12. Rotational Kinematics

Rolling Motion (Free Wheels)

# Speeds at points on a wheel

Patrick Ford

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All right. So here we have a car that accelerates from rest for 10 seconds. So the initial loss of the car zero and it takes 10 seconds accelerating. Okay, It's tires will experience eight radiance per second. So V is the speed of the car. Let's draw a little car here. V is the speed of the car. This is a really crappy car. Um, but W has to do with the wheels. If the car is moving that way, the wheel is spinning this way. Okay, um, I'm giving you the acceleration of the wheel, so I'm gonna put it separately here because this is linear. And I'm gonna make a column here for angular. Uh, Alfa equals eight. Uh, if the car is initially at rest, Omega initial is also zero. Um, the tires have a radius off 0.4 m, so I'm gonna write it down here that the radius of the tire is 0.4 m. You see that? Yes, you can. And we want to know what is the angular speed of the tires after 10 seconds. So, after 10 seconds, what is Omega Final Four? The tires. Okay, that's part a. So this looks like a motion problem. And it is. I got three motion variables here that are given, and I'm asking for one and one of them is ignored. What's ignored here is the number of rotations. Delta Theta is my ignore variable sad face. This tells me that I should be using the first equation or make a final equals Omega initial plus Alfa Tea or make initial zero Alfa is ate. Time is 10. The answer is 80 80 radiance per second. By the way, nothing new in this question. We've done stuff like this before. The part that's news Part B, Part B were being asked for the speeds at the top center and bottom of the tire. Well, the tire is a tire is a rolling is enrolling motion. Or you can think of it as the tire is a free access or free wheel. It's call it free wheel. So this means that on top of the other equations, we know we're also going to be able to use the three equations that we just learned. So the top will simply be, um, to our Omega. To the Radius is 0.4, Omega's 80. Um, the bottom of the center of mass in the middle is one our omega, and the bottom is zero always. Okay, so if you multiply the top, you get 64. The top is double what's in the middle. So the bottom the middle must be 32. And the bottom is zero. Okay, so that's it for this question. We have to find Omega find, which is old stuff. Then we have to find the top V C. M. And the bottom. And we got them. Okay, let me out of the way here. So you can see numbers. And that's it for this one. Hopefully, this makes sense. Let me know if you guys have any questions.

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