Circular Motion | Circular Motion Class 11 | Banking Of Roads physics Class 11 | Banking Of Roads | Banking Of Roads Physics | Banking Of Roads class 11 JEE | Banking Of Roads class 11 NEET
Circular Motion | Circular Motion Class 11 | Banking Of Roads physics Class 11 | Banking Of Roads | Banking Of Roads Physics | Banking Of Roads class 11 JEE | Banking Of Roads class 11 NEET
Explore the fascinating world of circular motion and the concept of banking of roads in physics class 11. Learn about the principles behind circular motion and the importance of banking roads for safe driving. This video tutorial will cover the basics and applications of these concepts to enhance your understanding of physics.
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Explore the fascinating world of circular motion and the concept of banking of roads in physics class 11. Learn about the principles behind circular motion and the importance of banking roads for safe driving. This video tutorial will cover the basics and applications of these concepts to enhance your understanding of physics.
#srbphysicskota #srbphysics #circularmotion #circularmotionjee #circularmotiononeshot #physics #banking #bankingofroads #aksir #aksirphysics #centrepetalacceleration
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Transcript
00:00see today we will start banking off-road
00:04what will we start? Banking off-road
00:07we will start, see in banking off-road
00:12we will first tell about banking off-road on smooth surface
00:17yes banking off-road we are telling that, banking off-road means
00:22when any body, car, bike or any vehicle
00:27turns on a turn, so the turn
00:31we call it banking, what do we call it?
00:34banking, the turn there
00:38the road there is inclined, how is it?
00:41inclined, it is on theta angle
00:44if you have any doubt, you can ask
00:46you?
00:48you can ask
00:51no
00:52see here
00:54when any vehicle, see here we have taken a car
00:57it will turn on a turn, when it will turn on a turn
01:00it will become a circular path
01:03when it will turn on a circular path
01:05so the road here is a little inclined
01:08it is on theta angle
01:12so what is the velocity?
01:14the relation between velocity and angle
01:16we will tell that in this
01:18so see here the diagram we have taken
01:21in this, there is a car
01:23when the car will turn on a turn
01:25so its velocity will be V
01:27and a normal reaction
01:31here I have taken its center C
01:33and the radius of this circular path
01:37will be R
01:39so here we will do its component
01:41what all forces will be applied on this?
01:45we are drawing FBD
01:47FBD of M
01:49draw it
01:53this will be MG
01:55in this direction
01:57we will take normal reaction
01:59this will be theta angle
02:03and
02:07we will do two components of this
02:09here this will be centripetal acceleration
02:11AC
02:13this is mass
02:15see here
02:17when we will do two components
02:19of normal reaction
02:21this will come in this direction
02:23n cos theta
02:25and in this direction
02:27n sin theta
02:29no problem
02:31n cos theta
02:33will be equal to MG
02:35because in this direction
02:37there is no acceleration in vertical
02:39so the equation
02:41n cos theta
02:43will be equal to MG
02:47and
02:49n sin theta
02:51is giving centripetal acceleration
02:53n sin theta
02:55is equal to
02:57M into
02:59AC
03:01we will write first equation
03:03and second equation
03:05if you have any problem
03:07you can ask
03:09see here n cos theta
03:11here normal reaction
03:13we will take perpendicular to plane
03:15this will be normal reaction
03:17in this direction n cos theta
03:19will be one component
03:21and here
03:23n sin theta
03:25n sin theta will provide centripetal
03:27acceleration
03:29so n sin theta
03:31will be equal to M into AC
03:33and n cos theta will be equal to MG
03:35no doubt
03:37now divide these two equations
03:39first and second
03:41divide
03:43in first equation
03:45divide in second equation
03:47n sin theta
03:49divided by
03:51n cos theta
03:53is equal to M into AC
03:55and this will be equal to MG
03:57see we will not teach much today
03:59we will teach little
04:01we will complete
04:03banking off road
04:05see here
04:07M is cancelled from M
04:09normal reaction is cancelled
04:11here sin theta upon cos theta
04:13what is it?
04:15tan theta
04:17is it or not?
04:19any problem?
04:21no
04:23AC is not value of
04:25centripetal acceleration
04:27P square by
04:29very good
04:31you tell
04:37P square by
04:39R
04:41this formula will be
04:43here this relation
04:45you have to remember
04:47in banking off road
04:49this tan theta value
04:51is
04:59tan theta
05:01is equal to
05:03V square by R
05:05you have to remember this formula
05:07in banking off road
05:09we will apply this
05:11any doubt?
05:13R is radius of circular path
05:15V is speed by which
05:17body is moving on circular turn
05:19that is V velocity
05:21and G is acceleration due to gravity
05:23any doubt?
05:25you don't have?
05:27no one?
05:29very good
05:31see we have taken
05:33this question
05:35this is the question
05:41you are not smart
05:43study
05:47you will not study
05:49very good
05:51you don't know smartness
05:53you did right or wrong
05:55radius of the curve
05:57road on
05:59national highway
06:01is R
06:03radius is
06:05R
06:07radius of third road
06:09is R
06:11width of road is
06:13B
06:15width of road is
06:17B
06:19outer edge
06:21of the road
06:23is raised by
06:25H with respect to
06:27inner edge
06:33so
06:35in respect to inner edge
06:37outer edge height
06:39will be more
06:41how much?
06:43H
06:45with respect to inner edge
06:47so that a car
06:49with velocity V can pass
06:51safely over it
06:53so if there is a car
06:55cross it safely with speed
06:57means
06:59there should not be accident
07:01so
07:03we have to find
07:05value of H
07:07so first of all
07:09we make a diagram
07:11this is road
07:13this is theta angle
07:19width of road is
07:21B
07:23height is
07:25H
07:27ok
07:29now
07:31in this car will pass
07:33its velocity is V
07:35from here we will make a car
07:37pass
07:41this car is passing
07:43its velocity is V
07:45so the radius of
07:47circular path of V
07:49will be
07:51radius R
07:53any problem?
07:55so here the formula is
07:57tan theta
07:59is equal to
08:01V square by
08:03RG
08:05this formula was told
08:07few minutes ago
08:09tan theta is
08:11perpendicular upon
08:13base
08:15tan theta is
08:17hypotenuse
08:19from here
08:21when we write tan theta
08:23how much we will write?
08:25tan theta is equal to H
08:27and base will be B
08:33from here question asked
08:35value of height
08:37so height H will be
08:39P square
08:43P
08:47upon
08:49RG
08:51you know
08:53you said
08:57so this height will be
08:59P square
09:01by RG
09:03I hope you will not have any problem
09:09move ahead
09:13ok this question
09:15you try to solve
09:23a guy holds a pendulum
09:25in his hand
09:27while standing at the edge
09:29of a circular
09:31platform of
09:33radius R
09:35rotating with an angular
09:37speed omega
09:41let us leave this
09:43for later
09:45let us
09:47study banking
09:49ok
09:51this was told
09:53on smooth surface
09:55now
09:57we are going to take
09:59rough surface
10:01what kind of surface?
10:03rough surface
10:05where
10:07there is friction
10:09here we will take
10:11rough surface
10:13motion of object
10:15on rough inclined circular
10:17path
10:19when a body moves
10:21on inclined circular path
10:23in that condition
10:25its motion will be
10:27velocity
10:29from minimum velocity
10:31to maximum velocity
10:33in range of velocity
10:35we can do anything
10:37clear
10:39rough
10:41circular path
10:43from minimum velocity
10:45to maximum velocity
10:47on any velocity
10:49you can pass any excuse
10:51no doubt
10:53no
10:55now you have to see
10:57what is the value
10:59to cross safely
11:01what will be Vmax
11:03maximum velocity
11:05will be when
11:07centripetal force
11:09friction centripetal
11:11force
11:13will be supported
11:15hello
11:17support
11:19here
11:21body is moving
11:23here
11:25friction is taken
11:27down the plane
11:29in two directions
11:31one is down the plane
11:33second
11:35up the plane
11:37away from the center
11:39and down the plane
11:41will be
11:43towards
11:45center
11:47that is why
11:49friction is taken
11:51down the plane
11:53because we need
11:55maximum velocity
11:57maximum
11:59yes
12:01towards center
12:03centripetal force is supported
12:05by friction
12:07yes
12:09normal reaction
12:11downward
12:13mg
12:15diagram
12:17fbd
12:19yes
12:21it is difficult
12:23difficult
12:25means
12:27it is difficult
12:29you have to pay attention
12:31this is
12:33theta
12:35this is normal
12:37this angle is also theta
12:39and here
12:41in this direction
12:43this normal reaction
12:45will have one component
12:47in this direction n
12:49cos theta and in this direction
12:51n sin theta
12:53I hope there will be no problem
12:55friction fs limiting
12:57we have put in this direction
12:59where we have put fs
13:01limiting
13:03this angle is theta
13:05so what will be this angle
13:07so according to this
13:09what will be this angle
13:11theta
13:13means fs
13:15friction's one component
13:17will come in this direction
13:19fs limiting
13:21cos theta will come in this direction
13:23and fs limiting
13:25sin theta
13:27will come in this direction
13:29and this
13:31will be in this direction
13:33centripetal axis
13:35this is a beautiful diagram
13:37n cos theta
13:39this is normal reaction's component
13:41will be in this direction
13:43n sin theta will come in this direction
13:45friction fs limiting
13:47down the plane
13:49one component of this
13:51fs limiting cos theta will come in this direction
13:53left
13:55second one will be downward
13:57fs limiting
13:59sin theta
14:01now here we will equal their components
14:03so here
14:05n cos theta
14:07will be
14:09mg plus fs limiting
14:11sin theta
14:13no problem
14:15n cos theta
14:17is equal to
14:19n cos theta
14:21is equal to
14:23mg plus fs limiting
14:25sin theta
14:27fs limiting value is
14:29mu s into
14:31n
14:33so this is mg
14:35plus mu s
14:37into n
14:39sin theta
14:43and here is n
14:45cos theta
14:47mu s into n sin theta
14:49transfer it
14:51left side
14:53so this will be n cos theta
14:55minus mu s
14:57into n
14:59sin theta is equal to
15:01mg
15:03here take n common
15:05we will get normal reaction's value
15:07cos theta
15:09minus mu s
15:11sin theta
15:13is equal to
15:15mg
15:17from here normal reaction's value will be
15:23mg divided by
15:25cos theta minus mu s
15:27sin theta
15:29if you have any problem, you can ask
15:31there is no question
15:33till now
15:35say
15:41now see second question
15:43n sin theta
15:45hello
15:53n sin theta
15:55plus fs limiting
15:59cos theta
16:01will be m into ac
16:03see
16:05n sin theta
16:07this component
16:09n sin theta and fs limiting
16:11cos theta
16:13this is centripetal acceleration
16:15m into ac
16:17no problem
16:19put the value
16:21n sin theta
16:23fs limiting's value
16:25mu s into
16:27n
16:29ok
16:31and
16:33this is
16:35cos theta
16:37this will be
16:39mg square upon r
16:41this is v max
16:43here take n common
16:49sin theta plus
16:51mu s
16:53cos theta
16:55this is m v max square
16:57upon r
16:59we have found value of n
17:01put it
17:05write it
17:07first equation and second equation
17:15pi equation first and
17:17second put it
17:19instead of n
17:21mg divided by
17:23cos theta
17:25what happened
17:29very easy
17:31cos theta minus mu s
17:33sin theta
17:35and above
17:37it is written
17:39sin theta plus
17:43mu s
17:45cos theta
17:47this is m v max
17:49square
17:51upon r
17:55cancel m
17:57multiply r
18:01maximum value of v
18:03maximum velocity
18:05body
18:07inclined plane
18:09maximum velocity
18:11no doubt
18:13you
18:15sawrab
18:17how much
18:19v max value
18:23under root
18:25r g
18:27in the bracket
18:29see this
18:31sin theta plus mu s
18:33cos theta
18:35sin theta plus
18:37mu s
18:39cos theta
18:41cos theta minus
18:43mu s sin theta
18:45this is the value
18:47ok
18:49see this is
18:51maximum value of v
18:53from this maximum velocity
18:55any body
18:57circular
18:59banking of road
19:01means curve
19:03from this maximum velocity
19:05any body can move
19:07from this maximum velocity
19:09any body can turn
19:11from this maximum velocity
19:13what will happen
19:15centipetal force
19:17required
19:19car will turn
19:21from this maximum velocity
19:23car will not turn
19:25that is why
19:27in highway
19:29it is written
19:31from this maximum velocity
19:33you should not drive
19:35more speed
19:37more speed
19:39keep driving
19:43ok
19:45take care
19:47ok
19:49ok
19:51ok
19:53ok
19:55ok
20:03ok
20:05write
20:07fast
20:09where are you lost
20:15done
20:17yours
20:19where are you lost
20:23now see
20:25find minimum velocity
20:27what to find
20:29v
20:31minimum value
20:33velocity
20:35circular path
20:37maximum
20:39minimum
20:41friction
20:43limiting
20:45away from the center
20:47down the plane
20:49friction
20:51up the plane
20:55centipetal force
20:57support
20:59for minimum
21:01we will make
21:03FVD
21:05what will be the force
21:07mg
21:09downward
21:11normal reaction
21:13perpendicular
21:15and the friction
21:17in this direction
21:19FS limiting
21:21ok
21:29ok
21:31ok
21:35ok
21:37ok
21:39ok
21:41ok
21:43ok
21:45ok
21:47ok
21:49ok
21:51ok
21:53ok
21:55ok
21:57ok
21:59ok
22:11ok
22:13ok
22:15ok
22:17for fs-limiting
22:21for fs-limiting
22:25for fs-limiting
22:29for fs-limiting
22:33for fs-limiting
22:37for fs-limiting
22:41for fs-limiting
22:45for fs-limiting
22:49for fs-limiting
22:53for fs-limiting
22:57for fs-limiting
23:01for fs-limiting
23:05for fs-limiting
23:09for fs-limiting
23:13for fs-limiting
23:17for fs-limiting
23:21for fs-limiting
23:25for fs-limiting
23:29for fs-limiting
23:33for fs-limiting
23:37for fs-limiting
23:41for fs-limiting
23:45for fs-limiting
23:49for fs-limiting
23:53for fs-limiting
23:57for fs-limiting
24:01for fs-limiting
24:05for fs-limiting
24:09for fs-limiting
24:13for fs-limiting
24:17for fs-limiting
24:21for fs-limiting
24:25for fs-limiting
24:29for fs-limiting
24:33for fs-limiting
24:37common
24:45sine theta
24:49sine theta
24:53sine theta
24:57mu s
25:31cos theta plus mu s sin theta
25:37and this is sin theta minus mu s cos theta
25:48this will be mv minimum square by r
25:52so from here m to am is cancelled
25:56multiply r here
26:00from here v minimum square value will be
26:05rg in the bracket
26:08sin theta minus mu s cos theta
26:14divided by cos theta plus mu s sin theta
26:22we will get v minimum and take its root
26:26shall we play?
26:34tell me what is the value of v minimum
26:37under root
26:40rg in the bracket
26:43sin theta minus mu s cos theta
26:50divided by
26:53cos theta plus mu s sin theta
26:59if you have any doubt, you can ask
27:03you?
27:05not at all?
27:07very good
27:11this is the minimum value of v
27:14on rough inclined plane
27:16and what we have found before is
27:18the maximum value of v
27:23means the overall value of v
27:26minimum and maximum
27:29is a range
27:32we will write it
27:35if you have noted it, then tell me
27:38have you done?
27:41so the value of v
27:53the minimum value of v is
27:56under root
27:59rg in the bracket
28:04sin theta minus mu s cos theta
28:11divided by cos theta
28:15plus mu s sin theta
28:19tell me
28:22what is the maximum value of v?
28:25under root
28:28rg
28:31sin theta plus mu s
28:34cos theta divided by
28:37cos theta minus mu s
28:41sin theta
28:49here
28:52sin theta plus mu s cos theta
28:55divided by cos theta minus mu s sin theta
28:58is v max or v minimum
29:01overall, the minimum value of v will be this
29:04and what will be the maximum value of v?
29:07between this minimum and maximum
29:10you can drive a car at any speed
29:13any vehicle, car, truck
29:16if you are driving on the road
29:19no problem, you will cross safely
29:22how will you cross?
29:25your vehicle will not turn
29:28even if you slow down, it will turn
29:31and
29:34even if you speed up, it will turn
29:37yes, the accident will be big in speeding
29:40but in slowing down
29:43yes
29:46overall, you have to remember
29:49whenever you turn on the road
29:52you have to drive between v minimum
29:55and v max
29:58you don't have to be over smart
30:01it is not a normal road
30:04you can drive at any speed
30:07any doubt?
30:11say that
30:14no?
30:17yes, that's all for today
30:20tomorrow we will start vertical circular motion
30:23we will start
30:26vertical circular motion
30:29we will discuss about it tomorrow
30:32that's all for today