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Gravitron test ride for the first time. The A brave 90 kg man_ named Steve. agrees give spin faster and . faster until it achieves Gravitron has diameter of [0 m: I...

Question

Gravitron test ride for the first time. The A brave 90 kg man_ named Steve. agrees give spin faster and . faster until it achieves Gravitron has diameter of [0 m: It iS programed speed: the floor of the Gravitron lowers itself; one revolution in 15 Once maintains this leaving the riders to hang midair. picture and an FBD of Steve as if you are looking at him head on at one point Draw time while the Gravitron is spinningDetermine Steve' speed as the Gravitron is spinningWbat is the centriret

Gravitron test ride for the first time. The A brave 90 kg man_ named Steve. agrees give spin faster and . faster until it achieves Gravitron has diameter of [0 m: It iS programed speed: the floor of the Gravitron lowers itself; one revolution in 15 Once maintains this leaving the riders to hang midair. picture and an FBD of Steve as if you are looking at him head on at one point Draw time while the Gravitron is spinning Determine Steve' speed as the Gravitron is spinning Wbat is the centriretal acce_eraticn of Steve?



Answers

The Gravitron is a carnival ride that looks like a large cylinder. People stand inside the cylinder against the wall as it begins to spin. Eventually, it is rotating fast enough that the floor can be removed without anyone falling. Given then the coefficient of friction between a person’s clothing and the wall is ?, the tangential speed is v, and the radius of the ride is r, what is greatest mass that a person can be to safely go on this ride?
(A) $\mu v^{2} /(r g)$
(B) $r^{2} v^{2} /(\mu g)$
(C) $r g /\left(\mu v^{2}\right)$
(D) None of the above.

We have acceleration requests whisker over our So our comes out to be whiskered over a which is 30 m per second square of that divided where acceleration is to G to dance 9.8 it was 8.62 middle. Now, over there we have m g and and both acting downward, which it was an the square over our. So we find the normal force to be m times V squared over r minus. G equals the school over our is to G so that it was mg on putting the numbers. Yeah, so that's the normal force in the

We're told that the loops are like teardrops. So I'm going to draw this kind of like a teardrop. That's extreme there. The biggest loop Is 40 m high. Maximum speed, 31 meters per second. The speed at the top 13 and the centripetal acceleration is to G. Was the radius at the top. Well, the radius at the top would be V squared over. R. So are would be V squared at the top over two G. So, V top clear this out. V top is 13 G. 9.81 V. T. Sure. Where'd over two G Would be the radius, which would be 8.61 meters. Okay, so that's a let's look at me. The total mass is m Well, the force, it's just going to be mass times acceleration. Uh Force is just going to be to GM two G. I am where M. Well, I guess that's the answer is just to GM two G is 19.6. Wait a minute though. We don't really have to know that big. Yes, we do. Because it would be V squared over R times M. So we do need to know um Okay. See what if it had a loop with a radius of 20 meters? I should with V at the top still being the same? What would be the centripetal acceleration? Well, this centripetal acceleration would be V. T squared over 20. So now if I just change if I change VT to uh No. VT is 13. All right. I'm just gonna put it down here. VT squared over 20 eight point 45 meters per second. What is the centripetal acceleration at the top? Well, that's less than gravity. So the normal force Would be less than zero. So, there have to be some artificial way of keeping it on the track. Thank you for watching.

All right. In the given problem, the ship off the roller coaster rights at us Specifics Place is in the form off a teardrop like this. The total height off this teardrop shape from bottom toe talk is given us 40 meter. If you consider the bottom most point B and the top most point to be a then the speed off the roller coaster car and the bottom most point is we be is whereto lucky 1.0 meter per second and that at the dock is given as 30 points 13 10 meter for second. No. In the first part of the problem, we have to find the radius. Oh, the are portion and that most point there, the speed has been given us 30 meters per second. So using the expression for centripetal acceleration which is given us wise of G at that topmost point. And the expression for it is we squared by, uh so expression for radios will become we a square invited by centripetal acceleration. So here it will become the squared off 13 divided by two times off G. So if we put these values, this is 1 69 the squad off 13 and this is two times off. 9.8. So this comes out to be 8.62 meter. This is the radius off the Arctic Ocean at the top most point. And this is the answer off. The first part of the problem? No. In the second part of the problem, we have to find the force exerted by the rails on the roller coaster car at the top. Most point. So here, if you consider all the forces acting on the car, this is, uh, rich off the car. MGI acting vertically downward. The centrifugal force and in the car is really outward given by And these were by up on their difference, believer a normal reaction exerted by the reels on the car. So here this normal reaction is in given by M times off the square. Invited by R minus M. G. Here. This is the centripetal acceleration at that almost point which has been given us Fuji. So this becomes m times off duty miners M times off G as the mass off the car has been given a skeptical in. So finally it comes out. Toby M times off g This is the normal reaction exerted by the reels on the car and the answer of the second part of the problem. No. In the third part of the problem, we have to find the centripetal acceleration at the top. Most point in case the ship off the track is perfectly circular, so give but track is circular. Then this height will be equal toe diameter, which is equal toe devise off radius. So why so furious is equal to 40 meter. So really, yes, we come out to be 20 meter. In that case, the centripetal exploration law at the top, most point will become These were divided by R or 30 square, divided by 20 which comes out of the 1 69 by 20 is equal to it. Wind for five, neater for a second display. So this is answer for the third part of the problem. No, taking this as the centripetal acceleration. Now we have to find the normal force. At that topmost point was the question is again the same, which is m the square by our in centrifugal force minus m Udoji. So if I put these values now, this is the centripetal acceleration. So this is end times off 8.45 minus em. Giants off 9.8. So definitely this value now will come out to be negative. So in case off teardrop model off roller coaster radius at the top will reduce so centripetal acceleration will increase, so normal reaction will be more than a week. And for that roller coaster to rename stick at the rails are the most point. The normal reaction should be positive means centripetal force. This should obviously more than the weight. Or we can say centrifugal force should always be more than the way. But here, in case off this circular track, the centripetal or centrifugal force is becoming less than the weight, so the car will not be able to remain stick over the rates. Hence taking teardrop model off the track is more beneficial and safe. Thank you.

So we have a barrel. Yeah, you have a barrel that it's spinning. So I'll drive with a narrow here to indicate that it's spinning and there are writers on the walls. So imagine that these air on the walls these people here are on the walls of the barrel and they're experiencing a force that is from the center of this barrel and is going towards the wall. And this makes the riders feel as if they're being pushed against the wall of this barrel. And so in this scenario, whenever the writers air pinned to the walls, okay, they are experience of force going in this way, so outside of the wall off the barrel. And they're also experiencing a force towards the center of the barrel in accordance to Newton's third Law. And similarly, whenever the floor drops away the experience of force that is going downwards and then according to his third law, they also experience of force that is going upward. And we say that this downwards force is a real force and the upwards force is an unreal fictitious force. And for the forces for the Wall, we say that the force going outside of the wall is a real force and the force going towards the center of the barrel is a fictitious horse.


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