Banked turn Physics Problems - By The Organic Chemistry Tutor
00:00 | in this video , we're going to focus on banked | |
00:02 | curves . So here's a question for you . A | |
00:05 | car is traveling on a frictionless banked curve radius 200 | |
00:09 | m . If the curve is banked at an angle | |
00:12 | of 15° , at what speed should the car travel | |
00:15 | in order for it not to slide up or down | |
00:21 | ? So let's try a picture . So banked curve | |
00:24 | looks like an incline , but it's different their circle | |
00:28 | emotion involved . So I'm going to draw box . | |
00:36 | Now , imagine this box as being the car . | |
00:41 | Let's call this the X . Direction , and let's | |
00:46 | call this the Y direction and the car is moving | |
00:50 | forward in the Z direction . Now , you want | |
00:56 | to make a distinction between this type of motion and | |
01:00 | a regular incline which we covered earlier , where Box | |
01:03 | simply slides down in the X . Direction , it's | |
01:07 | not moving in the Y . Direction . And so | |
01:09 | you want to make that distinction . Now , if | |
01:15 | the car is moving too fast without friction being present | |
01:22 | , it's going to slide off , it's gonna slide | |
01:25 | up the banked curve in that direction . If it's | |
01:29 | movement to slow , it's going to slide down towards | |
01:32 | the center so there is a certain speed at which | |
01:36 | the car will not slide up or down . Our | |
01:40 | goal is to find that speed and I want to | |
01:44 | make a distinction between the free body diagrams of a | |
01:47 | regular incline and the banked curve . So in a | |
01:52 | typical incline we would have a normal force perpendicular to | |
01:57 | the surface as usual and as a component of the | |
02:03 | way force that will accelerate the block downward . So | |
02:09 | this is the weight and here is a component of | |
02:16 | the way force decide is MG co signed data and | |
02:20 | this is MG sine data . Not on a regular | |
02:27 | incline . The normal force is equal two MG Times | |
02:32 | Coastline Data . So let's say if we had an | |
02:36 | angle of 45° Co signed , 45 Is .7071 . | |
02:44 | So that means that the number force supports 70.71 of | |
02:50 | the weight of the object in the regular inclined problem | |
02:57 | . Now on a banked curve , the situation is | |
03:00 | different . The normal force is still directed in the | |
03:09 | same direction , it's still perpendicular to the surface , | |
03:18 | however , it supports more than the weight of the | |
03:22 | object . The normal force has a component that is | |
03:28 | directed towards the center of the circle . So keep | |
03:30 | mine . This object is traveling in circular motion . | |
03:36 | It also has a white component . So I'm going | |
03:39 | to call this F . And X . And let's | |
03:46 | call this F . And why ? Now ? We | |
03:52 | still have to do with the weight of the object | |
03:53 | , which is M . G . Now it turns | |
03:58 | out that this angle here is equivalent to this angle | |
04:04 | . Now , if you want to see it , | |
04:06 | let's say that this angle Run out of space , | |
04:09 | but let's say it's 60 , that means this angle | |
04:11 | is 30 and this line is perpendicular to that line | |
04:16 | , Which means this angle must be 60 as well | |
04:21 | And the normal force is perpendicular to the surface , | |
04:23 | so that's another 90° angle , which means this is | |
04:26 | 30 And this has to be 60 . So these | |
04:29 | two are equivalent to each other . Hopefully you can | |
04:32 | see that because I am short in space . There's | |
04:38 | not much space to work there . Now we need | |
04:42 | to realize is that this force , the white component | |
04:50 | of the normal force is equal to the weight of | |
04:52 | the object when friction is not present . So F | |
04:58 | N . Y is equal to MG . Now S | |
05:02 | N Y is F N . Co sign If you | |
05:06 | call from circuit over . Hopefully you took trig . | |
05:10 | Cosign Theta is equal to the adjacent side divided by | |
05:13 | the ipod knows so that's F . N . Y | |
05:16 | . Over . FN . So F . N . | |
05:18 | Y . If you re arrange it by cross multiplying | |
05:22 | is F . N . Closing data . And so | |
05:25 | these two are equal to each other . So F | |
05:35 | . N . Co signed data is equal to MG | |
05:39 | , which means that the normal force is MG divided | |
05:43 | by co sign . So make sure you see the | |
05:45 | difference between these two equations in a regular inclined problem | |
05:50 | . The normal forces MG times coastline data . But | |
05:54 | for a bank curve it's MG divided by co sign | |
05:58 | . And if the angle is 45 , like in | |
06:00 | this example One divided by coastline 45 or one divided | |
06:05 | by .707 , 1 Is 1.414 . So the number | |
06:11 | four supports a 141.4 of the weight force at this | |
06:17 | angle . So clearly the normal force on a banked | |
06:22 | curve is doing a lot more work . Then the | |
06:27 | normal force on the incline , it's putting a lot | |
06:31 | more effort . You might be wondering why is it | |
06:34 | so different ? What's the reason ? Well , if | |
06:37 | you look at an incline , the normal force only | |
06:39 | has to support just a portion of the weight . | |
06:42 | The MG coastline part of the weight . The reason | |
06:46 | being is a portion of the weight force is used | |
06:49 | to cause the object to slide down the inquiry . | |
06:55 | Now in this example , the normal force has to | |
07:00 | support the full weight of the object . And not | |
07:03 | only that , but it has to provide the centripetal | |
07:06 | force necessary to keep the object traveling in circular motion | |
07:11 | . And so that's why the normal forces a lot | |
07:13 | larger on a bank curve then on a regular inclined | |
07:19 | . But now let's get back to the problem . | |
07:21 | So let me erase a lot of stuff so I'm | |
07:32 | going to redraw the incline . Mhm . Here's our | |
07:38 | vehicle , here's the normal force and this is the | |
07:45 | X . Component of the normal force and this is | |
07:48 | the white component . So keep in mind whatever angle | |
07:51 | we have here is equivalent to this angle . So | |
07:55 | the Y . Component of the normal force . As | |
07:56 | we said , it's F . N . Times coastline | |
07:59 | data which means that the X . Component has to | |
08:04 | be F . N . Signed data sign is associated | |
08:09 | with the opposite side . Co sign is associated with | |
08:13 | the adjacent side with respect to the angle that you're | |
08:15 | considering . Mhm . And let's not forget about the | |
08:20 | weight force . Now keep in mind if there's no | |
08:24 | friction , there is an exact speed for the car | |
08:27 | not to slide up the incline or the bank curve | |
08:31 | or down the bank curve . And this is important | |
08:36 | to understand if the car is moving too fast , | |
08:38 | if it's moving greater at that design speed , the | |
08:43 | car is going to slide up the incline if it's | |
08:45 | moving too fast , if it's moment to slow , | |
08:50 | it's going to slide down the bank curve . So | |
08:54 | just make sure you understand that because that's important . | |
08:58 | So we need to find the speed at which it's | |
09:00 | going to maintain its current position . How can we | |
09:04 | find that speed ? So let's focus on the forces | |
09:10 | in the Y direction . So we have the net | |
09:13 | force in the Y direction that's equal to this upward | |
09:16 | force and because it's going in a positive direction , | |
09:19 | it's going to be positive FN . Co sign . | |
09:22 | And this forces going in a negative Y direction . | |
09:24 | So negative empty . Now there is no net acceleration | |
09:30 | in the Y direction , The car is not being | |
09:33 | lifted off the ground and it's not going straight through | |
09:35 | the ground , it's maintaining its position in the Y | |
09:39 | . Direction . So therefore the net force in the | |
09:41 | Y direction we could say is zero . So I'm | |
09:46 | gonna add MG to both sides . And so M | |
09:49 | . G . Is equal to fn coastline data . | |
09:54 | And if you divide both sides by co sign , | |
09:56 | you'll see that the normal force on a bank curve | |
10:00 | when no friction is present , it's equal to MG | |
10:03 | divided by coastline . On a regular incline . It's | |
10:09 | MG times co sign . Now , what about the | |
10:15 | forces in the Y . Direction ? I mean not | |
10:17 | the right direction but the X . Direction . Let's | |
10:21 | see what we can come up with . The only | |
10:27 | force in the X . Direction is FN signed data | |
10:31 | . The X . Component of the normal force . | |
10:35 | Now there is a net force in the X . | |
10:38 | Direction because there is an acceleration towards the center Mhm | |
10:45 | . Of the bank curve . And that acceleration based | |
10:50 | on this diagram , it's pointed in a negative X | |
10:52 | . Direction . So this is X . And this | |
10:56 | is the Y . Direction . Now I should add | |
11:02 | a negative sign to F . Ensign data because this | |
11:05 | is going in a negative X . Direction . Now | |
11:08 | the net force is the mass times acceleration and that's | |
11:14 | the centripetal acceleration because it causes the the card to | |
11:18 | move within circular motion and it's going in the negative | |
11:22 | X . Direction . So I'm going to put a | |
11:23 | negative sign in front of it just so you can | |
11:26 | get used to the process of setting up these types | |
11:30 | of formulas . Now in this example the two negative | |
11:34 | signs will cancel and the centripetal acceleration is V . | |
11:39 | Squared over R . So this net force is really | |
11:42 | this centripetal force and that's equal to F . N | |
11:46 | . Signed data . Now what I'm gonna do is | |
11:51 | I'm going to replace the normal force with em cheek | |
11:55 | over coastline data because they are equal to each other | |
11:58 | . So let's take this and insert it into this | |
12:00 | equation . And so what we're gonna have now is | |
12:03 | M . V . Squared over R . Is equal | |
12:06 | to MG divided by co sign time sign . Now | |
12:14 | we can divide both sides by M . So we | |
12:16 | can cancel it and sign divided by co sign its | |
12:21 | tangent . If you remember your trick stuff so they | |
12:24 | squared over R . Is equal to G times tangent | |
12:28 | beta . Now our goal in this problem is to | |
12:32 | calculate the speed . So we need to get V | |
12:34 | by itself . So let's multiply both sides by our | |
12:39 | And so we can get rid of this . And | |
12:41 | so the squared is equal to RGs handed data . | |
12:44 | And now what we need to do at this point | |
12:46 | is take the square root of both sides . So | |
12:48 | here's the equation V . Is equal to . Let's | |
12:51 | just get rid of some stuff . Now let's just | |
13:01 | write this bigger . So V . Is equal to | |
13:06 | the square root of our G . Tangent data . | |
13:11 | So this equation allows us to calculate the speed at | |
13:16 | which the car can maintain his position in a bank | |
13:20 | curve without sliding up or down and when no friction | |
13:24 | is present , So the radius of curvature , it's | |
13:29 | 200 m G is 9.8 m/s squared And the road | |
13:36 | is banked at an angle of 15° , So that's | |
13:40 | going to be tangent 15 . So make sure your | |
13:42 | calculator is in degree mode . So the answer that | |
13:52 | you should have is 22.9 meters per second . So | |
13:59 | if the car maintains that speed it's not going to | |
14:08 | slide up or down , it's simply going to travel | |
14:12 | forward . Let me try that better . Let's say | |
14:18 | if the car speeds up and travels at 30 minutes | |
14:21 | per second , it's going to go up the incline | |
14:25 | . If it goes that 15 years per second it's | |
14:27 | going to go down the incline . So if you | |
14:31 | view it just from a flat surface like this , | |
14:35 | Yeah , If it travels at 30 it's gonna go | |
14:40 | up this way . If it travels that 15 is | |
14:43 | going to go down that way , But if it | |
14:45 | maintains a speed of 22.9 it's going to stay where | |
14:48 | it is . Now let's move on to part B | |
15:01 | . What angle should the role be banked For cars | |
15:05 | to travel at a speed of 30 m/s without sliding | |
15:08 | up or down in the absence of friction ? So | |
15:13 | if we can increase the angle we can increase the | |
15:18 | speed at which your vehicle can safely travel the bank | |
15:23 | curve without sliding up or down . So that new | |
15:27 | angle has to be greater than 15° because at an | |
15:30 | angle 15° the speed was 22.9 m/s . So if | |
15:35 | we want to increase the speed to 30 we need | |
15:37 | a much larger angle . So let's go ahead and | |
15:40 | calculate angle , let's square both sides . So the | |
15:43 | squared is equal to R . G . Tangent beta | |
15:46 | And now let's divide both sides by Archie . So | |
15:49 | V squared divided by RG is tangent . Now to | |
15:54 | get to the angle , we need to take the | |
15:57 | arc tangent of both sides . So the angle is | |
16:00 | going to be our tangent or inverse tangent of the | |
16:04 | squared divided by R . G . So this is | |
16:07 | the equation that you want to use if you need | |
16:09 | to find the banking angle when no friction is present | |
16:16 | . So it's going to be our tangent V squared | |
16:20 | or 30 square , divided by the radius , which | |
16:22 | is 200 Times 9.8 , 30 squared , or 30 | |
16:30 | times 30 , is 900 And 200 times 9.8 . | |
16:35 | That's 1960 . So this is gonna be our tangent | |
16:41 | 900 over 1960 . And so that's going to give | |
16:45 | you an angle of 24 point 66° or we can | |
16:50 | say 24.7 , let's surround it . So at this | |
16:55 | angle The car can safely travel at a speed of | |
17:00 | 30 m/s without sliding up or down the bench curve | |
00:0-1 | . |
DESCRIPTION:
This physics video tutorial provides plenty of practice problems on banked turns without friction. It explains how to set up the free body diagram to solve banked curve problems and how to derive the formula to calculate the speed of a car and the banking angle without static friction and how to calculate the minimum speed and the maximum speed that a car can travel without sliding up or down the banked curve as well.
OVERVIEW:
Banked turn Physics Problems is a free educational video by The Organic Chemistry Tutor.
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