(II) A potter’s wheel is rotating around a vertical axis throu...

Question

(II) A potter’s wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev/s . The wheel can be considered a uniform disk of mass 5.0 kg and diameter 0.40 m. The potter then throws a 2.8-kg chunk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, a uniform disk that is oriented horizontally is spinning about a vertical axis through its center at a rotational speed of 2.4 revolutions per second. The mass of the disk is 3.0 kg and its diameter is 0.60 m. While the uniform disk is rotating another smaller disk that is initially at rest and has a radius of 10 centimeters and a mass of 1.5 kg is placed on the center of the larger uniform disk. As a result, both of the disks begin to spin together about the same axis, determine what the final rotational speed of this particular system will be. So that's our end goal is we're trying to determine what the final rotational speed of this particular system will be awesome. We're also given some multiple choice answers. Let us know that they're all in the same units of revolutions per second. A is 2.1 B is 2.3 C is 13 and D is 14. OK. So first off, let us note that since no external torque is applied to the two disc system, the angular momentum will be conserved as a result. So let us also note that since the smaller disk is initially at rest, the initial angular momentum of the smaller disk will be zero. Let us use the subscript lowercase I to denote the initial values and use the subscript lowercase F to denote the final values moving forward when we're writing our equations. Thus, we could write that the initial angular momentum which is gonna be L subscript I is equal to the final, which is angular momentum. So once again, since we're saying that the angular momentum is conserved, the initial angular momentum is equal to the final angular momentum. Therefore, we can also write that I subscript or I should say capital I subscript lowercase I multiplied by Omega lowercase I or I'm just gonna say I to keep things short is equal to if multiplied by omega F which when we rearrange that equation to isolate and sulfur omega F, we will get that Omega F is equal to I subscript I. So I, I multiplied by Omega I which we're gonna then divide that by I F Awesome. So moving right along here. So now we need to consider the larger and smaller disks. So let us use capital L to denote the larger disk and S to denote the smaller disk. Therefore, we could write that Omega F is equal to IL where I haven't said yet. Just in case we're confused, we'll take a moment here and pause. So I denotes the moment of inertia. And Omega in this case will denote the angular frequency. That's what Omega stands for. So Omega I is our initial angular frequency and Omega F is our final angular frequency. So now we're considering I subscript capital L is the moment of inertia for the larger disk. And then we're gonna divide this by the moment of inertia for the larger disk plus the moment of inertia for the smaller disk. And then we're gonna multiply that whole division by Omega I which we can expand this and we can write that Omega F is equal to and we're gonna put a bracket and we're gonna say one half multiplied by the mass of the larger dis multiplied by the radius of the larger disk squared. And this is all divided by, we'll put another bracket, one half multiplied by the mass of the larger dis multiplied by the radius of the larger dis squared plus one half multiplied by the mass of the smaller disk multiplied by the radius of the smaller dis squared. So what we did is is we expanded our equation. So we wrote out what the equation values for the moment of inertia for the larger and smaller discs are. So we just wrote out what the equation for the moment of inertia is. And since we're dealing with a uniform disk, that's why we're using one half multiplied by the mass multiplied by the radius squared. Awesome. And don't forget that this entire equation or this entire division is all multiplied by Omega subscript I OK. So moving right along here, so we can say that Omega F is equal to ML. So we need to take the mass of the larger disk and we need to multiply it by the radius of the larger dis squared. And we're gonna divide this all by the mass of the larger dis multiplied by the radius of the larger dis squared plus the mass of the smaller disk multiplied by the radius of the smaller dis squared. And we're gonna multiply this all by Omega I. So we also need to note at this point that we now know that and let's write this little side note in blue, let's add a little star here that Omega is equal to two multiplied by pi multiplied by V. Thus, we could write that two multiplied by pi multiplied by VF which V in this case is the velocity. So VF is the final velocity. And we can say that this is all equal to the mass of the larger dis multiplied by the radius of the larger disk all divided by. And I'm just gonna save time and just say ML multiplied by RL squared plus MS multiplied by RS squared. And this is all multiplied by two multiplied by pi multiplied by V I. So now if we rearrange and isolate, so we need to rearrange this equation to isolate and solve for VF which will be our final answer that we're ultimately trying to solve for. So when we rearrange it just to solve for VF, we will get that VF is equal to M L multiplied by RL squared, all divided by ML multiplied by RL squared plus MS multiplied by RS squared. And this is all multiplied by V I which is our initial velocity. So now at this stage, we can plug in all of our known variables and solve for VF our final answer. So let's do that. So VF is equal to 3.0 kg means that's the mass of our larger dis multiplied by 0.30 m. And don't forget that uh that's our value for our radius squared. So we converted. So are given our radius. As we see, let's go back up to the top here. So we're given that the radius is 10 centimeters and then the diameter. So we're told the mass of the larger disc is 3.0 kg and the diameter is 0.60 m. So we have to just divide the diameter by two to get it into the radius. So the diameter divided by two gives us the radius. So that's where we got 0.30 m. Awesome. So moving right along here everybody. So now we need to divide everything by 3.0 kilograms multiplied by 0.30 meters squared plus 1.5 kg multiplied by these. That's the mass of our smaller disc multiplied by 0.1 m squared. And don't forget your bracket and don't forget that we have to multiply everything by 2.4 revolutions per second is that is the rota, that is the value of V I that we're given in the prom itself because that's the rotational speed that's given to us in the prom itself. So when we plug that into our calculator, we will find that VF is equal to 2.274 and it's gonna be revolutions per second, which is approximately equal to 2.3 revolutions per second when we round to one decimal place or when we round to two significant figures. And that's it we solve for this problem. Hooray. So looking at our multiple choice answers, the correct answer has to be the letter B 2.3 revolutions per second. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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