University of North Carolina at Chapel Hill
Normal Forces - Example 4


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Video Transcript

welcome to our fourth example. Video on normal forces in this video will consider a special problem, which is the loop the loop now in the loop The loop problem. What happens is you're writing a unit cycle or some other wheeled vehicle, preferably and you want to come and go all the way around the loop, the loop without falling on your head. Very difficult task and one for which we can do all sorts of fun physics. Now, the key to this problem is that notice here that when you're on your bicycle, you're always going to have a normal force pushing on you due to the ground beneath you. So, for example, when you are on your side over here, right in your your unit cycle, the normal force will actually be pointed towards the center of the circle. When you're above it, it will be pointed toward the center. When you to the other side, it will be pointed towards the center. Normal force is going to act kind of like a centripetal force. It's always center pointing now. It is not the only force that we have to think about here. Let's let's go ahead and assume that you are not exhibiting any force to make yourself go faster around. So you're not increasing your orbital speed at all. And but you are, That is so you don't have any tangential force, but you are attempting to go all the way around with some initial velocity that you had as you come in in this case, then you need to be in a situation. And so this is your most difficult position. You need to be in a situation where you're going fast enough so that you can coast all the way around. In order to do this, we need to make sure that we're still touching the top. At that point, let's go ahead and draw our free body diagram at the top. When we're at the top, we know we have force of gravity down, but then we also have a normal force down. If we were on the side, though, we would have had gravity down and normal force towards the center. If we were on the other side, the normal force would have pointed the other direction. So if we call these positions 12 and three, then this would be positioned to this would be position one and then positioned three. Well, look, something like this. So this is an interesting situation here. How How do we handle this? Well, uh, because this is the most difficult situation for us to handle. What we need to say is, OK, well, I have to force is pointing down. I have negative normal force and I have negative gravitational force and I need to set that equal to M A. Well. In this case, our acceleration is a centripetal acceleration. The center pointing acceleration Centrepoint acceleration that is equal to If you've watched the videos on rotational Kinnah Matics, we have v squared over r or it could be equal to omega squared times are so let's go ahead and plug that in then we have. And also I should point out that the centripetal acceleration at the very top is going to be pointed straight down as well. So we'll also refer to that as negative. So these air three forces or other two forces in a line that lead to an acceleration that's in the same direction is the two forces. That's not an unreasonable thing to say, though it does tend to throw people off. So here we have negative M V squared over art. Now the condition to not fall. So are condition to not fall Is that normal force is greater than zero. That's all we need. It just has to be greater than zero because that means we're still touching the ground. We've ceased to touch the ground. That's when we've fallen. So let's take a close look. Here we have our equation f n negative fn minus F G has to be equal to negative m b squared over r. We can cancel out all the negatives we say I need FN to be greater than zero. If n needs to be. Griffin needs to be greater than zero. Then we need M V squared for r minus F g to be greater than zero, which means we'll need to find that the has to be greater than the square root of f g times are over. M. Remember that F g is mg, so this is gonna be equal to the square root of our time. She and this is the minimum speed that you need to get around the loop. The loop

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