Banked turn Physics Problems - Free Educational videos for Students in K-12

Banked turn Physics Problems - By The Organic Chemistry Tutor

Transcript


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	.

Summarizer

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.

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