My professor did a very poor job of explaining this problem, your video was helpful and concise, a much better and less meandering explanation than I got before.
Dear Sir, static friction is a self-adjusting force and adjusts its value so that it is just sufficient to prevent relative motion between the point of contact of the wheel and the inclined surface. While rolling without slipping, it is often less than the maximum possible value of static friction which is called the limiting friction. In this case of rolling without slipping, we can't equate static friction with limiting friction. Static friction is actually equal to [mgsin(theta) / (1 + mr^2/I) where I = moment of inertia of rolling object about the axis of rotation passing through its center of mass. For a solid cylinder, I = (1/2)m.r^2. This gives static friction f=(1/3)mgsin(theta). From net force equation, we get mgsin(theta) - f = m x acm (eqn 1) where acm is linear acceleration of center of mass. Which gives acm = (2/3)gsin(theta). We obtain the expression for static friction by solving 3 simultaneous equations in 3 unknowns (alpha, acm and f) : net force equation (eqn 1), the torque equation rxf=Ixalpha (eqn 2) and the relation between angular acceleration alpha and linear acceleration of center of mass acm, which is given by alpha=acm/r (eqn 3)
I agree with your comment. People seem to often mistake the equation for the maximum available force of static friction (mgCos(theta)mu) as being the actual force of static friction which as you point out is not necessarily true. I worked out the equation for static friction to be F = mgSin(theta)k/k + 1 where "k" is the coefficient for the moment of inertia such as 1/2 for a solid disk.
Youre absolutelly right. Solving the problem by the way he did is only true for the limiting case where the inclination is so big that the static friction force is at it's limit.
I was trying to help my son with a similar problem and your video was excellent in helping us understand the approach. Thanks so much for clearing it up.
For 2:08, why is the torque of friction positive? With the right hand rule, my thumb would point into the page, indicating that friction would be negative. So what specifically let’s the torque of friction be positive? Thank you SO much.
In that equation we equated the magnitude of the torque to the product of the moment of inertia and the magnitude of the accelration. Magnitudes of vectors can never be negative.
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.
@@rephaelreyes8552 It has to be static friction because there is no relative motion in between the incline and the bottom point of the cylinder. Also, if there is any relative motion, the relationship of a = r α is not valid Professor Biezen's earlier tutorial video th-cam.com/video/suaJz-zWt3A/w-d-xo.html
A quicker way to solve the problem is to keep friction force without using f=uN equation. Follow exactly the same steps professor did on yo-yo case. th-cam.com/video/tkaZaQ-V4Bk/w-d-xo.html
There's a second way by viewing the adherent point as the spinning centre of a normal rotation. It uses 'sigma torque ext = dL/dt', and I'm trying to find an explanation of that because I don't understand how that makes mgsin(theta)R = I*alpha . (The I here being inertial momentum when the adherent point is the centre.
I'm having a massive case of confusion after self-learning about rolling friction, and then coming back to this question. I have a couple of questions, if someone could help me clarify, I would greatly appreciate it: 1. The coefficient of friction in this question (and the friction force in this question), are they referring to static friction, which is what I thought the friction always is when rolling without slipping, or is it rolling friction? I know we didnt solve for u here, but if we had solved for it, would we consider it to be the coefficient of static friction, or would it be considered the coefficient of rolling friction? 2a. If it is the coefficient of rolling friction, can one say that rolling Friction = coeff. of rolling friction x Normal Force? Also, are we just ignoring static friction then? 2b. If it is not rolling friction, then it must be static friction (which is what I've usually seen). If so, how can we know when rolling friction applies vs. when your classic Fstatic = us x normal force is what must be done? Because if this question does deal with only static friction, I know exactly how to handle it, it's the same old friction problem, but how in the world do I apply rolling friction to this? The object is rolling, so rolling friction must be involved somehow I guess. This is also leads to a similar scenario question I have as well: 3. I have a solid block on a flat surface, I kick it so that it has an initial velocity of 10 m/s. Given enough kinematics and force info, I can easily figure out the total displacement of the block, and I can also solve for the KINETIC coefficient of friction in this case (easy peasy). Let's just say, for the sake of simplicity, that the block traveled 10 meters before stopping. If I have a ball of same mass and material as the block that is on the same flat surface, and I kick it horizontally such that it also starts with a horizontal velocity of 10 m/s just like the block, it moves linearly and also rolls until it stops, let's say no slipping. Given enough kinematics and force/torque info, if I use linear kinematics, the linear displacement also comes out the same as the blocks displacement of "10 meters", but this doesnt seem right to me since some of the energy must have been used for rolling instead of translating. Would the linear displacement be the same for the rolling ball, or am I right in thinking that I am missing something? Regarding the force portion of the question, I can also solve for the friction force and/or the coefficient of friction. I would solve it considering STATIC friction only, with no consideration towards rolling friction (because that is all that I have been taught). But, by doing so, am I actually solving for the rolling friction force (and coefficient) or am I right in saying that I am solving for the static friction force (and coefficient)? If I am right about static, it again brings up the point of how would rolling friction apply here? THANK YOU for your time and effort, future responders!
Final answer is correct. Indeed a=2/3*g*sin(theta) BUT the explanation has a mistake. On minute 1:22 he says "friction = myu * N" and this is not correct, need to be left as friction (f), no myu involved, no N involved. Still, final answer is correct a=2/3*g*sin(theta). Apologize in advance in case I'm wrong...
This was the one thing I couldn’t figure out (because they didn’t want us to use conservation of energy) but then this video made me realize that torque was a thing so that’s cool… thank you so much, I love when it makes sense!
I came back to this Video, Michel. Always interesting to see the simplicity of a problem when you show us how it's done. I notice that if Theta is equal to 90 degrees, then a = 2/3 g which is Free Fall... as shown in your video with a Yo Yo... :) Physics is AMAZING.. thanks.
Awesome Video! But since the force of static friction can vary from 0 to mgcosthetaU(max) one should note that in the above approach static friction is assumed to be at the maximum. But regardless the answer would still be the same if you were to redo the whole problem by replacing mgcosthetaU with Fs
The video is indeed a simplified version of what is actually going on, including ignoring rolling friction, but that is necessary at first to introduce the basic principles in physics. Note that the friction does not need to be at the maximum level. Friction can be anywhere between 0 and mg cos(theta) mu
Isn't your static frictional force supposed to be an unknown since we can't assume that the static frictional force is at its maximum. All we know is that static frictional is just right for the body to roll smoothly down the ramp without sliding. At any angle, the ball cylinder would roll without slipping, so its static frictional force isn't at a maximum. If it were a box sliding, you could say fsmax is mew*mgsin(theta) since we know its a maximum static frictional force just before the box slides, right?
I'm still a little bit confused since my book says otherwise about the static friction. Also would the net torque be a negative value since your static frictional force causes a clockwise rotation of the object?
If the solid cylinder rolling down an incline starts off with an initial torque=Ixalpha then i) does that initial torque value remain a constant all the way down the incline, and ii) is the largest that the initial torque scalar-value can be ONLY as large as the static Friction force (in the opposite direction)? It seems to be clear and make sense if so. [Huge fan! Thanks in advance.]
I am teaching this right now to Eng Tech students . I really enjoy your explanation but have a question/comment this may be semantics but I am thinking of the work-energy equivalent. If friction is creating the torque and it is in the opposite direction to motion doesn't this contradict the concept of work? Wouldn't the upsetting force (the component of mg acting perpendicular to contact) be causing the torque and doing the work? If you look at the work required to roll the cylinder up the hill it becomes a bit more obvious. I would appreciate your thoughts since I am not physicist but a lowly engineer.
That is a good observation. Typically we assume there is enough friction to cause the wheel (cylinder) to rotate, otherwise the wheel would slide down the hill. But the friction force is small enough to ignore it in the energy conservation equation. (Like ignoring the wind resistance in our kinematics problems.)
When objects roll, the surface of the rolling objects and the surface of the incline do not slide relative to one another and therefore we use static friction.
a solid sphere is rolling down aN INCLINED PLANE WITHOUT SLIPPING IF THE INCLINED HAS A INCLINATION THETA WITH HORIZONTAL THEN THE COEFFICIENT OF FRICTION BETWEEN SPHERE AND THE INCLINED PLANE SHOULD BE... PLEASE SOLVE ANS OF PROBLEM IS FRICTION COFFICIENT >=2/7TANTHETA
a = (mgsin(theta) - mg cos(theta) mu) / m = g sin(theta) - g cos(theta) mu Torque = I alpha mgcos(theta) mu R = (2/5) mR^2 (a/R) Then substitute for a in the second equation and solve for mu
You get a 99% in my book. The 1% is because you mentioned coefficient of static friction, but really solved the problem of kinetic friction "given a light push". To solve the problem using both static and kinetic friction, I believe we would have to introduce energy equations. But you're use of torque and moment arm explanation of friction was brilliant.
There is no kinetic friction present in this problem. When an object rolls down an incline then there is no relative motion between the surface of the incline and the surface of the wheel. (They are static relative to one-another).
As Michel already mentioned, there's no kinetic friction. The cylinder is rolling without any sliding. Of course new points of the cylinder get in contact with the inclined plane every moment, but none of them actually slide.
one little confusion in here from the derivation we are able to find the coefficient of friction also but it seems(not sure about that) the coefficient of friction depends on the surface roughness of the materials, isn't it?
The interesting part is if you sub (a) back into (u) giving u=(1/3) tan (theta). This is one third the standard static friction giving a rough estimate that the rolling static friction is 1/3 the static friction.
If we take the center of the cylinder as a reference point, so relative to this point the contact point between the cylinder and the inclined is in motion (not at rest). My question what is the expression of the work done by the friction between the cylinder and incline relative to the center of the cylinder?
@@MichelvanBiezen But , in general , the work done by any force depends on the force vector and the displacement of the point of application of this force. The displacement is determined relative to a frame of reference , so the work done by a force (friction force or other force) should depend on the frame of reference of certain origin. if we multiply the moment of the friction force by the angular abscissa of the disk (It is the work done by the friction relative to the center of the disk) we found that it is equal to the rotational kinetic energy of the disk. What is your comment on this equality? Thank you.
The work done is a SCALAR quantity and as such cannot be associated with any particular position or direction. The work done is simply a dot product (= scalar product) of the Friction force and the displacement.
@@MichelvanBiezen yes , that's right the displacement vector AB of the point of application of the force remains constant in any frame of reference, of origin O, and having any direction. Let's study this special case: if the origin O moves with the same velocity with the point of application of this force , in this case the displacement vector AB relative to this frame becomes zero, so the work done by this force becomes also zero. What's your comment and thank you again.
I have to conduct a research question for physics, I’m thinking on rolling motion, something in relation to a ball rolling down an inclined plane, any ideas? (How does?....what is the effect of.....)
Hi, Wondering if you can help solving an old exam problem... where a Cylinder a rolled carpet that start rolling down an incline so the mass is variable , the question: what is the speed of carpet edge when the carpet ends unfolding.
4:50 could have just replaced gcos(theta) with a/2 in the right equation insteas of solving for the friction constant. Great video. I didn't understand how the friction interacts with the cylinder in my own lecture, so this was very helpful.
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. I'm just more confused now!
The moment of inertia of a sphere is always 2/5*(mR^2). Therefore it is impossible for two spheres with equal mass and radius to have different moments of inertia.
But I suppose you mean two different objects with different moments of inertia. In that case I completely understand your confusion. In this video the method used applies only when the object is standing still. Once in motion the friction acts only as torque and not as an actual force, so F = m* a is different than the initial condition. You are in fact now calculating the acceleration it would have if the objects starts moving.
@@raquelsimonsabsay7743 I've had some time to give this a lot of thought and I know now where I was going wrong. Different spheres can have a different moment of inertia depending on how their mass is distributed throughout the object. A solid sphere does not have the same moment of inertia as a hollow sphere. The force of static friction acting on an object that is rolling down an incline without slipping points up the incline. It acts as both a force and a torque. It doesn't matter if a force acts through the centre of mass or not, it will still have the same effect. The problem I was having is that I thought of both a hollow sphere and a solid sphere as having the same magnitude force of static friction acting on them because they had the same friction coefficient and thought that they ought to accelerate at the same rate regardless of their mass distribution which I knew couldn't be right. I forgot to account for the fact that the equation for static friction tells you the maximum force possible before slipping occurs, not what the force of fiction actually is for a given situation. Turns out that the force of static friction which also acts as a torque increases for a higher moment of inertia coefficient (such as 1/2 for a solid disk or 2/5 for a solid sphere). It's actually proportional to k/k+1 and the acceleration is proportional to 1/k+1 where "k" is the coefficient.
Because when the cylinder rolls down the incline, the surface of the cylinder (when in contact to the inclined plane) does not move (linearly) with respect to the incline.
I have a question in my university textbook which I couldn't solve A sphere rolls down a plane inclined at 30° to the horizontal. calculate the velocity and acceleration of the sphere after it has moved 5m from rest along the plane (I = 2MR²/5) I'm confused cus I can't figure out the force perpendicular to the radius which is causing the ball to roll Any help sir
Solve the problem exactly the same way as shown in the video, but use the sphere moment of inertia instead. Once you have the acceleration you can calculate the velocity by using the equation if kinematics: V^2 = Vo^2 + 2 a x solve for x.
Hello, I am currently using this wonderful video as the theory behind a simple experiment we are doing as part of a high school research assignment. For his final solution, it seems that pnu is not a factor?! Does the frictional coefficient of the inclined surface not matter? This leaves me so, so confused.
The friction will cause a very slight slow down, which is typically ignored, as it is minor compared to the moment of inertia effects. If you roll different object of the same shape, but made of different materials, you will find that they accelerate at different rates due to the differences in the friction forces (as part of the reason).
@@MichelvanBiezen Thank you for the quick and apt reply. I am will consider rolling friction as the main culprit, as the incline surfaces are quite deformable (i.e. carpet/rubber). I should have said it before, but your videos are excellent - having resources like this for free is amazing. Thank you.
Yes, we are ignoring the friction force (other than causing the rolling motion). If you want to calculate the work done by the friction force you use the equation: W = F x d
YOU ARE A LIFE SAVER. I was intending on doing a simple project on finding the coefficient of friction down an inclined plane which is extremely easy but I was so stupid that i chose a cart which used wheels to move. Hence my data did not line up with the standard model of coefficient of friction and I was so confused. Thanks to your video, I won't fail my project now XD
It is assumed the object is rolling not sliding. In fact there are 3 cases: 1. μ > (1/3)tan(θ) : object stands still (?) 2. μ = (1/3)tan(θ) : object rolls (a = 2/3 * g * sin(θ)) 3. μ < (1/3)tan(θ) : object slides while rolling (a > 2/3 * g * sin(θ)) My guess is that object doesn't abruptly go from 100% rolling to 100% sliding: Sliding portion gradually increases as θ increases for a given μ.
sir I am a bit confused regarding the friction force won't it be in negative as the normal force is upward but the equation mgcostetha is for the weight which is downward force?
The parallel axis theorem is used to find the moment of inertia relative to another point away from the center of mass. We are not trying to do that here.
Nice video! The static friction force, however, isn't mu (static) times the normal force, since that is the maximum static friction force and there is no indication that static friction is pushing at its maximum value. You get the same result, though, since you eliminated mu. But if you leave the force of static friction as Fs you get the same result: From the torque: R*Fs*sin 90 = I * alpha = (1/2 m R^2) * (a/R) so Fs = 1/2 m a From the Forces in the down-the-ramp direction, mg sin (theta) - Fs = ma, so... mg sin (theta) - [1/2 m a] = ma So a = 2/3*g*sin(theta)
It depends on what they are asking. If they ask for the acceleration, then it is better to use the technique shown in the video. But if they ask for the velocity at the bottom of the incline, it is better to use the conservation of energy equation.
Just like the forces that allow you to walk, (your foot pushes backwards against the floor and the floor pushes back in the the direction of your direction of motion, the wheel pushes backwards against the incline and the incline pushes the wheel forward. Also note that the rolling friction is something different which is explained in our mechanical engineering videos.
Can someone explain to me why coefficient of friction depends on angle? I noticed if you set the equations for "acceleration" on the torque side and the Force side equal, you will get an equation like this { u = 1/3 * tan(theta) }. And I do not understand this result. Isn't coefficient of friction a quantity that's only related to the nature of the contacting surfaces? How is it changing while the surfaces are still the same two? Or it tells something about the resistance in rotational motion rather than the static friction?
The coefficient of friction does not depend on angle. But the friction force does depend on angle since it depends on the normal force component of the weight of the object.
Can you please tell me how you recorded the video? I can see you using a collar mike. How do you connect the collar mike's audio to the video recording? I record my lectures using a phone and there is a lot of background noise. I'm not sure where to purchase the collar mike and how to use it with my phone to record good quality video like yours. Please help!
We used to use a wireless microphone, but we had a lot of interference. Now we use a wired microphone and I am directly connected to the camera through a wire.
If you are looking for the acceleration, it is better to use this method, however you should be able to find the final velocity using conservation of energy, and then find the acceleration using the equations of kinematics.
@@MichelvanBiezen thanks so much! I'm supposed to find time for a hoop to reach the bottom of the Incline and I used energy conservation to find the final velocity to solve, but my friend told me to use the method in the video so I was slightly confused.
I don't believe in god but sir your are the god of knowledge 😭😭😭 almost every time I search for my school work I see brilliant explanations from your videos😭😭😭
The static friction force is calculated (without needing to calculate alpha). As seen with the equation near the upper left corner). The static friction force doesn't slow down the wheel, it just provides the torque to make it roll. (otherwise the wheel would slip).
@@MichelvanBiezen Thank you for the reply. But what if I don't have the friction coefficient u? I'm working on a problem that is similar to this, and I only have the mass, the radius of the solid and the angle theta to work with.
They probably assume the friction is there to make it roll (instead of slide). Just work out the problem assuming that there is sufficient friction as shown in the video.
this is an interesting example of how a mistake cancels itself out during the derivation, so that we still end up with the correct result even after using a wrong assumption. The wrong assumption here is that static friction equals limiting friction. Since static friction is numerically equal to (ma)/2 in this case, and only that result is finally used to arrive at the expression for acom, everything works out fine in the end.
So acceleration of the cylinder is independent of the moment of inertia? I find that very interesting because my original intuition was that the a cylinder with lager moment of inetia would accelerate slower. Thank you for the help.
Plz sir explain how to take components on inclined plane..... i don't understand.how to know its mgcos is perpendicular to the inclined line plz elobrate...........from INDIA
Michel van Biezen .........sir you are great......i know you have lack of time still you reply as soon as possible.......this shows your passion and greatness about teaching.........thanx
The difficult part is to figure out what is the force causing the torque that gets the rod to roll. If you figure out , this is the friction force by the bank to the rod, the rest would be solvable.
If solid sphere and hollow sphere are moving down an incline with a *constant velocity* . Which will reach the bottom first? Please reply. Just tell me the answer, I'll automatically understand the solution. I have an entrance exam on Sunday. I need to know the answer as I can't find it anywhere else.
The solid sphere will reach the bottom first. (It has a smaller moment of inertia which means that more mass is distributed closer to the point of rotation, which in this case is at the center of the ball)
so the acceleration is NOT dependent on the radius of the cylinder? That does not seem right. So a small cylinder and a large cylinder will go down the ramp at the same rate? No dependency on the moment of inertia? Hmmm. I will have to work this out on my own. That does not seem right. Free fall of course is mass and size independent, but the rolling motion... hmmmmm
Of course, this result is limited to the domain where a friction force obeys F(fr)=N*mu. If one makes theta = 90 degrees in a=(2/3)*g*sin(theta), we get a = (2/3)*g. Yet, we know that it has to be a=g (object simply falling).
That is indeed true, but while reading and using this equation, you have to take into account, that this only works for cylinders pure rolling. If theta is 90 degree, it wouldn't roll, or atleast wouldn't roll that well, and therefore not being pure roll. If anything in this message is incorrect, PLEASE let me know. I want to learn as much as possible.
hi, could you explain uphill roller gravity defying conditions so we have a perpetual motion forever for everyone for millions of trillions of uncountable years and endless supply of energy and electricity without any pollution solution become easy
If you were to do an experiment with a few cylinders (both solid, hollowed, and (centered?)) rolled down a slope, would you measure the velocity of the various cylinders at the end of the slope and afterwards calculate the moment of inertia? (mgh=½mv^2+½I*w^2 , w=v/r) I'm about to write a rather large paper on this subject, but I am not quite sure about an experiment. (You would of course know the angle of the slope, the mass of the cylinders and their radii) I apoligize for any spelling mistakes, I'm not native. Thanks.
My professor did a very poor job of explaining this problem, your video was helpful and concise, a much better and less meandering explanation than I got before.
Dear Sir, static friction is a self-adjusting force and adjusts its value so that it is just sufficient to prevent relative motion between the point of contact of the wheel and the inclined surface. While rolling without slipping, it is often less than the maximum possible value of static friction which is called the limiting friction. In this case of rolling without slipping, we can't equate static friction with limiting friction. Static friction is actually equal to [mgsin(theta) / (1 + mr^2/I) where I = moment of inertia of rolling object about the axis of rotation passing through its center of mass. For a solid cylinder, I = (1/2)m.r^2. This gives static friction f=(1/3)mgsin(theta). From net force equation, we get mgsin(theta) - f = m x acm (eqn 1) where acm is linear acceleration of center of mass. Which gives acm = (2/3)gsin(theta). We obtain the expression for static friction by solving 3 simultaneous equations in 3 unknowns (alpha, acm and f) : net force equation (eqn 1), the torque equation rxf=Ixalpha (eqn 2) and the relation between angular acceleration alpha and linear acceleration of center of mass acm, which is given by alpha=acm/r (eqn 3)
I agree with your comment. People seem to often mistake the equation for the maximum available force of static friction (mgCos(theta)mu) as being the actual force of static friction which as you point out is not necessarily true. I worked out the equation for static friction to be F = mgSin(theta)k/k + 1 where "k" is the coefficient for the moment of inertia such as 1/2 for a solid disk.
Youre absolutelly right. Solving the problem by the way he did is only true for the limiting case where the inclination is so big that the static friction force is at it's limit.
Really important distinction, thanks for sharing!
I wish I could thumbs-up this more than once. You are seriously a life saver and a very good instructor
I was trying to help my son with a similar problem and your video was excellent in helping us understand the approach. Thanks so much for clearing it up.
this respectable man has the best explanation in the entire world!!!
For 2:08, why is the torque of friction positive? With the right hand rule, my thumb would point into the page, indicating that friction would be negative. So what specifically let’s the torque of friction be positive? Thank you SO much.
In that equation we equated the magnitude of the torque to the product of the moment of inertia and the magnitude of the accelration. Magnitudes of vectors can never be negative.
@@MichelvanBiezen Oh, okay. By default, should we use the magnitude of the torque? Thanks.
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.
He might have meant kinetic friction. Could you still use the f=uN equation if that was the case?
@@rephaelreyes8552 It has to be static friction because there is no relative motion in between the incline and the bottom point of the cylinder. Also, if there is any relative motion, the relationship of a = r α is not valid Professor Biezen's earlier tutorial video th-cam.com/video/suaJz-zWt3A/w-d-xo.html
A quicker way to solve the problem is to keep friction force without using f=uN equation. Follow exactly the same steps professor did on yo-yo case. th-cam.com/video/tkaZaQ-V4Bk/w-d-xo.html
Yes you are correct
There's a second way by viewing the adherent point as the spinning centre of a normal rotation. It uses 'sigma torque ext = dL/dt', and I'm trying to find an explanation of that because I don't understand how that makes mgsin(theta)R = I*alpha . (The I here being inertial momentum when the adherent point is the centre.
I'm having a massive case of confusion after self-learning about rolling friction, and then coming back to this question. I have a couple of questions, if someone could help me clarify, I would greatly appreciate it:
1. The coefficient of friction in this question (and the friction force in this question), are they referring to static friction, which is what I thought the friction always is when rolling without slipping, or is it rolling friction? I know we didnt solve for u here, but if we had solved for it, would we consider it to be the coefficient of static friction, or would it be considered the coefficient of rolling friction?
2a. If it is the coefficient of rolling friction, can one say that rolling Friction = coeff. of rolling friction x Normal Force? Also, are we just ignoring static friction then?
2b. If it is not rolling friction, then it must be static friction (which is what I've usually seen). If so, how can we know when rolling friction applies vs. when your classic Fstatic = us x normal force is what must be done? Because if this question does deal with only static friction, I know exactly how to handle it, it's the same old friction problem, but how in the world do I apply rolling friction to this? The object is rolling, so rolling friction must be involved somehow I guess.
This is also leads to a similar scenario question I have as well:
3. I have a solid block on a flat surface, I kick it so that it has an initial velocity of 10 m/s. Given enough kinematics and force info, I can easily figure out the total displacement of the block, and I can also solve for the KINETIC coefficient of friction in this case (easy peasy). Let's just say, for the sake of simplicity, that the block traveled 10 meters before stopping.
If I have a ball of same mass and material as the block that is on the same flat surface, and I kick it horizontally such that it also starts with a horizontal velocity of 10 m/s just like the block, it moves linearly and also rolls until it stops, let's say no slipping.
Given enough kinematics and force/torque info, if I use linear kinematics, the linear displacement also comes out the same as the blocks displacement of "10 meters", but this doesnt seem right to me since some of the energy must have been used for rolling instead of translating. Would the linear displacement be the same for the rolling ball, or am I right in thinking that I am missing something? Regarding the force portion of the question, I can also solve for the friction force and/or the coefficient of friction. I would solve it considering STATIC friction only, with no consideration towards rolling friction (because that is all that I have been taught). But, by doing so, am I actually solving for the rolling friction force (and coefficient) or am I right in saying that I am solving for the static friction force (and coefficient)? If I am right about static, it again brings up the point of how would rolling friction apply here?
THANK YOU for your time and effort, future responders!
Thanks for adding this topic as today is my Final exam of Class 11. So thank you very much
You have taught us🔥🙏🙏....
May God bless you 🙏
I have completely understood everything
Thank you. Glad you found our videos! 🙂
Final answer is correct.
Indeed a=2/3*g*sin(theta) BUT the explanation has a mistake. On minute 1:22 he says "friction = myu * N" and this is not correct, need to be left as friction (f), no myu involved, no N involved.
Still, final answer is correct a=2/3*g*sin(theta).
Apologize in advance in case I'm wrong...
This was the one thing I couldn’t figure out (because they didn’t want us to use conservation of energy) but then this video made me realize that torque was a thing so that’s cool… thank you so much, I love when it makes sense!
Yes, once you see how its done, it becomes relatively simple.
I came back to this Video, Michel. Always interesting to see the simplicity of a problem when you show us how it's done. I notice that if Theta is equal to 90 degrees, then a = 2/3 g which is Free Fall... as shown in your video with a Yo Yo... :) Physics is AMAZING.. thanks.
Hi Philip. Yes, taking problems to the limit as you did, does reveal some interesting insight.
Awesome Video! But since the force of static friction can vary from 0 to mgcosthetaU(max) one should note that in the above approach static friction is assumed to be at the maximum. But regardless the answer would still be the same if you were to redo the whole problem by replacing mgcosthetaU with Fs
The video is indeed a simplified version of what is actually going on, including ignoring rolling friction, but that is necessary at first to introduce the basic principles in physics. Note that the friction does not need to be at the maximum level. Friction can be anywhere between 0 and mg cos(theta) mu
@@MichelvanBiezenso the most correct way is to use rolling friction?
Isn't your static frictional force supposed to be an unknown since we can't assume that the static frictional force is at its maximum. All we know is that static frictional is just right for the body to roll smoothly down the ramp without sliding. At any angle, the ball cylinder would roll without slipping, so its static frictional force isn't at a maximum. If it were a box sliding, you could say fsmax is mew*mgsin(theta) since we know its a maximum static frictional force just before the box slides, right?
The static friction is as shown in the video.
I'm still a little bit confused since my book says otherwise about the static friction. Also would the net torque be a negative value since your static frictional force causes a clockwise rotation of the object?
If you want to express the torque as a vector quantity, yes. If you express the torque as a magnitude only, no.
If the solid cylinder rolling down an incline starts off with an initial torque=Ixalpha then i) does that initial torque value remain a constant all the way down the incline, and ii) is the largest that the initial torque scalar-value can be ONLY as large as the static Friction force (in the opposite direction)? It seems to be clear and make sense if so. [Huge fan! Thanks in advance.]
So according to the equation, any two different solid cylinders, regardless of weight or radius will accelerate at the same rate down the ramp?
That is correct. Radius and mass do not matter.
@@MichelvanBiezen thanks a lot!
I am teaching this right now to Eng Tech students . I really enjoy your explanation but have a question/comment this may be semantics but I am thinking of the work-energy equivalent. If friction is creating the torque and it is in the opposite direction to motion doesn't this contradict the concept of work? Wouldn't the upsetting force (the component of mg acting perpendicular to contact) be causing the torque and doing the work? If you look at the work required to roll the cylinder up the hill it becomes a bit more obvious. I would appreciate your thoughts since I am not physicist but a lowly engineer.
That is a good observation. Typically we assume there is enough friction to cause the wheel (cylinder) to rotate, otherwise the wheel would slide down the hill. But the friction force is small enough to ignore it in the energy conservation equation. (Like ignoring the wind resistance in our kinematics problems.)
this is absolute elegance
Best explanation ever for this topic, thank you for sharing your amazing lectures with us, God bless you!
How did you take static as limiting friction?
When objects roll, the surface of the rolling objects and the surface of the incline do not slide relative to one another and therefore we use static friction.
a solid sphere is rolling down aN INCLINED PLANE WITHOUT SLIPPING IF THE INCLINED HAS A INCLINATION THETA WITH HORIZONTAL THEN THE COEFFICIENT OF FRICTION BETWEEN SPHERE AND THE INCLINED PLANE SHOULD BE...
PLEASE SOLVE ANS OF PROBLEM IS
FRICTION COFFICIENT >=2/7TANTHETA
a = (mgsin(theta) - mg cos(theta) mu) / m = g sin(theta) - g cos(theta) mu Torque = I alpha mgcos(theta) mu R = (2/5) mR^2 (a/R) Then substitute for a in the second equation and solve for mu
You get a 99% in my book. The 1% is because you mentioned coefficient of static friction, but really solved the problem of kinetic friction "given a light push". To solve the problem using both static and kinetic friction, I believe we would have to introduce energy equations. But you're use of torque and moment arm explanation of friction was brilliant.
There is no kinetic friction present in this problem. When an object rolls down an incline then there is no relative motion between the surface of the incline and the surface of the wheel. (They are static relative to one-another).
As Michel already mentioned, there's no kinetic friction. The cylinder is rolling without any sliding. Of course new points of the cylinder get in contact with the inclined plane every moment, but none of them actually slide.
you are doing all what i want thank you professor.
You are welcome. Glad you found our videos. 🙂
Only one word : OUTSTANDING TEACHING !!
Totally loved it. No doubts at all.! 👍❤️🐼
Thats two words dumbass
thats 2 words
one little confusion in here from the derivation we are able to find the coefficient of friction also but it seems(not sure about that) the coefficient of friction depends on the surface roughness of the materials, isn't it?
The one we get from this equation is the minimum coefficient of friction required for rolling without slipping to happen
You are correct. Saying friction = Myu * N in this case is wrong
The interesting part is if you sub (a) back into (u) giving u=(1/3) tan (theta).
This is one third the standard static friction giving a rough estimate that the rolling static friction is 1/3 the static friction.
If we take the center of the cylinder as a reference point, so relative to this point the contact point between the cylinder and the inclined is in motion (not at rest). My question what is the expression of the work done by the friction between the cylinder and incline relative to the center of the cylinder?
The work done by the friction force is not relative to any particular location.
@@MichelvanBiezen But , in general , the work done by any force depends on the force vector and the displacement of the point of application of this force. The displacement is determined relative to a frame of reference , so the work done by a force (friction force or other force) should depend on the frame of reference of certain origin. if we multiply the moment of the friction force by the angular abscissa of the disk (It is the work done by the friction relative to the center of the disk) we found that it is equal to the rotational kinetic energy of the disk. What is your comment on this equality? Thank you.
The work done is a SCALAR quantity and as such cannot be associated with any particular position or direction. The work done is simply a dot product (= scalar product) of the Friction force and the displacement.
@@MichelvanBiezen yes , that's right the displacement vector AB of the point of application of the force remains constant in any frame of reference, of origin O, and having any direction. Let's study this special case: if the origin O moves with the same velocity with the point of application of this force , in this case the displacement vector AB relative to this frame becomes zero, so the work done by this force becomes also zero. What's your comment and thank you again.
I have to conduct a research question for physics, I’m thinking on rolling motion, something in relation to a ball rolling down an inclined plane, any ideas? (How does?....what is the effect of.....)
Hi, Wondering if you can help solving an old exam problem... where a Cylinder a rolled carpet that start rolling down an incline so the mass is variable , the question: what is the speed of carpet edge when the carpet ends unfolding.
It is very easy
Taught in india in 11th class
4:50 could have just replaced gcos(theta) with a/2 in the right equation insteas of solving for the friction constant.
Great video. I didn't understand how the friction interacts with the cylinder in my own lecture, so this was very helpful.
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. I'm just more confused now!
The moment of inertia of a sphere is always 2/5*(mR^2). Therefore it is impossible for two spheres with equal mass and radius to have different moments of inertia.
But I suppose you mean two different objects with different moments of inertia. In that case I completely understand your confusion. In this video the method used applies only when the object is standing still. Once in motion the friction acts only as torque and not as an actual force, so F = m* a is different than the initial condition. You are in fact now calculating the acceleration it would have if the objects starts moving.
@@raquelsimonsabsay7743 I've had some time to give this a lot of thought and I know now where I was going wrong. Different spheres can have a different moment of inertia depending on how their mass is distributed throughout the object. A solid sphere does not have the same moment of inertia as a hollow sphere.
The force of static friction acting on an object that is rolling down an incline without slipping points up the incline. It acts as both a force and a torque. It doesn't matter if a force acts through the centre of mass or not, it will still have the same effect. The problem I was having is that I thought of both a hollow sphere and a solid sphere as having the same magnitude force of static friction acting on them because they had the same friction coefficient and thought that they ought to accelerate at the same rate regardless of their mass distribution which I knew couldn't be right. I forgot to account for the fact that the equation for static friction tells you the maximum force possible before slipping occurs, not what the force of fiction actually is for a given situation. Turns out that the force of static friction which also acts as a torque increases for a higher moment of inertia coefficient (such as 1/2 for a solid disk or 2/5 for a solid sphere). It's actually proportional to k/k+1 and the acceleration is proportional to 1/k+1 where "k" is the coefficient.
Why is static friction as opposed to kinetic friction acting as the cylinder rolls down the incline?
Because when the cylinder rolls down the incline, the surface of the cylinder (when in contact to the inclined plane) does not move (linearly) with respect to the incline.
I have a question in my university textbook which I couldn't solve
A sphere rolls down a plane inclined at 30° to the horizontal. calculate the velocity and acceleration of the sphere after it has moved 5m from rest along the plane (I = 2MR²/5)
I'm confused cus I can't figure out the force perpendicular to the radius which is causing the ball to roll
Any help sir
Solve the problem exactly the same way as shown in the video, but use the sphere moment of inertia instead. Once you have the acceleration you can calculate the velocity by using the equation if kinematics: V^2 = Vo^2 + 2 a x solve for x.
@@MichelvanBiezen thank professor, I now understand better
Thank for making knowledge free 🙏
When we get feedback that the videos are helping students around the world, it is all worth it.
Is this formula applicable to a vehicle of a certain load? Where does drag comes into the calculation?
Hello, I am currently using this wonderful video as the theory behind a simple experiment we are doing as part of a high school research assignment. For his final solution, it seems that pnu is not a factor?! Does the frictional coefficient of the inclined surface not matter? This leaves me so, so confused.
The friction will cause a very slight slow down, which is typically ignored, as it is minor compared to the moment of inertia effects. If you roll different object of the same shape, but made of different materials, you will find that they accelerate at different rates due to the differences in the friction forces (as part of the reason).
@@MichelvanBiezen Thank you for the quick and apt reply. I am will consider rolling friction as the main culprit, as the incline surfaces are quite deformable (i.e. carpet/rubber). I should have said it before, but your videos are excellent - having resources like this for free is amazing.
Thank you.
What would be the minimum coefficient of friction required for the disk to roll without slipping?
Ask miss Google. I m sure she will help 🐼
@@lenael4747 goofy lil thing huh
@@EricaLuvsYella
= 2/7tan theta
th-cam.com/video/SIRu76QV96M/w-d-xo.html
thanks alot sir . How much is the work of friction force in this example ?
Yes, we are ignoring the friction force (other than causing the rolling motion). If you want to calculate the work done by the friction force you use the equation: W = F x d
@@MichelvanBiezen thanks alot
YOU ARE A LIFE SAVER. I was intending on doing a simple project on finding the coefficient of friction down an inclined plane which is extremely easy but I was so stupid that i chose a cart which used wheels to move. Hence my data did not line up with the standard model of coefficient of friction and I was so confused. Thanks to your video, I won't fail my project now XD
Finally !!! This video help me a lot ! Thanks!
So why de difference with hollow tubes and not hollow. Even with equal mass
While if you drop anything it will fall down equal.
Falling down, yes it does not make a difference. But when objects roll down an incline it DOES make a difference.
So is the acceleration the same when we change the size of the solid cylinder. If yes,, why?
Yes, since the acceleration only depends on the angle.
Thank you sir
It is assumed the object is rolling not sliding. In fact there are 3 cases:
1. μ > (1/3)tan(θ) : object stands still (?)
2. μ = (1/3)tan(θ) : object rolls (a = 2/3 * g * sin(θ))
3. μ < (1/3)tan(θ) : object slides while rolling (a > 2/3 * g * sin(θ))
My guess is that object doesn't abruptly go from 100% rolling to 100% sliding:
Sliding portion gradually increases as θ increases for a given μ.
could you not use the law of conservation of energy for this problem?
Try it and see if you get the same answer.
sir I am a bit confused regarding the friction force won't it be in negative as the normal force is upward but the equation mgcostetha is for the weight which is downward force?
The direction of the friction force is typically opposite to the direction of motion.
Hello sir, what is the reasoning behind not applying the parallel axis theorem in this example?
The parallel axis theorem is used to find the moment of inertia relative to another point away from the center of mass. We are not trying to do that here.
Nice video! The static friction force, however, isn't mu (static) times the normal force, since that is the maximum static friction force and there is no indication that static friction is pushing at its maximum value. You get the same result, though, since you eliminated mu. But if you leave the force of static friction as Fs you get the same result:
From the torque: R*Fs*sin 90 = I * alpha = (1/2 m R^2) * (a/R) so Fs = 1/2 m a
From the Forces in the down-the-ramp direction, mg sin (theta) - Fs = ma, so...
mg sin (theta) - [1/2 m a] = ma
So a = 2/3*g*sin(theta)
Totally agree !!
could we use conservation of energy for this question?
It depends on what they are asking. If they ask for the acceleration, then it is better to use the technique shown in the video. But if they ask for the velocity at the bottom of the incline, it is better to use the conservation of energy equation.
why does the component mgsintheta cannot produce the torque on the cylinder??
Because it acts through the center of mass, which is the center of rotation.
Thank you!
The force of static friction that causes the torque is directly proportional to mgSin(theta) though.
a fantastic video a life saver in my class.
Only doubt the direction of Friction Force... As rolling friction acts in the direction of rolling
Just like the forces that allow you to walk, (your foot pushes backwards against the floor and the floor pushes back in the the direction of your direction of motion, the wheel pushes backwards against the incline and the incline pushes the wheel forward. Also note that the rolling friction is something different which is explained in our mechanical engineering videos.
If there was no friction, would the cylinder roll or slide?
The cylinder would slide.
Thank you!
Can someone explain to me why coefficient of friction depends on angle?
I noticed if you set the equations for "acceleration" on the torque side and the Force side equal, you will get an equation like this
{ u = 1/3 * tan(theta) }.
And I do not understand this result.
Isn't coefficient of friction a quantity that's only related to the nature of the contacting surfaces? How is it changing while the surfaces are still the same two? Or it tells something about the resistance in rotational motion rather than the static friction?
The coefficient of friction does not depend on angle. But the friction force does depend on angle since it depends on the normal force component of the weight of the object.
Michel van Biezen you didn’t understand my question sir
Can you please tell me how you recorded the video? I can see you using a collar mike. How do you connect the collar mike's audio to the video recording? I record my lectures using a phone and there is a lot of background noise. I'm not sure where to purchase the collar mike and how to use it with my phone to record good quality video like yours. Please help!
We used to use a wireless microphone, but we had a lot of interference. Now we use a wired microphone and I am directly connected to the camera through a wire.
Thank you so much!
We love you man, keep making these awesome videos.
We hope we can. 🙂
thank you so much, This problem was killing me, I realize now that my prof doesn't explain anything.
You are welcome. Glad you found our videos 🙂
best channel ever
Thank you. Glad you found our videos. 🙂
is it possible to use energy conservation to solve this problem?
If you are looking for the acceleration, it is better to use this method, however you should be able to find the final velocity using conservation of energy, and then find the acceleration using the equations of kinematics.
@@MichelvanBiezen thanks so much! I'm supposed to find time for a hoop to reach the bottom of the Incline and I used energy conservation to find the final velocity to solve, but my friend told me to use the method in the video so I was slightly confused.
I don't believe in god but sir your are the god of knowledge 😭😭😭 almost every time I search for my school work I see brilliant explanations from your videos😭😭😭
How is the friction force opposing the linear acceleration when it is only causing rotation?
how did it become 3/2 (a) from a + 1/2 (a) please explain. thanks.. noob here
1a + (1/2)a = (2/2)a + (1/2)a = (3/2)a
So if I want to calculate the static friction force, I still need to calculate alpha right?
The static friction force is calculated (without needing to calculate alpha). As seen with the equation near the upper left corner). The static friction force doesn't slow down the wheel, it just provides the torque to make it roll. (otherwise the wheel would slip).
@@MichelvanBiezen Thank you for the reply. But what if I don't have the friction coefficient u? I'm working on a problem that is similar to this, and I only have the mass, the radius of the solid and the angle theta to work with.
They probably assume the friction is there to make it roll (instead of slide). Just work out the problem assuming that there is sufficient friction as shown in the video.
is this answer in m/s^2? not angular acceleration right?
That is correct. (The answer is given in terms of g)
Yes ty man my professor just wrote down the top 2 equations ty for showing how to resolve
Glad you found our videos and you found them helpful. 🙂
You are an amazing teacher..thank you
Thank you. Glad you found our videos. 🙂
There was no need to express the frictional force in terms of μ! Let the friction be expressed in terms of m and a....
Totally agree !!
Fucking legend. People like you make the world a better place!
You sir, have an incredible gift to teach
this is an interesting example of how a mistake cancels itself out during the derivation, so that we still end up with the correct result even after using a wrong assumption. The wrong assumption here is that static friction equals limiting friction. Since static friction is numerically equal to (ma)/2 in this case, and only that result is finally used to arrive at the expression for acom, everything works out fine in the end.
I still think that the lagrangian approach is simpler and elegant, but i like this explanation too
a=gsintheta/(1+c) where c is the moment of inertia's differentiating constant
So acceleration of the cylinder is independent of the moment of inertia? I find that very interesting because my original intuition was that the a cylinder with lager moment of inetia would accelerate slower. Thank you for the help.
That is not correct. If the moment of inertia is larger, the acceleration will be slower. (The fraction in the answer will be smaller)
Wawoooooo that is cool .... great work
Glad yoy liked it! 🙂
This video definitely save me in todays final!
Glad it was helpful
Great video helped me so much !!!
Plz sir explain how to take components on inclined plane..... i don't understand.how to know its mgcos is perpendicular to the inclined line plz elobrate...........from INDIA
Definition of cos(theta) = ratio of the adjacent side of the angle divided by the hypotenuse Therefore adjacent side = mg cos(theta)
Michel van Biezen .........sir you are great......i know you have lack of time still you reply as soon as possible.......this shows your passion and greatness about teaching.........thanx
Well done! Makes physics fun!
The difficult part is to figure out what is the force causing the torque that gets the rod to roll. If you figure out , this is the friction force by the bank to the rod, the rest would be solvable.
Which chp in class 11th physics
If solid sphere and hollow sphere are moving down an incline with a *constant velocity* . Which will reach the bottom first?
Please reply.
Just tell me the answer, I'll automatically understand the solution.
I have an entrance exam on Sunday. I need to know the answer as I can't find it anywhere else.
The solid sphere will reach the bottom first. (It has a smaller moment of inertia which means that more mass is distributed closer to the point of rotation, which in this case is at the center of the ball)
@@MichelvanBiezen thank you so much. That helped a lot.
Helllo, what happened to mgsin(theta)?
mg sin(theta) does not cause a torque and therefore does not cause an angular acceleration.
Thank you so much for the video, im 4 years late, but this was explained very clearly.
Better late than never.
i did a question where the mass= 1.7kg and is on an incline at 37 degrees. i used your working and got the wrong answer but i'm not sure why
Did the problem include the moment of inertia?
heavier cyclists roll down hills faster than lighter cyclists.. if mass cancels out, then how can we explain this phenomenon?
In the real world there is wind resistance, rolling friction, internal friction with the moving parts, etc. which have an effect on acceleration.
Ossum way to descrive👌👌👌
very clear explanation, thank you very much
Glad it was helpful! 🙂
Pls an object inclind up
That is a good suggestion. We need to add some examples of objects moving up an incline.
Another great video... !!! As ALWAYS :)
Thank you. Glad you enjoyed it!
brilliant. thank you so much
so the acceleration is NOT dependent on the radius of the cylinder? That does not seem right. So a small cylinder and a large cylinder will go down the ramp at the same rate? No dependency on the moment of inertia? Hmmm. I will have to work this out on my own. That does not seem right. Free fall of course is mass and size independent, but the rolling motion... hmmmmm
It turns out, it does NOT depend on the radius.
Of course, this result is limited to the domain where a friction force obeys F(fr)=N*mu.
If one makes theta = 90 degrees in a=(2/3)*g*sin(theta), we get a = (2/3)*g. Yet, we know that it has to be a=g (object simply falling).
That is indeed true, but while reading and using this equation, you have to take into account, that this only works for cylinders pure rolling. If theta is 90 degree, it wouldn't roll, or atleast wouldn't roll that well, and therefore not being pure roll. If anything in this message is incorrect, PLEASE let me know. I want to learn as much as possible.
Cảm ơn thầy.
Thanks teacher
Awesome explanation sir! Thank you!
Glad it was helpful!
hi, could you explain uphill roller gravity defying conditions so we have a perpetual motion forever for everyone for millions of trillions of uncountable years and endless supply of energy and electricity without any pollution solution become easy
The laws of physics cannot be broken. You cannot get energy from nothing.
@@MichelvanBiezen but experiment shows it's possible, and i challenged isro, nasa, and every scientist of the universe
can somebody help me to get the distance travelled
Once you calculate the acceleration, then d = (1/2) a t^2
Thank you so much for this. My lecturer Ralph is really bad
If you were to do an experiment with a few cylinders (both solid, hollowed, and (centered?)) rolled down a slope, would you measure the velocity of the various cylinders at the end of the slope and afterwards calculate the moment of inertia? (mgh=½mv^2+½I*w^2 , w=v/r)
I'm about to write a rather large paper on this subject, but I am not quite sure about an experiment.
(You would of course know the angle of the slope, the mass of the cylinders and their radii)
I apoligize for any spelling mistakes, I'm not native. Thanks.
Thanks sir for this lecture.
Thanku very much sir for such a nice explanation .
Glad it was helpful.