The 10 mg bead in FIGURE CP8.69 is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ω꜀, the bead sits at the bottom of the spinning loop. When ω ω꜀, the bead moves out to some angle θ. a. What is ω꜀ in rpm for the loop shown in the figure?

Question

The 10 mg bead in FIGURE CP8.69 is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ω꜀, the bead sits at the bottom of the spinning loop. When ω > ω꜀, the bead moves out to some angle θ. a. What is ω꜀ in rpm for the loop shown in the figure?

Diagram of a rotating wire loop with a bead, showing angle α and radius R.

Accepted Answer

Welcome back, everybody. We are making observations about this toy right here. Now, just to be clear, this is a sphere that is free to rotate about this vertical axis. And on the inside you have a pearl that is connected to a string connected to the center of our sphere here. Now, we are told a couple of bits of information about this entire system. We're told that this are right here. The distance between the center of the sphere to the pearl is equal to 35 centimeters or 350.35 m. We're told that the mass of the pearl is eight g or 80.8 kg. And a child who is playing with this toy makes an observation that there's a certain angular velocity threshold which we will call omega a. And if you spin this sphere fast enough, then passed this threshold, the pearl will start sliding up the side of the sphere. And we are tasked with finding what that threshold is. Well, before we start trying to find a formula for this thing, let's highlight some important forces at play here. Of course, on the pearl, we have the force of gravity. But then we also have the normal force which is going to act in the direction along the line of the rod or string connecting to the pearl and the center of the sphere. So I'm gonna label that our normal force right here. So first and foremost, let's take, let's apply Newton's second law in the radial direction here. So we have that the sum of all forces in the radial direction is equal to the mass times the radial acceleration here. So what's acting in the radial radial direction? Well, we have the normal force but it's not just the normal force is normal force times the sine of our angle alpha right here. This is going to be equal to our mass times are angular acceleration. But I'm going to thumb in uh the formula V squared over R as that is equal to angular acceleration. Another substitute I'm gonna make is that we can sub in our times our omega is equal to V and we can put this right here. This gives us a mass times. Let's see here, our radius times our omega squared. Now, I want to point out something important here. This radius is not the same thing as this radius. What this small R is referring to is as we rotate this based on the position of the pearl. There's gonna be another circle made right here. And this is the are that we are referring to just to be super clear about that. So now what I'm gonna do is I'm gonna divide both sides by the sign of our angle alpha. And you will see that our N is equal to our normal force is equal to this value right here. I'm going to call this equation one. So now we should find another equation that we can set equal to that to solve for some of these variables. So this time I will apply Newton's second law in the vertical direction. So this is just equal to mass times acceleration. So what does this translate to? Well, in the vertical direction, we have our normal force this time times the co sign of our angle minus our force due to gravity. And what is this equal to? Well at this threshold speed? This pearl is not going to be moving up or down, meaning that it's not going to have an acceleration in the vertical direction, meaning this side is zero, adding our force due to gravity, the both sides and then dividing both sides by R cosine of our angle. You will see that we get that our normal force is equal to MG times the cosine of our angle alpha. And this I will call equation to. So now what we can, we can do since we have found two equations for the normal force is we can set these two equations equal to one another. Let me scroll down here just a little bit. To give us some, some room here. So let's go ahead and do that. I'm gonna change colors for this. So we have that our mass times our small, our times our omega squared divided by our sign of our alpha is equal to MG divided by the cosine of our angle alpha. I'm gonna move this over just a little bit here. All right. So what can we do from this point? Well, if you see we can divide by M on both sides and the mass cancel cancels out here. And what we need to do is we need to isolate this term right here. So what we can do is we can multiply both sides by sine of alpha divided by our little our, but I'm actually gonna make it another substitution here. You'll notice looking at our little are here when this rotates this R is the same thing as our angle alpha or the sign of our angle alpha times our radius of our sphere. So just to write that out, we have that our little R is equal to R big R times the sine of our alpha. So I'm gonna multiply by that on both sides as well. This cancels out with, with this, on this side and this cancels out with this on this side, but we need to do this to both sides. So then we have sine of alpha divided by R sine of alpha. So this means that we have our omega squared is equal to this sine of alpha on the top and bottom cancel out. So then we're just left with our acceleration due to gravity times our radius of R C R times the co sign of our angle alpha. To finally get rid of this power, we just take the square root of our side. And then we have found a formula that we can work with here. Now, we can simplify this even further. If we just think about this conceptually here, the ideal threshold for this will be when our co sign of our angle alpha is equal to one. So what we can do is we can just sub this into here and then just simplify the omega that we are look To the square root of our acceleration due to gravity divided by the radius of our sphere. Let's go ahead and plug in some values here and that is the square root of 9.81 divided by our radius of our sphere of .35 giving us 5. radiance per second. But we want our answer in revolutions per minute. So let's go ahead and convert here. We know that there are 60 seconds in one minute. And we know that there's two pi radiance in one revolution. So when you multiply straight across, we get that our final angular velocity or are critical angular velocity is equal to revolutions per minute. Corresponding to our answer choice of c Thank you all so much for watching. Hope this video helped. We will see you well, in the next one.

Verified Solution

Key Concepts

Centripetal Force

Angular Velocity (ω)

Critical Angular Velocity (ω꜀)