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Two uniform rods, each of mass $m$ and length $l$…

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Problem 71 Easy Difficulty

Two uniform rods, each of mass $m$ and length $l$, are attached to gears as shown. For the range $0 \leq u \leq 180^{\circ},$ determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.

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Physics 101 Mechanics

Vector Mechanics for Engineers: Statics and Dynamics

Chapter 10

Method of Virtual Work

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Video Transcript

here we have two rods each of mass M that are attached to gears of equal radius, as shown in this diagram, which is clearly not to scale. Uh, that's we have Rod A B and Rod CD. We've defined fada as from the left horizontal here and from the vertical on the left, on both with Rob serve length L and MASS M. Um, we will. And we want to figure out what the positions of equilibrium of the system are and figure out in each case whether the delivery amiss, stable, unstable or neutral. And the way we're going to do that is we're gonna identify which forces can potentially do any net virtual work on our system. Uh, we were going to express those forces in terms of their potential energy. Also, that we can use the equilibrium petition that is the first derivative of the potential energy of the system with respect to fada is equal to zero. So we're gonna want to express all of these, uh, potential energies in terms of a single variable fada. So let's go ahead and do that, and we're going to go ahead and treat the potential energies of these rods pretty much just like you would treat a Ah, wait, um, potential energy due to gravity. So it's just we have effectively to forces w and w from the centers of masses of the rods going vertically downward, both of which article to just the mass of the Rod Thomas gravity. So as you may recall, the general form of ah, potential energy due to weight is just that Wait times the y coordinate of said, uh, object center of mass s so we can go ahead and write the total expression for our potential energy. Because we only have these two forces acting on the system. The gears turning aren't going to do any net displacement or work on the system. And there are no over forces present. Aside from gravity on the weights of these two rots. So our total potential energy is just again. The weight will do the left one first, and we need Thio. Define a bit of a coordinate system here for just what are y coordinate here is and so we're gonna go ahead and define horizontal access through be seeing the centers of Iran's such that we can go ahead and say this that this height right here, which is gonna be the why here is gonna be equal to, uh l over to sign Fada with Fada being as defined earlier. And we're gonna go ahead, make this negative because it's below are defined the origin. So it's gonna be negative l over to sign Fada, and then we're gonna add the weight of the second one. So again, just the same force W times again just have to rod for the center of mass l over two. And this time, because fate is up here, this would be we would want cosine of Fada because that's the hype here. So cosine veda. And that just simplifies thio w l over to cosign fada minus sign feta. And that is our general expression for the potential energy of this system. So we can go ahead and take the first derivative with respect to theta. Uh and that is just quickly again. These constants just go out first. And this is super easy because it's just trig, So it's just negative. Sign Fada minus coastline Fada, the derivative of coastline fade a minus sign Fada. And since we're going to go ahead and be checking the equilibrium conditions. Eventually, we're gonna go ahead and take the second derivative with respect to Fada. Because, as you know, as you may recall, the way we check if the equilibrium is stable, unstable or neutral is by evaluating the relationship of, uh, the second derivative to zero. If it's less than zero, it's unstable. If it's greater and zero, it's stable. If it's zero for all derivatives, it's a neutral equilibrium. So again, just taking a never derivative with respect to Fada again, these constants just stay out front and again. This is just pretty simple, because it's trig that now negative sign Fader becomes native coastline Fada and a coastline faded becomes sign Fada. So we'll just flip it for simplicity's sake, sign Fada minus cosign. Fada. All right, so next we wanna actually find that equilibrium and to find the equilibrium we're gonna use again that equilibrium condition of setting the first review for prospective ADA equals zero. So that tells us that we need Thio. Have ah, sign Fada equal to negative cosign Fada from just setting this above equation equal to zero and solving which in our terms is tan Fada equals negative one, which you can easily solved numerically to get to solutions. Fate of one equals negative 45 10 degrees and Fada two equals 135 10 degrees. Eso. Now we want to go ahead and check the stability of these two equilibrium points. Um, so again, we're gonna just plug these to fade us into our second derivatives. So we'll start with fatal one of negative 45.0 degrees, which again, makes sense. Um, it is physically possible in our set up, but again, this is the second derivative with respect to fada equals W l over two of sign of negative 45 degrees, uh, minus co sign of 45 degrees because co sign up negative. 45 degrees equals co sign of 45 degrees. Ah, on that is just equal to again W l over two of negatives. Rad to over two plus are sorry minus rad to over two. And even without plugging in these constants, we know there are positive. This is this is less than zero because this number is negative, so we can go ahead and declare that this negative 45 degree angle is unstable. And that makes physical sense. Because if we go back to our diagram, if you imagine you know these bars up here it I can't imagine there. I can imagine there being an equilibrium point there, but it would definitely be unstable. One Jocelyn, it's gonna fall down to this other one, which makes more sense if you know gravity and everything. So let's go ahead and check number two, just to be sure, 135 0 degrees again. Second derivative of the potential energy with respect to theta is equal to W L over to and this is gonna be just sign of 135 degrees, uh, minus co sign of 135 degrees. And as you may know, that's just gonna be equal to the positives of what we had last time. Rad to over two plus rad to over two, which again this time is just obviously positive. So we know that this equilibrium is stable and as a reminder, if it had equaled zero for all successive derivatives with respect to theta, that would be neutral. But we just quickly found the equilibrium point and the stability of said equilibrium points just by identifying which forces could do any virtual work on our system, expressing their potential energies and using the equilibrium condition on that's that.

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Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

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