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Problem 92 Hard Difficulty

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that $a_x$ and $a_z$ are approximately zero and $v_x$ and $\omega_z$ are approximately constant. Rolling without slipping means $v_x = r\omega_z$ and $a_x = r\alpha_z$ . If an object is set in motion on a surface $without$ these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass $M$ and radius $R$, rotating with angular speed $\omega_0$ about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is $\mu_k$. (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations $a_x$ of the center of mass and $a_z$ of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially $\omega_z = \omega_0$ but $v_x =$ 0. Rolling without slipping sets in when $v_x = r\omega_z$ . Calculate the $distance$ the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

Answer

a. $\frac{2 \mu_{\mathrm{k}} g}{R}$
b. $\frac{R^{2} \omega_{0}^{2}}{18 \mu_{\mathrm{k}} g}$
c. $-\frac{1}{6} M R^{2} \omega_{0}^{2}$

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Video Transcript

it's kind of an interesting problem, Kind of a somewhat difficult problem. Um So I kind of read through it here. So when we have an object rolling without slipping, the rolling friction forces much less than the friction force when the object is sliding. Because there's really no relative motion here. Because we're assuming at this point down here when ruling without slipping is stationary, so there's really no relative motion. Um And so they say a silver dollar rule on its edge much farther than it will slide on a flat surface. Which we can, you know um Just observe frilly quote fairly readily. When an object is rolling without slipping on a horizontal surface, we can approximate the friction for us to be zero again. And that's because there's basically no relative motion between the what's rolling in the surface so that the um um that the accelerations are approximately zero and that the velocities. So the the acceleration and the angular acceleration how about zero. So this is this is close to zero. So if we were you know, just rolling without slipping with no other, we're not accelerating that just ruling at a constant velocity. Um and that the velocity and the angular velocity are approximately constant so that then rolling without slipping means that the velocity equals r times the angular velocity and the acceleration equals r times the angular acceleration. So that means again, that's basically assuming that this point is stationary at this point here is instantaneously is not moving. So we said, if an object is set in motion on the surface without these equalities, sliding friction will act on the object as it slips along until the rolling without slipping is established. So what we have here, because we have a cylinder okay with the mass capital and what the radius capital are that's rotating with an angular speed of omega not. So it's basically spinning here. No, no friction force, It's just kind of spinning there. Um Kind of like say say you have a you know like a tire spinning right? That it's, you know, it's um just spinning freely. I'm slipping on the pavement. Um So it's angular velocities. Omega not about. Access to the Senate is set on a horizontal surface for which the kinetic co affection is um UK. So now we have this thing spinning but it's set on a surface but we actually have we have some friction here, so we don't have no friction here but we have a little bit of friction. And the direction of this friction force is kind of going to change as we go through this problem. But I just drew up to the right as positive for now. So what we have then, So basically we have this thing spin and we set it down on something and it kind of spins spins and it moves because there's friction here. Um But it's um it's slip is still slipping. But then all of a sudden it gets up to a speed such that um it gets up to a speed that's at this point here has a um zero as the over Lasky which means that it all of a sudden starts rolling without slipping. Um Because basically it's the static coefficient of friction than that comes into play for that case which is going to be greater. So we have the normal force equals mass times gravity. We have the friction force is the static coefficient are the kinetic coefficient of friction times the weight or the normal force. Okay so again we have slipping here. So this kinetic that means that the only friction for the only force here is the friction force. Um And so that has to be the mass times acceleration and that gives us that a equals mu k times G. So that's what that's what when we set this thing down spinning um it starts to accelerate and that acceleration is the kinetic coefficient of friction times gravity. Now, if we do a moment balance on here, um we get that minus the friction force times are equals the mass moment about the center of mass here, Which is 1/2 are squared times the angular acceleration. Now. Um let's see here the friction force, you know, is this value here which tells us the angular acceleration um is let's see here. Did I think I missed the sign in here? Yeah, well yeah, I missed the sign. This should be plus there should be plus that's what I did. Yeah. So I am I basically changed the signs on this when I from my notes and I didn't copy everything down. So again, obviously, clearly Alpha is minus to a u K G over our so we have some angular acceleration but it's actually in this direction because this force is um is um in the opposite, you know, is in this direction here. So you know, it's basically saying that you know, it's acting to slow things down, right? Because if it's spinning this way and we start having a force on it this way, it's acting it's gonna act to slow things down right now. The final velocity when we start rolling without slipping is you know we have an acceleration here and then we have some time. So you know after some time we don't know what this is yet. We start rolling without slipping. And then we also know that when we start rolling without slipping that that velocity is going to be our omega. Yeah, we know omega is going to be omega not plus Elfatih, but it's going to be slower because alphas alpha is negative. It's gonna it's gonna slow down, right, this is going to slow down, but then it's gonna, you know, it's gonna some of that energy is going to go into causing it to move forward. So that says at the time that it takes again assuming that these are you know, constant accelerations. The time that it takes is omega not Times are all over a plus alpha time. eight over a um And again I should this is a minus sign in here. Yeah, yeah, that's a minus sign. So let's see here we have solving for tea. We get um what we have there. Yeah, so a minus our alpha. Yeah. There we go. Now. We can plug things in. So we have we can plug a N. We can plug Alpha in and then we get after we simplify we get T equals a mega, not are over three mu kg. So this is the time it takes to go from you know when it's spinning and slipping to then when it starts to roll without slipping. So it slows down enough. The angular velocity slows down enough and the angular speed of a lot and the linear velocity grows enough so that now we have rolling without slipping. And from that point on like I said, it should rule without slipping because mu static is what's creating it's creating the friction with rolling without slipping. And that's going to be greater than you kinetic. All right. So they ask us how far will it go? Well, we know that it has um you know, the distance, the distance that it goes is just one half times the acceleration times t squared Because it initially started with zero plenty of velocity. Um, we know a from here and we know t from here so we can plug everything in and we get um d equals a mega, not square times are squared all over 18 Mieux Que Times G. So, what, what can we say about this and this? Same? No, um you know, he goes up as a mega, not goes up. So, as if it's spinning faster, it's going to take longer to start rolling without slipping and it's going to go farther without ruling without before it starts rolling without slipping, which makes sense. It's spinning faster and faster. Asthma UK goes down the time goes up. And so there's the distance because basically it takes longer to get that translate that rotation because we have a very slippery surface. And in fact, um if the UK goes down, if BK goes to zero, then this goes to infinity. Because basically if U k is zero, then it's never gonna basically just gonna sit there and spin right, because there's no friction to cause it to, you know, try to accelerate. So it's just gonna sit there and spin, which means that he's going to go to zero, oh he's gonna go to infinity and um d is going to go, let's see here, he's also going to go to infinity. Well, basically, basically says it would take an infinite distance because it takes an infinite time. So it's kind of a degenerate case there, But you can make this makes sense. He goes to infinity of UK goes to zero and then basically, you know, this would have to go to infinity but it's not gonna move so it's never going to get there. Um Is that all makes sense? I think obviously units work out um you know, G comes in here and because that's just because of how the friction force works here. Um So that that that makes sense now they also ask us um let's see here, calculate the work done by the friction for us on the cylinder as it moves from where it was set down to where it begins to rule without slipping. So um you know this forest, this forces moving because as this thing starts to move, that forces moving so it's going to do some work and how we can figure out what that work is is by looking at the change in kinetic energy. So we have our initial kinetic energy, It's 1/2 times J kinds of meghan, a square and our final kinetic energy. So it was it was not translating. So just rotational kinetic energy. And when it starts, when by the time I started to roll without slipping and has some translational and rotational kinetic energy. Now we see that we have values, we know that we is eight times t. So this would be VF. I call this on mega F over here. Um I probably should have called this omega F basically the final the angular velocity when it start starts rolling without slipping. So let's call that make enough. Um So then we have translational and rotational kinetic energy in that state. Now this thing we know is eight times T. And make A F. Is just VF over our and doing some algebra out here, we just get the KF is 34 M A. T squared, right? Um Well If we plug in 80 here, right? And then we can add these up and then, so now we know we know what A. Is and we know what T. Is from before. You know how long it took to start rolling without stepping. And we know the acceleration it has until that state. No, we um we can plug those in and so we get our final kinetic energy and after some algebra hear lots of things cancel out is 1/12 M. R squared omega not squared. Now we can see that we can take the difference in those things and that's going to be the work that this friction force did. And so we have um 1/12 M. R. Squared. America's not squared minus 1/4 M. R. Squared. Omega not square. So that's minus 1/6. M. R. Squared or mega not squared. And it's negative because basically we're doing work on the, What do you want to say the 4? We're basically we're doing work on the environment, the disk is doing work on the environment in getting this up to speed. So we've we've, what you can see is we've lost energy and going from this initial state where it's just spinning on the surface to where it's rolling without slipping. We've lost some energy and that energy has gone out into the environment through that friction force, which we would expect right friction. You know, you know, friction basically dissipates energy in our system. So we would expect that we've lost some energy and we've lost this amount here and it depends on the mass, the radius and the initial angular velocity. So an interesting problem. Um And again it comes out it's not a difficult problem actually, but we get interesting results here and again, you can see how the answers come out to be fairly simple um And it's it's not all that intuitive basically what's going on here, because it's kind of hard to think about to imagine what the system is going through. But you know, the fact that m comes into here, Well, that basically comes in because the bigger the larger the mass, the larger the friction force is going to be, so it's going to dissipate more energy. Why are squared comes in here. You know, this this looks like v squared right, like a velocity square. And so we kind of get, you know, the change with an energy to the initial, you know, this isn't the initial velocity, but it looks like a velocity squared. So we can kind of make that makes some sense that we get a velocity like term here.

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