(III) A long vertical hollow tube with an inner diameter of 1....

Question

(III) A long vertical hollow tube with an inner diameter of 1.00 cm is filled with SAE 10 motor oil. A 0.900-cm-diameter, 30.0-cm-long 150-g rod is dropped vertically through the oil in the tube. What is the maximum speed attained by the rod as it falls?

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, an elongated cylindrical container has an internal diameter of 0.80 centimeters and is filled with Glycerin, a cylindrical rod that has a length of 25.0 centimeters and a diameter of 0.70 centimeters is dropped vertically through the Glycerin. The rod has a mass of 100 g, determine the maximum velocity the rod will achieve as it falls through the Glycerin given that the viscosity of Glycerin is 1.49 newtons multiplied by seconds per meter squared. So it appears that our end goal is we're trying to figure out what the maximum velocity that this specific rod will achieve as it falls through the Glycerin given all the conditions provided to us by the problem itself. So ultimately, we're trying to figure out what this maximum velocity of the rod is awesome. So with that in mind, let's read off our multiple choice answers to see what our final answer might be. And let us note that they're all in the same units of meters per second. So A is 4.9 multiplied by 10 to the power of negative two, B is 5.4 multiplied by 10 to the power of negative two C is 6.0 multiplied by 10 to the power of negative two and D is 6.5 multiplied by 10 to the power of negative two. OK. So first off, in order to solve this problem, we will need to recall and use the principles of fluid mechanics. Specifically, we will need to focus on the balance between the viscous drag force and the gravitational force acting on the rod itself. So now at this point, we need to write down all of our known variables. So we know that the inner diameter of the container and let's call this capital D. Once again, this is the inner diameter of the container is equal to 0.80 centimeters. Awesome. So capital D, the inner diameter of the container is 0.80 centimeters. We also know the diameter of the rod which we'll call lowercase D and the diameter of the rod is equal to 0.70 centimeters. We also know the length of the rod and we're gonna call the length of the rod H and this is equal to 25.0 centimeters. And this is also equal to 0.25 m. So I've gone ahead and given you the conversion, but you can use dimensional analysis or you can quickly look up this conversion yourself. But in order to save time, I just given you the conversion, we also know the mass of the rod, which we're gonna call lowercase M is equal to 100 g, which is also equal to 0.10 kg. And we know that the vico viscosity of Glycerin, which we're gonna call this value Ada and Ada is equal to 1.49 newtons multiplied by seconds per meter squared. And finally, we know that the acceleration due to gravity which we're gonna call G is equal to 9.81 and its units are meters per second squared. So now moving right along here, we now need to draw a diagram to help us better visualize this problem. So creating a diagram of this prom, we will draw our cylindrical shaped container which we have our cylinder in the center. And so the cylinder and then we have our rod in the center. So we have our walls which just to keep things simple, we'll just pretend like we cut off the tops of our cylinder. So we just have like a rectangular shape. So inside of our cylindrical container, we have our rod which is also a cylinder shape. And the of course, we have R which is the radius of our rod here. And then we have the force of the viscosity pointing in the upwards direction. And then we also have the weight force pointing, pushing downwards, pulling downwards, which is the mass of the rod multiplied by the acceleration due to gravity. Because when you drop it into this container, we have the weight of the rod pulling it down towards the bottom of the container. And then we have some length lowercase L which is the distance between where the rod is inside of our container to the outer wall of the container. So it's from where the rod is to where the edge of the container is. And then of course, we have H which is the length of our rod. Awesome. So now that we have a visualization of this part of this problem, we now need to calculate the thickness of the fluid layer, which is L in this case. So that's the thickness, that's the distance between our outer edge of the container to where the rod is. So that's the thickness of the fluid layer. And the radius R value is, we do not know what the radius value is. So let's start solving for a lowercase L first. So we need to recall and use the following equation to solve for lowercase L. So let us note that L is equal to one half multiplied by capital D minus lowercase D. So we can now plug in all of our known variables to solve for the numerical value of L. So we need to take one half and we need to multiply it by our value of D which is 0.80 centimeters. And we need to subtract it from lower case D which is 0.70 centimeters. So let me plug that into our calculator. We will find that L is equal to 0.050 centimeters which is also equal to 0.050 multiplied by 10 to the power of negative 2 m. Awesome. So now we need to recall and use the equation to help us solver the radius. So let us note that the radius is equal to the diameter D which we're using lowercase D divided by two which plugging in our known values, we take 0.70 centimeters and we need to divide it by two which will give us 0.35 centimeters, which is also equal to which do give us a little bit of space. We can also say that R is equal to 0.35 multiplied by 10 to the power of negative 2 m. Fantastic. So now at this point, we need to set up the balance of forces when the rod reaches the terminal velocity. And let us recall and note that the vis force which is F subscript vis is balanced by the weight which let's recall the equation for weight, which weight W is equal to the mass multiplied by the acceleration due to gravity. We also need to note and recall that ada multiplied by capital A multiplied by V divided by L is equal to M multiplied by G where capital A in this case is the area of the cylinder. And as we should recall, the area of a cylinder is equal to two multiplied by pi multiplied by R multiplied by H. So now at this point, we need to rearrange this equation to isolate and solve for V which is our final answer. So when we do that using a little bit of algebra, we will find that V is equal to M multiplied by G multiplied by L all divided by ada multiplied by capital A. We now need to plug in all of our known variables to solve for V, the numerical value of V. So let's plug in all of our known values. So we know that the mass is equal to 0.10 kg. We also know that G the acceleration due to gravity is equal to 9.81 m per second squared. We also know that the value of L is equal to 0.050 multiplied by 10 to the power of negative 2 m. And the reason why we had to convert all of our centimeter units to meter units is because we're trying to find our final answer in meters per second. So we have to make sure we have everything converted to the proper units. And we need to divide this by our a of value which are a of value is equal to 4.9 and its units are newtons multiplied by seconds per meter squared. Awesome. And now we need to multiply this by the area of our cylinder. So we need to note that it's two multiplied by pi multiplied by zero point 35, multiplied by 10 to the power of negative 2 m. And we need to multiply this by and I running out of room here. But let's see if we'll be able to fit it. 0.25. So it's multiplied by 0.25 m. Awesome. And don't forget your bracket. So let me plug that into our calculator. We will find that V rounding to one decimal place is equal to 6.0 multiplied by 10 to the power of negative 2 m per second. And that's it we've solved for this problem. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter C 6.0 multiplied by 10 to the power of negative 2 m per second. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

(III) A long vertical hollow tube with an inner diameter of 1.00 cm is filled with SAE 10 motor oil. A 0.900-cm-diameter, 30.0-cm-long 150-g rod is dropped vertically through the oil in the tube. What is the maximum speed attained by the rod as it falls?

Key Concepts

Viscosity

Terminal Velocity

Drag Force

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