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A 1000 kg rollercoaster car begins 100 meters above the ground. After the car rolls down to 20 meters above the ground, it enters a circular loop with a diameter of...

Question

A 1000 kg rollercoaster car begins 100 meters above the ground. After the car rolls down to 20 meters above the ground, it enters a circular loop with a diameter of 40 meters across Assume friction and drag are negligible What is the KE of the car when entering the loop? b_ What is the velocity of the car at the top of the loop? What is the direction and magnitude of the normal force on the car from the track at the top of loop?

A 1000 kg rollercoaster car begins 100 meters above the ground. After the car rolls down to 20 meters above the ground, it enters a circular loop with a diameter of 40 meters across Assume friction and drag are negligible What is the KE of the car when entering the loop? b_ What is the velocity of the car at the top of the loop? What is the direction and magnitude of the normal force on the car from the track at the top of loop?



Answers

A roller coaster car (mass $=988 \mathrm{kg}$ including passengers) is about to roll down a track. The diameter of the circular loop is $20.0 \mathrm{m}$ and the car starts out from rest $40.0 \mathrm{m}$ above the lowest point of the track. Ignore friction and air resistance. (a) At what speed does the car reach the top of the loop? (b) What is the force exerted on the car by the track at the top of the loop? (c) From what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop? CAN'T COPY THE FIGURE

He'll be. No, I don't have any little Okay. On center, we have reduced our no word for Skipped it energy three diddled acting in the downward direction the production If God we know there, that is more. The miners just wanted me and the normal force capital and it equals to the gravitational force stuff. We also know that the acceleration due to gravity his G if it was to 9.81 meter per second we have to complete this. Get off the car at the top of the group Pollution Don t do that. Muscles got music was smaller and speed at this is one. Now the radios off the loop is well are supposed to be by grand Be meet it. Hence the centrifugal force experienced by the car as part of the course with MV squared off on us Now the gravitational force and the car is the blue is close to Mt. And don't normal force is capital and of course F R minus And now equating the gravitational force. And Norman fourth we have Maiga is close to Don't and m G is the worst. Yes, but what is this? And Jeanne Thanks. We got two mg. I'm the squirrel of one party. Just centripetal force time now for being. For we we have he is the questo under root off don't more g not substituting the values we have the postal under root off to and to but at 39.81 meter per second, multiply 20 meter They only have 19.8 meters for second. The speed off the guard at the totals, the loop is 19.8 meters.

So basically, we have given the cost to start having a mass of uh, 1200 Kg and it has a it is moving on the subplot here already, a little. So what is the manager interaction of the force on the track? On the, uh, the track on the car? Uh, at the top of the hill, if the car is spirit 11. So basically the release of the, that track that again, we're So when, uh, in the first case, when the speed age 18 m 11 m/s, then not only the force of the top of the track, that will be anything but n G you don't want. And the minus of mps square is upward. That was obviously so that the distributed ports do on due to the car on the track. So the ports on the card track without drugs. And so the F B I think about M G minus of envy is quite upon our like mm demands of MBS, Carolina's OMG 1202 G If you were taking this as a team or. Yeah. And then point out as well as the 9.8 On the Ps but we eat 11 squared up on 18. So this comes out to the f. Yeah, That is coming out to be 3.723. Okay, so this force obviously Egypt This force obviously is in the upward direction, right? This this force will be obviously in the upward direction since we have taken apart at the positive side. Okay. In the next part we have given that the city uh 14 represent. Okay, so in this case what we're gonna do it again that when you think about MG, So 10 years got up on that, Right? So um 1202 G 10.8 minutes off the square 14 square upon our are age 18, So it will be -1.321 of three. One of 1.3 into 10 to the power of. Okay, so this means that the pores port uh do do to do to car will be in the other direction itself. So that means that this will be in the uh don't worry that it's okay. Thank you.

So here we're gonna choose positive why to be upwards. And in this case, we can say that gravity points down. Of course, however, we can say that the direction of the centripetal acceleration is down as well. But more important, more specifically, rather, it is going to be towards center of loop. So here we can sit at the force normal minus the weight would be equal to negative M V squared over R and four part eh weaken saw for force normal. This would be equal to mg minus m v squared over r, where V is equal in 11 meters per second and therefore forced normal here would be approximately 3.7 times 10 to the third Nunes Now for part B, we know that here f forced normal points upward. And we know that because we chose positive wide to be upwards and this is a positive number. So if we chose positive wide to be upwards and we got a positive value for force normal, we know that of course, the normal force points upwards for C V is equal in 14 meters per second. We're going to use the exact same equation forced normal equals mg minus m V squared over r. And at this point we know that solving. We're getting negative 1.3 times 10 to the third Nunes. And we can say that for a party because F event is less than zero. Ah, we can say that and we chose plus why to be upwards Because everyone is less than zero Nunes f event points downward and this would be our final answer for a party That is the end of the solution. Thank you for watching.

Here in this problem I can write F X is equal to m B squared by us now need vertical force Acting on the roller is equal to zero so I can write some medicine of FBI is equal to zero or and minus M D is equal to zero so n is equal to mg and I can ride. Need force f is equal to root under f X squared plus any square f x squared plus n it's square so going forward and putting the value so I can have physical to root under MB Square by are all the square plus mg square, so I get F physical too. AM Route ended B to the powerful by Ari Square plus the square, so just putting the value 3. 20 kg multiplication under root of 16 m pers 16 m purse again for to the powerful by 35 square plus 9.8 m per second squared on simplification. I get 3913.15 Newton. Now the value of Dante to can be written age and by FX. So he did physical to 10 in bus and by FX, it can be further return edge, tanning booth empty by and the square by on for the simplification. I can write physical too. Then in bus RG by B Square. So finally I can write physical too. Standing buzz. 35 m multiplication. 9.8 m per second square by 16 m per second. Holy square on solving I get T T is equal to 53.26 degrees as our answer.


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