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8. A Ferris wheel has diameter of 90 m and period of rotation of 65 passenger weighs 300 point) What is her apparent weight at the lowest point of the Ferris wheel?...

Question

8. A Ferris wheel has diameter of 90 m and period of rotation of 65 passenger weighs 300 point) What is her apparent weight at the lowest point of the Ferris wheel?198.2 N DJIzAN D350.8 N 400.6 N

8. A Ferris wheel has diameter of 90 m and period of rotation of 65 passenger weighs 300 point) What is her apparent weight at the lowest point of the Ferris wheel? 198.2 N DJIzAN D350.8 N 400.6 N



Answers

An $80-$ fi-diameter Ferris wheel rotates once every 24 s. What is the apparent weight of a $70 \mathrm{kg}$ passenger at (a) the lowest point of the circle and (b) at the highest point?

In this problem. We'll talk about circular motion. Consider that we have an object that is moving in a circular motion. And we want to describe the dynamics of this most and assume that the radius of the motion is our, um so at this point here that I'm marking in red, we can composed exploration of the object into two components. One of them is the centripetal component that is, it would shoe the linear velocity where their speed squared over two or the angular speed squared times are actually I sat over to before, but it's actually over. The radios of the trajectory are, And this a centripetal acceleration points towards the center of the trajectory and accounts for the change in direction of the project. The tangential exploration, on the other hand, accounts for the changing speed. And is it which is the derivative of the speed with respect to time on in our problem, we have a person riding a Ferris wheel and the math of the person, the the weight of the person. I'm gonna call it f uh, actually, I'm gonna call it w is 520 Newtons. So this is the gravitational force that acts on the person on. We know that at the top of the trajectory, the trajectory is a circle like this. And at the top of the trajectory, Mhm the difference between the weight of the person and her apparent weight. And we'll call it that, will you? A is equal to 1.5 Newtons. Yeah. Also, this is the magnitude of the different. And our goal is to find in question they whether the apparent way is 518.5 Newtons or 521.5 Newtons. Okay, so in order to find this first, we need to notice that the apparent weight of person is equal to the normal force exerted by seat on the person. So basically, I'm gonna draw everybody diagram of the person we have the gravitational force. That is the weight. Then we have the normal force. I'm gonna call it parent. Wait, W A Yup. I noticed that W minus w A must be equal to this interpreter. Four. So that's MV Squared over two, and I noticed that I'm over our I'm sorry. Not over to and, um MV squared over r is positive and pointing, and and the typical forces pointing towards the center of the tragedy. No Notice that Omega. I'm sorry. W minus mv squared over R is it what you omega? I'm sorry. To W A So w a is smaller than w. So the apparent weight is smaller than the real weight of the person. So the only option is that if current weight people too 518.5 million, then in question be we have to find what would be the apparent weight on the person on the bottom off the fairies again, I'm gonna draw a free body diagram. We have the parent way. Then we have the weight pointing downwards. And now, since the tribunal force points towards the center of the trajectory, we have that different weight. Mine is the weight of the person is M v r squared. Overall, eso w A is equal to shoot am Omega Square over our plus that will you know we have that that we are actually should be so w a is greater than that were you Which means that w A in this case is 512.518 point five Newton's. And finally, given that the angular speed omega is equal to zero point 0 25 radiance per second, our goal is to find the radius of the various well are so notice that am omega Square times are is equal to the magnitude of the difference between omega Start between W N w A. You know this to be true to 1.5 million. So our is it. Which 1.5 Newtons divided by the mass times Omega Square Notice that mg the weight is it what you 520 Newtons? So am is the weight of 520 Newtons divided by 9.8 m per second squared. So em is equal to 53 106 kg. So our is 1.5 Newtons divided by 53.6 kilograms times 0.25 radiance per second squared. Which leads us to our being equal to 45.2 m and this concludes our exercise

So here in this question, the masses given So the monster is it 70 kg rooted once every 24 2nd. So time period is given. He is equal toe 24 2nd. And what is the print rate office? 70 kg passenger. Okay, so the mask is 70 kg I and the radios is deep by toe end diameter is given 80 ft. So on the edited by two we are getting for different so we can convert it into meter on converting it into a meter. We are getting it is we're went to meter? No, a print weed in a part. The value of capital and normal reaction at the lowest point. It's bean Oscar, right? So from mhm the given data, we can find the value off he speared as to buy our over time period. So if you substitute these values, the value off why is 3.1 for the value off? Our is 12.2 and time period is 24 24. So on solving the value, we are getting the value off we as three point 19 meter per second. Okay, So, in order to solve for part a that is Enloe. In at the lowest point, we can write net force at Louis Point. The Net foods that is net centripetal force will be provided by n minus W. So we can make it equal to envy Square by our So from here, the value off is it a blue Pless M V square by our So the video off word. We can write it as M g. So the expression for and will become m g plus M V square by our. So if we substitute the values here, the value off M is 70. The value of G is 9.8 70. The value off we is three point and 19 Esquire and the value off our is 12.2 m. So on solving it, we are getting the value off. And as 7 14 Newton are the lower most point. So this is the answer for a part. Now let us all for be part. So for me part, that is, we need to find the value off end. And I used point so at I used to point the value off and uh huh, M G minus and will provide necessary centripetal foes. So from here, the value off and comes out b m g minus MV squared by our. So now if you substitute the values here, the value off M 70 the value off G is 9.8 again, the value off M 70 the value off we is 3.19 Is Squire in the value off our is 12 point. So if we solve it, we will get the value off. Normal reaction is 6 13. So this is the answer for be part. Thank you.

Be too quick. Three by the diagrams. The circles represent the trajectories, and then we have the point at the top of the trajectory of the top of the curb and then the point at the bottom of the curve to separate free by the diagrams. We can then apply Newton's second law to each diagram, and we can then say that here, first on the top mg minus force normal at the top would be equaling two m v squared over r. And then for the second free body diagram, we can say force normal at the bottom. Bot will be rather minus mg. This will be equaling two m v squared over our and so we can say that here, the fairest movie, this fierce movie, The Fair's Wheel. My apologies is steadily rotating. This means that the that the value F sub c is gonna be equaling two m b squared over r. And here we can say that this would be this this this would be the same everywhere. So the apparent weight of the student is given by the force normal. Wait, A parent would be equaling the force normal, and so we can say that at the top for party, we know that the force normal at the top is equaling. 556. Newton's and MG. The weight is equaling 667 Newtons. This means that here the seats is pushing up with the force that is smaller than the students. Wait so we can say, of course, F and some top. This would be, of course, the force exerted by the seat on the student, and this is smaller. Then the students wait mg, which means that we say that the student experience a decrease in his apparent weight at the lightest point. So he says, Yes, we can say a decrease in apparent weight and you can say one feels lighter from part A four part B. We can find the centripetal force, have subsea to be equaling M V squared over are equaling mg minus the force. Newton forced normal. My apologies, Force normal at the top. This is equaling. 667. Newton's minus 556. Newton's giving us a centripetal force of 111 Newtons. We can say that then the normal force at the bottom for Part B would be equaling M v squared over r. This is a plus mg. This would be equaling to the centripetal force plus mg. This is equaling. Ah, 111 Newton's plus 667 Newtons. This is equaling. 778 Newtons. 778 Newtons here. This is greater than the students Wait and at the bottom, the student is going to feel heavier than they actually are because of this, uh, that because of this sum increased normal force four parts, See, if the speed has doubled is doubled. Now we can say the centripetal force prime would be equaling to the mass multiplied by two times the in visual velocity squared, divided by our and so this would be equaling 24 multiplied by 111. Newton's giving us 444 Newtons. So at the highest point we have that the force normal at the top crime would be equaling two mg minus the centripetal force crime. This is equaling 667. Newton's minus 444 Newtons giving us 223. Newton's answer for Part C and then for part deep, the force normal at the bottom prime would be equaling. Two. The centripetal Force Prime plus MG. So we have 444 Newtons, plus 667 Newton's giving US 1100 and 11 Newton's. This would be our final answer for Part D. That is the end of the solution. Thank you for watching.

All right, So, we've got our student here and they're riding around on this ferris wheel And we are told that the weight of the student is 667 newtons. And when they are at the top of the ark, when they're at the top of the circle on this steadily rotating ferris wheel, They're the normal force that they feel so a normal force that they feel is going to be 500 And 56 mutants. And part A of this problem is asking us, does the student feel lighter or heavier at the top of this arc and two to understand this, What we can think of is just when we're standing on level ground, her weight force is balanced by the normal force. So you feel the reaction force from the ground on you. So The weight force than the normal force have to be equal and opposite according to Newton's 3rd law. In this case, the normal force is actually less than the weight force. So this person on the ferris wheel all feel like they weigh less, They'll feel like they're lighter than they normally would because the normal force is the force that they feel. So they will feel later at the top. And ah for the same reason they'll feel heavier when they're at the bottom. So that's part a part B. We need to figure out what the magnitude of the normal forces at the lowest part. So what is their normal force down here? So we're going to have their weight force down and that they have a normal force up and to do this first thing we're gonna do is analyze the top part because we're given what the normal forces there. And so the some of the forces, the net force that they feel is going to be mass times acceleration and acceleration in this case is going to be the centripetal acceleration. Since this as a steadily rotating thing, they're not going to be speeding up radio. So that's their net force. And there are some of the forces are the rate force down and the normal force up. So if we subtract those, this is for the top at the top, they're going to have a net force of 111 Newton's now at the bottom. This is still going to be true. Their net force, their radio force, the centripetal force Is still going to be 111 Newton's. So but at the bottom, the forces are going to be hating a little bit differently when they're at the bottom. Now, instead of the rate force being towards the center in the normal force being away from the center. Now, the weight forces away from the center in the normal forces towards the center. So this is for top. Now let's look at bottom. So at the bottom, mm sfc it's still going to equal 1 11. But now, instead of F G minus normal force, it's going to be normal force minus F G. Because now the normal forces towards the center and the weight forces away from the center. And so this is still 1 11. Normal forces what we're looking for now. And wait forces still 6 67 newtons. And so if we add this to both sides, we end up with a normal force on the bottom That is going to be 778 newtons. So they're going to feel much heavier when they get to the bottom. Now let's look at part C. Now, in part C. It asks what the magnitude of the normal force will be when we double the speed. And for this we can look at this equation. And a CFC centripetal acceleration is also given as a V squared over R. So this is actually M V squared over our and we're not doing anything to our but we are doubling this fee. So if we double the V, what happens to our net force? Well, the V is squared, so if we double the V, this is going to become to the squared over R. So we've got a square that too. And so our new centripetal acceleration is going to be four times the old centripetal acceleration, doubling the speed, quadruples the acceleration. So our new MSNBC is actually four times are old. M.A. So it will be 444. Uh Oops, no, that's right. So at the top, what's our normal force going to be? Well, we have uh 4 44 4 44 here, Our weight force is still the same. 667. So if we solve this for the normal force, so here, I'm just going to go top the normal force, it's going to equal the wait force minus the centripetal force minus and sfc. And now this is our new a subsea. So it's four times as big as the old one. And so the normal force at the top is going to be 667 -444, Which is to 23, Newtons at the top. And at the bottom we'll do the same thing here where we have our normal force at the bottom, it's going to equal the wait force plus the centripetal force. So we have 4 44 plus 6 67. So at the bottom, our normal force is going to be equal to 1000 or 1.11 times 10 to the third newton's. So the the normal force at the top begins much less. The normal force on the bottom becomes much greater, so they're going to feel very light at the top and very heavy at the bottom.


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