PHYS 201 FK4 FK5 SK3 Solution S2026

追加: 2026-05-15

[00:00:00]Okay, let's go over today's quizzes.

[00:00:03]FK4A. A blender is turned on. Motor starts from rest. So, omega initial is zero. And it has a constant angular acceleration, alpha. It reaches an angular velocity of 21 radians/s after rotating through 30 radians. What is alpha? What's the angular acceleration?

[00:00:21]So, we'll use a cube formula. Omega f^2= omega initial plus 2 alpha delta theta.

[00:00:28]Omega finals 21 radians per second.

[00:00:30]There is no omega initial 2 alpha 30 radians is delta theta and you just solve for alpha. Once you do that alpha is 7.35 radians/s squared. FK4B is similar. All right it's this time it is running at a constant speed of 19 radians per second and then it turns off and slows down. Um stops at rotating through 50 radians. What is alpha? So same cute equation. Omega final is now zero. Omega initial is 19 radians per second. Square that plus 2 alpha * 50 radians your delta theta. Right? Solve for alpha and you get -3.61 radians/s squared.

[00:01:16]EK5A. Okay. We're going to compare the angular velocity and the linear velocity. Linear velocity being the tangental velocity of these three dots on this spinning rotating wheel. Since everything is on the same wheel, then the omegas are actually the same. A= B= C. But remember that your tangential velocity depends on R * omega. So the further out the radius, the higher the tangental velocity. We talked about this in class with like longer legs versus shorter legs. Since A has the shortest radius, A is the smallest. B and C have the same radius. So VC equals VB. And both of those are greater than BA.

[00:02:00]Similar for FK5B. All right. All the omegas are the same 1 2 3 cuz same wheel, same spinning, but three is on the outside and one or two on the inside. So three has the largest tangential velocity. Two and one are equal because they're at the same radius.

[00:02:19]Okay, I was inspired by the homework for this one. So, problem 65, you have a centrifuge rotating at 4,000 RPM and the test tubes are 12 cm long. So, we want to find out what the centrifal acceleration, what alpha is um at the end of these test tubes. So, first we're going to convert to radians per second.

[00:02:40]So, 4,000 RPMs converts to 418.88 radians/s. So here's a conversion basically time 2<unk>i / 60. Centrifal acceleration is r omega^ 2. So the radius is.12 m 12 cm needs to get converted. Multiply that by the radians per second. So 21,55.25 m/s squared. That's a lot of acceleration. But hey, that's what you need to separate out these compounds and precipitates and whatnot. All right, so you got the sp centrifuge. And now we're going to turn it off. The acceleration is50 radians/s squared. How long does it take to stop? So we're going to use omega final omega initial plus alpha delta t. So 0= omega initial plus alpha delta t. We're going to solve for delta t. So minus omega initial over alpha. So we have our radians per second. Our acceleration. The time it takes to stop is 8.38 seconds.

[00:03:43]Lastly for SK3B similar problem except this time 3500 RPMs. We're going to convert that to 366.52 radians/s. So RPMs * 2<unk>i / 60. Centrial acceleration in this problem the radius is 9 cm. So 009 m* the radians/s that gives us 12,090.32 m/s squared. still quite a bit but you know a lot less than at 4,000 RPMs.

[00:04:14]This question is the centy is turned off. We have his acceleration is50 radians per second squar how many radians does it go through to stop. So we're going to use the Q equation for this omega squ= omega initial plus 2 alpha delta theta 0 = omega initial square plus 2 alpha delta theta. We're solving for delta theta. Why can I not say this? Uh omega i^ 2 / 2 alpha. We plug in our radians/s square that / 2 *50. And del delta theta is 1343.37 radians. Okay. So that's what we have here. And that was today's quiz. It was a little long, but hopefully you feel good about