(a) Above what speed does a 3.0-mm-diameter ball bearing in 20°C water experience quadratic drag?

Video transcript

Hey, everyone. So this problem is dealing with drag forces. Let's see what they're asking us. A sphere of radius 2.5 cm is dropped with an initial speed of V in sunflower oil at room temperature. And they give us that the viscosity of the oil given by Ada oil equals 45.4 times 10 to the negative three pascal's per second. And the density Of that oil is 916 kg per meter cubed. They ask us to determine the minimum speed of the sphere at which the sphere will encounter quadratic drag. Our answer choices here are a 0.83 m per second. B 0.99 m per second. C 1.66 m per second or D 1.98 m per second. So for this problem, the first thing that we need to do is recall that quadratic drag indicates a Reynolds number of at least 1000. So the minimum speed would be at a Reynolds number equal to 1000. So we're going to set our Reynolds number equal to 1000. And that looks like we can recall that the equation for Reynolds number is R E equals grow the D over A to. And so from the problem, we have Our density kg per meter cubed RV is what we're solving for R D is our diameter. So we can recall that the diameter is two times the radius. The radius was given to us as 2.5 cm. I'm gonna rewrite that 0.025 m to keep us in standard units. So that's gonna give us a diameter of 0.05 m. and our viscosity was given as 45.5, 45, sorry 45.4 Times 10 to the - pascal's times seconds. And then from here it's just a plug and chug. So 1000 equals 916 kg per meter cubed multiplied by our Diameter of 0.05 m over our viscosity of 45.4 times 10 to the negative three Pascal's time second and all of that multiplied by our speed. So when we solve for the speed, We get a speed of 0. meters per second. And that aligns with our answer choice B So that's all we have for this problem. We'll see you in the next video.