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New ride an amusement park starts with & car at rest at a height h It then rolls down frictionless track where passes through circular sections with radius R as...

Question

New ride an amusement park starts with & car at rest at a height h It then rolls down frictionless track where passes through circular sections with radius R as shown in the Figure below;Give an expression for the speed of the car when it as the top of the circular section (the point indicated by the star above). (7 points)b) What is Ihe maximum heighi E such thal the car will nol come off the tracks as i goes over the hill? Express your answer terms the radius the circle R reminder that an

new ride an amusement park starts with & car at rest at a height h It then rolls down frictionless track where passes through circular sections with radius R as shown in the Figure below; Give an expression for the speed of the car when it as the top of the circular section (the point indicated by the star above). (7 points) b) What is Ihe maximum heighi E such thal the car will nol come off the tracks as i goes over the hill? Express your answer terms the radius the circle R reminder that an object in circular motion with speed nas centnpetal acceleration which directed towards the center of the circle with magnitude of VIR:) points)



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A car in an amusement park ride rolls without friction around a track ($\textbf{Fig. P7.42}$). The car starts from rest at point $A$ at a height $h$ above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of $h$ (in terms of $R$) such that the car moves around the loop without falling off at the top (point $B$)? (b) If $h =$ 3.50$R$ and $R =$ 14.0 m, compute the speed, radial acceleration, and tangential acceleration of the passengers when the car is at point $C$, which is at the end of a horizontal diameter. Show these acceleration components in a diagram, approximately to scale.

So we start off with Hiroko's through it point a wave. Yeah, but off each from the ground Gonna find a point B. What is the minimum? Very off hitch Sussed it. The rollercoaster doesn't fall off at point B. So it was to stick onto the trick as there It's moving across Point B. No, my point b. The vehicle still would experience only be wait for it sold Wait. And for it to Not for it must be moving at a certain speed. A sort of velocity. Such debt. The entire weight is used to be the centripetal X elevation Such a beetle force required to keep the roller coaster in circular motion. And so this means that in some commotion we know that the centripetal acceleration is p squared over r We're RSD radios off the circular motion. This must be equals toe the acceleration due to gravity right at B which is just G and so the must be equals toe hard g discredit Sotus is d minimum velocity That is quite so. He must be great. So then or equals two Arch now to get this amount of velocity, it means stets from point a to point B de change in potential energy from point A to point B must be converted into kinetic energy and this kinetic energy must provide sufficient velocity. So this change in height is h minus to our right Since the height of B is two times the radius right times MGI, this will be just give us the gravitational potential energy that was committed into kinetic energy. So half and v square, you cannot energy cancel the end. You've rearranged recreation a bit to finalize hitch. So this one of a to g times V square us to and remember that's all. V must be greater or equals two skeletal farty sri Instead, no hitch must be greater equals two Archie over to cheap plus two are this leaves us with it's very half bus off our plus to our just five to, uh so each must be greater than five over to our for your next party question. Argumentative hitch is 3.5 are and, uh, ages tree point Fife are and our is 20 meters compute the speed as well as the video acceleration tangential acceleration. 50 passengers IHS in a car is at sea. So at point C. That is in the loop at the site. Right? Is this D position of the car and this moving Don't let's so we do the same same steps as before, right? I remember that the speed is governed by the kinetic Energy Day has and a kinetic energy that has dependent on the conversion off the gravitational potential energy from point A to point C. So the difference in gravitational potential energy is mg times, uh, tree boy five R minus are e created to half MV square. Sophie must be close to two times off. 2.5 are just five times g t t skirt off the entirety, so give us the one point tree toast for a second. What do you read? Your acceleration. It's simply centripetal acceleration, which is three square over. Our really found our V Square. All right, I'll be doing country squared. You gotta buy radios, which is 20. Excuse us for a night test. Bullets against square for the tangential extraditions Tangential direction and he cries. Over here is Don. What's this detention show direction? The only force acting on the card. A stall? What's its gravity, right? It's old mess. And so he oniy separation. It's tangential. Must be closed toe cheese, which is nine point it now to show all these acceleration components in the small diagram. So the I mean, this is ah, point off interests descent of mess off the car, then the acceleration due to the radio direction the redirection it's it's such or if he tangential exploration is such And so the nets acceleration in this direction Accomplice ing both the centripetal acceleration s well, esty, uh, gravity.

Using conservation off energy, we can write kinetic energy at point A that is kinetic energy off court Appoint a plus potential energy off car at point a equal to kinetic energy off car at point B plus potential energy off car at point Be yes, car starts from rest from point A so we have kinetic energy at point a equal toe zero. So this gives us zero plus potential energy. I've got a point. Is mess off car times gravitation Exploration times hide off guard at my A, which is h equal toe kinetic energy off car at point B. That is half times mess off car times. So get up, cpt. Off car at Quan be plus potential energy or card by me, which is massive car times Gravitational extradition times Height off car at point B A. Site up. Find bees equal toe diameter off circle that is two ice off radios off circle. Now we can write. This is mass times gravitation Exception Times h is equal toe half I am so get off last year, B But as m. G okay into to our now we can cancel mess off car from both sides off security. So this gives us G h equal toe half into, so get off. Will be with us. Duty are Yeah, it might be. We can write normal reaction force on guard. Plus, which, of course, this boat are producing. Contributed force on a centripetal force is equal toe. I mean, toe, I mean to So get off velocity. I've got a fine beat. You added by radio here there is self circles. Are yeah okay. Yeah, for H That is far as to be minimum. We have the minimum value off we be so will be will be minimum when and is equal toe zero. That is there is no any normal reaction force on car at point B. So we have wet equal toe and so get off We'll be divided by our as we head off. God is m times G. That is equal toe m we be soccer divided by are here mass and the mass will be canceled. So we have so get off Maybe equal toe j r. Now, by putting these values this value off, we we secure here. So we have g h g in tow h equal toe half into value off secure off baby is we are plus to J R. So here g will be can sell from both sides of the equation. So this gives us equal to are divided by two plus are so we have a equal toe. 5.2. Uh, so this is the minimum height off car at point? Yeah. In order to roll the car in the lost now in part B. Yeah, we can again, right? The kinetic energy at point. A alas, kinetic energy off. Sorry. Potentially. Najaf guarded. Quite a It is equal toe kinetic energy off car at point C plus potential energy off car at point. See now, by plugging defend values here, kinetic energy off car 8.0 s car starts from rest, plus potential energy at point is mass off car times gravitation expression into hide off car at pine is given. That is three bite 50 times radius off circle that is equal toe Connecticut Your policy that is half times mess off car times. So get off. Separate off car at point C or less. Potential energy and piracy, that is massive car into gravitation. Expression in tow. Hide at piracy s thanks by m c is horizontal to the center off circuit. So we have hide off fancy equal toe already yourself circle So this gives us J into 3.50 are minus g r equal toe hop into so get off supported by M. C. This gives us Sakharov Peter Pan sequel Toe do in tow Yeah 2.50 g r G R G r Nobody putting different values here to into 2.50 into geese accreditation expression that it's nine point and major parts can scare on our is very soft loof which is 20.0 m This gives us so get up zip it at prime C equal to 900 anti meter sucker divided by second scream So we have speed at times C equal to 31 point zero Yeah, meter bar second Yeah, Now we can calculate radial exploration Yeah, by using the relation radial expression equal toe So carob Sleep it off course at point c divided by release a blue as to get upset. PD's 900 anti meters to care for skin squared divided by various sub group is 20 point 0 m this gives us radial exploration. Equal toe 1.56 meter party can sucker. Now that in general exploration at point. Yeah, see, that is at point C. Yeah. We have to. Intentional expression. Equal toe expiration due to gravity, as the body is instead off free, fall at one C. Now we can draw this components of exploration. So this is required to see on radial expression is towards center off Luke. And we have This is a radial exhibition and we have conditional expression that is done worse. So yeah. Nobody using Director Edition, we can show Thank you. The net exploration at Quincy. So this one is the the injection expression on this one? Yes. Radial exhibition here. We have giant the tail off radial expression with head off the additional exploration. And this victor Okay, shows the net. Exhibition off. Car Air point. See

Okay, considering in chapter 10 problems 53. So it's very poorly drawn, but this is a roller coaster track that there is frictionless, and so we're starting at some rest. Height h, it falls down and into a valley. A circular valley of radius R immediately into a circular hill of radius. They are, um, So what we want to figure out is, how high can we start the car so that it doesn't fly off at this point right here at the very top of the hill. Okay, so this is kind of like a circular motion and energy conservation problem. So the first thing let's figure out the ve max for the hill. And how do we figure this out? Well, it's we notice a circular motion. And so at the very top, the only force is the force of gravity pointing down been pointing towards the center. So if we know that it's centripetal force pointing to the center needs to be in p squared over our That's the max. It could be without you falling off that circular path. So that means this must be equal to the force of gravity or in G at this point. So the V max then is are G square. You didn't because that is the maximum speed that you could stay moving in a circular motion at that top point of the curve. Okay, So if that's the maximum speed you can have, let's now figure out what speed you're gonna have their. So if we look at it, this height at that point should be, um, h at the curve. I'm gonna see. Should be h starting minus two are Okay. So we need the gravitational potential that you've gained from going this amount. Distance Thio equal the kinetic energy at the maximum speed. So I said a lot of things. Let's start writing it down. So the gravitational change in gravitational potential was negative change in kinetic energy. So this is mg I'm gonna is gonna put Delta age for how far we go. Um, actually, let's call this Delta age. So this is the age and this is the age in that you start at okay. Cool The missing equal one half and be max squared. Okay, so let's plug that in what we can plug in the next square. You know what that is's would have been orangey. This equals in she that was plugged in. Got the H H. Macs minus two are So now we can do all of this. We could just solve for each maximum. So he's in Angie's cancel. And that means we have one. Half are plus two are That is, um sorry. Plus too hard up. Five halfs are because h max Awesome. So now we can just go to part beat. That was all part A Part B says, what is H Max when articles 10 m. Well, that is just 25. Yes. Bypass times 10. Play by Awesome. That was a quick one.

Imagine a mess being released from a height in this case. 14 inches, moving down complaint thing and then thanks. And you, the height of which was 20 meters. How did we woke up? What the velocity would be at the top off the hook. So for this, we can look at the conservation of energy. We know that the mechanical energy before is equal to the mechanical energy pasta. So there's some off the potential. The initial potential connectivity is equal to this. Some off the potential energy come in and get G. Austen's. So, in this case, Michelle potential age, he's a giant. This case is the emotional height. It, uh, what you need, The kinetic energy zero as the object stops a wrist and moves down, can't you and reaches a maximum height? Yes, it's a big energy is 1/2 him square. If you re arrange this, we get yeah, is equal to you. Squared is according to into that change height each eye. But yeah, and some said, you know that use we know already used for the giant and Egypt we get. Hey is equal to 19 or eight second. So that's the value of velocity when the cart reached the top of the look. Now let's look at the next part in the next spot, you want to know what is the force acting under Scott at the top of the new? Well, in this case, that is the moment for us. In fact, in what was Qatar, there are two forces both backing down with the normal, which is a track or the onto the cot plus M g, which the weight off the car acting down. And we know that Scott is motion. It's a centripetal force, anything on the car. So if we use Newton's second law for me, too, we know you, me centripetal force acting in the car. The result in such a political IHS the same as the normal Nolan loss. That's the weight both active. Don't we know it made just the script. That's normal force. That's the weight. If you rearrange this equation, we consult in the force at the track is exerting onto the car. When we get in support in, he spread over harder minus, and since we know that mess the cot, we can work this out 29 kilo nutrients 29,000 U tix. Okay, so, finding How did he find what? The minimum height above the looking so that the cock does not use contact with a check. Well, in this case, we can issue that there is no contact with the caught to the check. So and he's equal to zero. And this tells us from the PDS a creation that him he squared over r minus G is a question zero. We just got that from the creation. Okay, you re arranged this. We get, he squared. He's in court too. Are are they? And this is true, Teoh, which I brightness. Yeah, we got this. So first off the problem. So therefore the minimum height, which the card can be the least fell so that it does not lose contact with Look, is officer rearranging are over too less each. If we substitute these values and we get in minimum height, what's 25


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