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QUESTIONAsyouride on Ferris whezpparencwe ghtis different the [OP Fnar a1-h coton Calculate your apparent weightatche IOdanc botzom of Ferris wheel given that the r...

Question

QUESTIONAsyouride on Ferris whezpparencwe ghtis different the [OP Fnar a1-h coton Calculate your apparent weightatche IOdanc botzom of Ferris wheel given that the radius Ofthe wheee 7.2m, comdieres one revciucion 2vec and your mass Answerz [ne top actnebotom

QUESTION Asyouride on Ferris whez pparencwe ghtis different the [OP Fnar a1-h coton Calculate your apparent weightatche IOdanc botzom of Ferris wheel given that the radius Ofthe wheee 7.2m, comdieres one revciucion 2vec and your mass Answerz [ne top actne botom



Answers

(a) As you ride on a Ferris wheel, your apparent weight is different at the top than at the bottom. Explain. (b) Calculate your apparent weight at the top and bottom of a Ferris wheel, given that the radius of the wheel is $7.2 \mathrm{m},$ it completes one revolution every $28 \mathrm{s}$, and your mass is $55 \mathrm{kg}$.

All right, Question 51 in the textbooks. Asks a CZ You ride on the Ferris wheel, your apparent weight is calculated. So it is different at the top. Then at the bottom. Explain. So this is our figure. Here the blacks trickle is Art Harris feel the two red rectangles are our relative positions on this parents bill and then stating that these two apparent weight that we feel in both these places are different. So first, if we kind of explain what, uh, wait is if you think if we're just standing on a on a road going for a walk, your parent rate is the weight we feel relative due to gravity. So if we're just sitting here, of course, I've been normal for us pushing us up off the sidewalk Saturday into the sidewalk and we're balanced through the force of gravity is so the net force here. Obviously, we're not excited me up or down, I'm gonna put why their force of gravity in the wind direction asked me zero. So if he's the only two forces acting on us than our normal weights is the weight we feel the gravity or also just knows the normal for us. So this is what we want to solve for each of these scenarios. So if you take a look, what that would feel like if you're through, got a diagram? If both these positions at the top on dry underneath, obviously there will be a normal, normal force that the person they're every experience because we're sitting on a seat. So the normal force would be pushing us off of that seat. We feel gravity. And because we're traveling in a circle, we do experience some triple acceleration towards the center. So the force that we feel it would be opposing that motion on the bottom. Same idea. We have gravity R C position to change. So we have the normal going up. However, again with the counter act that central for accelerations be hell a force pulling us down this arrow's mg so that both these scenarios, if we look at the um, the net results similar to having found in our example, our normal force or apparent weight from the top scenario has to be mg minus the four students said terrible acceleration. And on the bottom it's the normal force is the course due to gravity, plus the same triple acceleration. So we do see that there are differences in our parent waits that both these positions and the more enforcement support you both your weight and provide this insert of acceleration around the circle. So they are different. And we do, uh, well, for this, we observe that thes Central Force will provide a increased to her net weight are apparent rate starting at the bottom of the Where's Bill party? This question asks to calculate the apparent right at the top and bottom of the Ferris wheel. Given that the races, we'll give him these parameters that won't complicate first at the top. We now know we have expressions for our parent. Wait or normal for us in both the scenarios with N G Plus and the squared over r Actually, negative actually was just every sense, because the bottom it is less sorry. The top. It is less as he found trimmer. First question here and for the bottom similar, but it is just the sign of these Internal forces just reversed N G plus and the spread over our So the Sirio were given radius forgiven. That's, however, we don't have velocity, which is I'm missing term. We are given, period. So it's fairly straightforward to calculate the, um, philosophy from that. So if you think we're traveling in a circle, the distance traveled in that circle is given by the circumference, which is two pi r. And if we're giving a certain amount of period of rotation of 28 seconds is period. So if it's our distance over time, that would be given. That gives us our velocity. So our velocity here is just too cool into two pi r over the by our over the period of travel. And if we're to calculate that it would be 1.6 meters per second. Great. So again, we have mass, we have velocity, we have radius so we can substitute values. And for each of these, and it's all for a normal force, we're apparently able scenarios, um, rounding our values to two significant figures. We do see that, um, the centripetal acceleration. Sorry. Yeah. This insurance, such a centripetal acceleration. Excuse me, decreases in the boat top and increases the bottom. So our numbers to be really, really similar as the again receded. That is true. So our net Sorry. Apparently at the top of 0.52 killings, 10 to 26 figs and at the bottom of 0.56 Killer Newtons. These air there too apparent rates. And again, as expected, this is our value at the bottom is larger than at the top.

For this problem. We want to determine what the apparent weight of a person is in a Ferris wheel ride at the bottom and the top of the Ferris wheel. So if we were to stand them inside that compartment with the scale, uh, their weight would be different. Have different positions on the Ferris wheel. So first, let's investigate the bottom of the Ferris wheel. So at the bottom are free. Body diagram will look something like this. Be experiencing a wheat, um, from the force of gravity energy. And there will also be a normal worse with greater magnitude, then gravitational force. And we know this because the net acceleration of this person must be radial e inward because of the fact that they're undergoing uniform circular motion. So in our coordinate system, we will define Radio Lee inward as positive and away from that as you get it. So if we were to do the symbol of forces Newton's second here in the radial direction, we would have and it's toward the center. Nine s m g equals I m. A. Where is the radio acceleration and we're gonna express the radio exploration as Omega squared are so the apparent weight of the person at the bottom, it's well state. And at the bottom members that normal force corresponds to their way is going to be MGI Waas and Omega Squared are Or we could factor out that m tried it. His lower case I'm gonna capitalize. It now is to be all well and good is m times G plus Omega squared. Cool. So now, um, the first observation will make about this is that there? Apparently it is actually greater. Then just the force due to gravity at the bottom of the first meal. Now the top. You have a similar situation, except MGI is pointing towards the centre. Uh, the Ferris wheel No. And m g is actually gonna be greater than the normal horse. We know that again because we're experiencing uniform circular motion and the overall acceleration needs to be radio lee inward. So over the top, the acceleration should be pointing down. So in that case, and is pointing away from the inside, an MG is pointing towards him. So if we apply Newton's second long in in the radial direction, what will end up with James org Wales and equals and B squared. Those are sir M Omega Squared are since we want to do everything in terms off omega and our and the maps. So if we solve now for end in this case, which will want to make sure to call and talk to distinguish between the at the bottom, we can add it to the other side of the tracks. The implemented squared are we'll end up with and G minus Omega squared are. So the observation here is that the apparently is actually less then that is just the force of gravity. So at the top of the Ferris wheel, we will. I feel like we wait last and at the bottom we will feel like we wait more. If you think about it for a moment, that might make a little a little bit of sense. So there you have it we've determined are apparent weight at the top and the bottom. Ah, Ferris wheel undergoing uniform circular motion

So here we give an example of a fair steel which would be an example of a vertical circle. And with a vertical circle we have certain object and it's important to remember that mainly we have two forces that are occurring. We have been forced to gravity essentially the weight and the normal force. So all these forces are going to be contributing to the central plaza uh force which points towards the center of the circle. So the centripetal force in this case, let's define this direction as positive, would be equivalent to the weight minus the normal force. Were given that the normal force is 1/4 of the weight. So our centripetal force is 3/4 of the mass times the force of gravity. And we know that the centripetal force is equivalent to the mass times the velocity squared divided by the radius. So we're given information that we have uh Angular velocity of 0.100 revolutions per second. So we first have to convert this to S. I. Units. So in one revolution There are two pi radiance. So we need this in terms of radiance per second. And we can find that are Angular rotation. 0.6- eight gradients per second. Yeah, so we know since it's at the highest point of the circle, the tangential velocity, which is what we're interested in is just the radius times the angular velocity Masses Cross out. So 3/4 force of gravity is equivalent to omega squared times the radius. So now our radius is equivalent to three G over four omega squared. And now we can determine the radius, mm hmm. Mhm. Yeah. Okay. And we can find that the radius is equivalent to 18.6 m, and this gives our final answers.

So for party, let's start off by finding the velocity. We know that for a knob decked in uniform circular motion, the velocity is going to be equal to two pi r Divide about period. So this is going to be two pi times the radius of 11 meters divided by the period of 12.5 seconds and we find that the velocity is five 0.5 to 9 meters per second for part B. We want to find the ratio of the apparent rate apparent weight to the rial. Wait s so we can draw everybody diagram here going straight up is forced normal and then coming straight down his force of gravity. Here we're going to say that that the sum of forces equals the mass times, the excess and chipotle acceleration. This will equal mg minus of of that event. Eso this will equal empty squared over are here. The force of gravity is towards the center of the Ferris wheel. That's why it's positive. We can say the Mumbai are and so we can say that the force normal is simply gonna be equal to M G minus M v squared over or the ratio of the apparent wait to re away. So we can say that wait, a parent divided by weight riel would be equal to M G minus, and these squared over r and then this will be divided by mg. And so this will be equal to G minus. B squared B squared over R, divided by G, canceling out all the mass terms and this is going to be equal to one minus 5.5 to 9 squared, divided by 11 times 9.8. And so this ratio is going to be 0.716 So the apparent weight is 71.6% of the real way. This would be your answer now report and rather rather my apologies. This's part A. This was not party. This is just We just need to calculate the velocity before we move on. So this would be part a. And then for part B, we had draw another three party A diagram. Now here. Forced normal is going up towards the center. Highland force of gravity is going down and we can say that force normal now is positive. Minus M g E equals M V squared over R and we know that after then is going to be equal to MG plus b squared over R. So we can say that the apparent wait, divided by the real way, would be equal to one plus. And then again, 1 5.5 to 9 squared, divided by 11 meters, times 9.8. So the only difference between the two is actually just the sign in part A. This is one minus. And then here it's one plus, given that here it is on the bottom of the Ferris wheel. Here it's the top of the Ferris wheel, and we find that the ratio was going to be 1.284 So here the apparent weight is 28.4% greater than the real way, and this would be a final answer. That is the end of the solution. Thank you for watching


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