rotational kinetic energy which is what I predicted. So I seemed to show that the work done on an object by a force (at least in this case) does not matter if it's through its centre of mass or not because friction isn't and it still works. But then I thought that if that is true then why can't I use F(net)=ma to find the acceleration? Does the work done not always result in a given speed (kinetic energy) when both linear and rotational acceleration are occuring? It appears not. Two spheres that are identical except for their moment of inertia will have the same force acting through the same distance (and hence the same work is done on each) but will have different kinetic energy at the bottom (translational, not rotational). So perhaps you can see why I'm getting confused over this seemingly simple problem! Any input you could add to help clarify anything would be greatly appreciated. Also one extra thing. Since it's static friction that is causing the torque, how do we know that it's at its maximum value? Shouldn't the equals sign be changed to a "less than or equal to" sign? Thanks again.
@@MichelvanBiezen I would also like to add one more thing. In the example of the two identical spheres (except for their moment of inertia) where they both have the same net force acting in them, the one with the larger MoI will take longer to reach the bottom because it is not accelerating as quickly. But in terms of momentum and impulse, the same net force for a longer period of time should result in a larger change in momentum but clearly that's not true here as it is going slower when it reaches the bottom due to it having a lower acceleration through the same distance. This doesn't make any sense to me either. Any ideas?
@@MichelvanBiezen after giving this some thought I have realised that the force of static friction is larger for objects with a larger coefficient for their moment of inertia. My mistake was to think of the force of static friction as being fixed for all objects which it isn't. The static friction force is equal to mgSin(@)K/(K+1) where "K" is the coefficient for the moment of inertia such as 1/2 for a solid disk. Using that with Newtons second law F=ma you find that the acceleration for any object rolling down an incline is a = gSin(@)/(K+1). Using your own two examples yields the same answer that you got only this equation works for all rolling objects.
when I was a kid.. my parents used to accuse me of having an INERTIA Problem.... because I was "apparently" too lazy... now that I know about Moments of Inertia.. I would tell my parents that my problem was with my MOMENT OF INERTIA.. meaning that, at the MOMENT.. I have NO INERTIA... (lol..) Go Figure...
I completely understood this. Thank you so much!
You sir are a legend, an absolutely bloody legend.
Greatly exaggerated, but we appreciate the comment. :)
This is so fascinating! I hope to become just as proficient as you!
No slow learner with best explanations from this lectures now we are all genius (THAnKS GOD)😊✌
Glad to hear that
The friction force is static friction. You cannot use the f=uN formula until the break away moment. Having said so, you will get the same result without using f=uN formula.
I like your way of teaching, you are the best
You continue to help me. Thank you so much!
We are glad you continue to find our videos helpful. 🙂
sir,your effort for the video is mind blowing i am a big fan of your channel and if you dont mind i am in love with your purple shirt and the black bow sir thank you so much for this videos
He should do an example of two cylinders, and leave the R dependency in there so we can see how the acceleration is affected by the radius of the cylinder.
Finding one of your videos on my question is like spotting a lighthouse in a storm sea. I can make it to shore!
Glad you find our videos helpful. 🙂
hmm.. after knowing this, I can asume that the acceleration can be written as a = gsin theta / (k+1) whereas k = the coefficient from I. Thank you very much
Your a geneius dude haha!!! Thats awesome you were able to figure it out, I checked it, and it WORKS!!!
awesome tutorials maestro
I love the background sound effect...
Based on final formula, if we increase friction(mu) , the acceleration will not change. Seems weird for me, i expect if we increase mu , acceleration will decrease.
That is because the object doesn't slide down the incline but rolls instead.
@@MichelvanBiezen Thank you dear Professor for reply. I greatly appreciate this. Assuming the theta and mass is constant, in the formula on right top of the board(beneath the f=ma), when i decrease the "mu" on the left side of equation, the "a" on the right side should increase.
No, but if mu is larger it would allow for a greater torque which would allow for a greater angular acceleration.
Mr. Michel, I defined clockwise to be a positive torque direction; on this basis I made force of static friction to be positive and the vertical component of the Force of gravity to be negative, while I seem to be having sign discrepancy in my problem. What should I do?
We often get hung up on negative signs. This can be remedied by simply caluculating the magnitude (which is always positive). The sign becomes important when we work with vector quantities.
@@MichelvanBiezen However, if we take an absolute value of every single force then it will affect our equation and hence the value?
What if the ball was filled with some volume of fluid, say water? would that change the centre of mass of the ball and cause a different acceleration down the ramp?
Just a small question, lets say we were given the radius and mass of the ball and there is some kinetic friction force present on the table, then, what is the acceleration of the same round object?
The wheel needs to slide in order to experience kinetic friction.
@@MichelvanBiezen then do they feel any sort of static friction?
Thank you so much as always🎉
You are so welcome! 🙂
Great explanation. Thank you!
Glad you liked it.
Why you Take up ward direction for friction? the contact point of ball will try to move back so friction direction on cylinder should be downward?
The direction of the friction force is in the opposite direction of motion of the object without friction. (Note that if there was no friction, the wheel would slide and not rotate). To make the wheel rotate in the correct direction, the friction must be directed upwards.
Mr. Professor does moment of inertia afect graviti, since it is much easier to pick up haevy rotating object?
No, the moment of inertia will not affect gravity in any way.
@@MichelvanBiezen thank You.
Great video! Thanks for the help!
Great explanation, thanks!
My head is in your feet sir 🙏🙏🙏
No need. Glad you find the videos helpful. 🙂
if the ball is slipping and not rolling, does {a=\alpha r} still hold true?
No, then it becomes just a sliding object and it will not rotate.
Sir, I think friction force should act along the direction of motion not opposite because the point of contact is going backward due clockwise angular velocity
The direction in the video is correct. Think about it this way: If there was no friction the wheel would slide and not rotate. With friction, what direction does the friction need to be in order for the wheel to rotate in a clockwise direction?
Sir,then it means if wheel was made to move up the inclined plane then friction would assist the torque and its motion would be along the direction of motion!!
Am i right ??
Friction works against the direction of applied force so even if the ball went up it would still be effected by frictional force working against it
Great video, thank you so much!
great video!
Glad you enjoyed it
Hello Professor Beizen, Will Fnet equation change if the ball is rolling up? If the ball is rolling up, would the friction be directed downwards? Will both mgsin(theta) and friction point downwards? Will the equation become mgsin(theta) + F = ma if ball is rolling up?
this was very helpful
We appreciate your comment.
does the x component of weight not provide a torque????
The torque is caused by the friction force (making the ball roll instead of slide)
Thank you.
What about solving with Moment of Inertia and Sliding down on an inclined plane?
How come the acceleration with a box on the same slope is faster ( a=g*sin theta) than a cylinder or hollow ball (a=3/5*g*sin theta)?
With a box, all of the potential energy is converted to translational kinetic energy and heat (due to friction). With a hollow ball (or any ball), the potential energy is converted to both translational AND rotational kinetic energy as well as heat (due to friction).
Sir. Could you please answer me? What happens if frictional force(kmgCosA) equals to mgSinA? Net force will be zero but there will be still net torque due to frictional force. What will happen?
This is just an approximation. If you want to learn more about rolling friction, take a look at the videos in this playlist: MECHANICAL ENGINEERING 11 FRICTION
@@MichelvanBiezen Thank you for your reply. But i did not understand it. What are we approximating? Could you please give some more explanation? Meanwhile I have watched all your videos related to rolling motion. I am looking forward for your answer. Thank you so much.
Just like we ignore wind resistance in kinematics, we ignore rolling friction and just approximate it with the coefficient of friction, which is not the whole friction effect. (see the mechanical engineering videos on rolling friction).
thank u very much sir
Maybe you have fulfilled your God-given purpose by making these vids for us
Maybe. We pray that we are fulfilling our purpose in life. 🙂
why is a equal to Rmew?
arigato gozaimasu!
god bless. thank you
Sir you have used a=r*alpha ,indicating pure rolling but you have taken kinetic friction that indicates slipping..?
Take a look at: PHYSICS 11 ROTATIONAL MOTION ruclips.net/p/PLX2gX-ftPVXXp-ceOBqew_PiJ-nQnuwOl or PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 ruclips.net/p/PLX2gX-ftPVXUV3FuKaomSwPLowv4qoOth
How is F=ma valid here when it doesn't account for the moment of inertia? You could have two spheres with the same radius and mass but different moment of inertia but they will both have different linear accelerations even though according to F=ma they should have the same acceleration. If F=ma is still valid as you claim then why not just use a=F/m to find the acceleration? Clearly you can't which suggests to me that it's not valid here. I'm just more confused now!
I agree there is a certain amount of "mystery" regarding this problem. But the problem is correct as shown in the video, and F=ma still works here as well. But in this case the force is doing 2 things: 1) it accelerates the object down the incline AND 2) it causes to object to rotate, both which require energy, and thus work, and thus the force is used to do both.
@@MichelvanBiezen thanks for your response. I started thinking about this problem recently when asking myself if I could prove that it doesn't matter if a force acts through an objects centre of mass or not.
I figured that if I found the net force acting on the ball (the component of gravity pulling down the incline minus the force of friction pushing up the incline), multiplied by the distance (the hypotenuse of the triangle) and added it to the torque times the angular distance the ball rolls through as it goes down the slope then it should be equal to mgh which it is, and this supports what you said about gravity doing all of the work - both linear and rotationally speaking.
Also the force of friction pushing up the slope multiplied by the distance also gives the amount of rotational kinetic energy which is what I predicted.
So I seemed to show that the work done on an object by a force (at least in this case) does not matter if it's through its centre of mass or not because friction isn't and it still works. But then I thought that if that is true then why can't I use F(net)=ma to find the acceleration? Does the work done not always result in a given speed (kinetic energy) when both linear and rotational acceleration are occuring? It appears not. Two spheres that are identical except for their moment of inertia will have the same force acting through the same distance (and hence the same work is done on each) but will have different kinetic energy at the bottom (translational, not rotational). So perhaps you can see why I'm getting confused over this seemingly simple problem!
Any input you could add to help clarify anything would be greatly appreciated.
Also one extra thing. Since it's static friction that is causing the torque, how do we know that it's at its maximum value? Shouldn't the equals sign be changed to a "less than or equal to" sign?
Thanks again.
@@MichelvanBiezen I would also like to add one more thing. In the example of the two identical spheres (except for their moment of inertia) where they both have the same net force acting in them, the one with the larger MoI will take longer to reach the bottom because it is not accelerating as quickly. But in terms of momentum and impulse, the same net force for a longer period of time should result in a larger change in momentum but clearly that's not true here as it is going slower when it reaches the bottom due to it having a lower acceleration through the same distance. This doesn't make any sense to me either. Any ideas?
@@MichelvanBiezen after giving this some thought I have realised that the force of static friction is larger for objects with a larger coefficient for their moment of inertia. My mistake was to think of the force of static friction as being fixed for all objects which it isn't. The static friction force is equal to mgSin(@)K/(K+1) where "K" is the coefficient for the moment of inertia such as 1/2 for a solid disk.
Using that with Newtons second law F=ma you find that the acceleration for any object rolling down an incline is a = gSin(@)/(K+1).
Using your own two examples yields the same answer that you got only this equation works for all rolling objects.
😎
Glad you liked the video. 🙂
@@MichelvanBiezen 🙂
when I was a kid.. my parents used to accuse me of having an INERTIA Problem.... because I was "apparently" too lazy... now that I know about Moments of Inertia.. I would tell my parents that my problem was with my MOMENT OF INERTIA.. meaning that, at the MOMENT.. I have NO INERTIA... (lol..) Go Figure...
hahaha, that was funny!