A 2.20-kg hoop 1.20 m in diameter is rolling to the right without...

Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom.

Accepted Answer

Welcome back everybody. We have a tire that is rolling along a flat path or flat road. And we are told a couple different things about this tire, we're told that it has a mass of 9.8 kg That it has a radius of 289 mm or .289 m. We are told that it has a constant angular speed or velocity of 15.3 radiance per second. Now it is rotating in the clockwise direction, meaning that it has this velocity that is going off to the right here. Now we are asked to find the resultant velocity vectors at three different points right here, at the bottom where it comes in contact with the road on the left, middle side and at the very top. Well, we already established that it has this constant velocity moving to the right here. So we know that that's going to be part of our consideration in all three points. So, let me actually go ahead and draw that out here. If RV here, we have our V here and we have RV here and I will label them such, what else do we have here? Well, since we're dealing with rotation, we're going to have also a V. E or a tangential velocity. Let me change my color here, A tangential velocity. Now this tangential velocity is going to be equal in magnitude to our normal velocity, but but it is going to be perpendicular to the radius at each of these points in the direction of motion. So at the very bottom here, we're going clockwise. So we have to make sure that our VT goes clockwise as well, but it is perpendicular to the radius. So this is our V. T. Down here, our V. T in the middle is going to look something like this and then our VT right here. Well, it's just going to be in the same direction as our normal V. Now you see why we are asked to find the resultant velocity at all. Three points. So let me go ahead and label our points of part one are too in part three. Let's go ahead and start with part one right here. Now, just to make sure that we're on the same page here, I'm going to have the right direction, be the positive X axis, the upwards direction, be the positive y axis. Since we are adding components. That will be important here. So what is the resultant velocity at our art one? Right here at this bottom point? Well, we have our velocity going off to the right and then we have plus our velocity going off to the left. But that's just going to be minus our tangential velocity, filling in our values here while we know that our velocity is equal to our tangential velocity and we're told that it is. Well, actually we have to calculate that. Let's calculate that real quick. We have a formula for our tangential velocity is given by a radius times our angular acceleration. So we have these numbers, let's go ahead and plug them in. We have point times if see radiance per second, which gives us a velocity and tangential velocity of 4.42 m per second. Great. So now let's go ahead and plug this value in. So our velocity 4.42 minus are tangential velocity of 4.42. We have that a resultant velocity at 0.1 is zero m per second because they cancel each other out. Great. So now let's move on to part two. What? We have a velocity going in this direction. Always to the right, but then we have our tangential velocity going upwards. So a resulting velocity is going to be in this sort of direction. So we're actually gonna use Pythagorean theorem or pythagoras theorem right here. So korean theorem states that we are going to have the square root V squared plus b t square To find the magnitude of the resultant vector. This is going to be equal to the square root of 4.42 squared plus 4.42 squared. Which when you plug into your calculator, we get that our results into philosophy vector for part two is 6.25 years per second in that general direction. Right there. Great. And now moving on to Part three, we are going to add in the X direction. Our two vectors this time they're facing the same way. So we will truly add them. So we will have 4.42 plus 4.42, giving us a result in velocity vector of 8.84 m per second. So now we have found our result in velocity at all three points, which corresponds to our answer choice of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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