00:00you
00:30.
01:00we will start horizontal circular motion
01:03which one we will start?
01:05horizontal circular motion
01:07what happens in horizontal circular motion?
01:10when a body moves on horizontal circular path
01:14so this kind of motion we call as horizontal circular motion
01:18we have told about circular motion
01:20before this in last lecture
01:22we have asked questions about uniform and non-uniform circular motion
01:27we have taught in last lecture
01:29so you must have studied
01:31have you studied?
01:33very good
01:35very good students
01:43see here I have taken a horizontal smooth table
01:48what kind of table is this?
01:50smooth horizontal table
01:58what happened?
02:00hello, what is the problem?
02:02tell me
02:04no problem
02:06I have made a smooth horizontal table
02:10I have made it with pen
02:12ok
02:19see here on horizontal table
02:22on a smooth table
02:24the body of m-mass
02:26we are rotating it by connecting it with thread
02:29how we are rotating it?
02:31by connecting it with thread
02:32you can rotate it by holding it with thread
02:34tension
02:36tension will be applied
02:38not torque, tension
02:40see we will draw FBD of m-mass
02:43we will get the equation of m-mass
02:45so here tension will be there
02:47it will provide centripetal force
02:50see here we will draw FBD of m-mass
02:54FBD of m-mass
02:58have you drawn FBD?
03:00I have told you about FBD
03:04here we will show the force applied
03:07it is a circular motion
03:09we will make an arc
03:11what kind of force will be applied?
03:14mg will be applied downward
03:19where will it be applied?
03:21downward
03:22normal reaction will be applied upward by the surface
03:25and the force applied towards the center will be tension
03:29thread
03:31yes, upward
03:33apply tension upward
03:35apply T here
03:37not velocity force
03:39we make force and acceleration in FBD
03:44ok?
03:45I have told you
03:47keep in mind
03:49in the first diagram we show the force
03:54and in the second diagram we show the acceleration
03:56I have told you
03:58this is called cause diagram
04:00this is called effect diagram
04:02I have told you the names
04:04remember?
04:06very good
04:08we keep giving good and very good to everyone
04:11see now
04:15see here
04:17normal will be equal to mg
04:19mg will be applied downward
04:21and the tension will be
04:23m into
04:25this can be written as centripetal acceleration
04:27AR
04:29can you see?
04:31oh my god
04:33write AC
04:37centripetal acceleration is also called radial acceleration
04:40centripetal
04:42absolutely
04:44your duty is good
04:52see there will be no problem
04:54see N is
04:56equal to mg
04:58whose is equal?
05:00tension was applied here alone
05:02so this will be m into AC
05:04now see here we also explain centripetal force
05:07what do we explain?
05:09centripetal force
05:11see the centripetal force
05:13the tension here
05:15T towards center
05:17where is it?
05:19towards center
05:21hello
05:27see here the tension
05:29is towards center
05:31so this is called centripetal force
05:33next page
05:39yes
05:41so the force
05:43which will be applied towards center
05:45will be called centripetal force
05:51yes
05:53this T
05:55this is called CP force
05:57CP means
05:59centripetal force
06:01so see the biggest
06:03goodness of students
06:05they study centripetal force
06:07they study centripetal force
06:09see this relation
06:11centripetal force
06:13is equal to m into AR
06:15whose is equal?
06:17m into AR
06:19yes
06:23acceleration
06:29here we take proportionality constant
06:33see this
06:35centripetal force is this
06:37and its relation with this
06:39is equal to
06:41clear
06:43m into AC
06:45and the value of centripetal acceleration
06:47we have told
06:49in uniform circular motion
06:51how much is it?
06:53tell me
06:55T square by
06:57R is equal to
06:59T square upon R
07:01R omega
07:03T square
07:05and V omega
07:07we have told three
07:09do you remember?
07:11at least you remember what we told
07:17yes T
07:19instead of T we have written T
07:21we have put different values of
07:23centripetal acceleration
07:25we hope that
07:27none of you have any problem
07:29we have never
07:31told you
07:35we have told you
07:43absolutely
07:49what?
07:55see
07:57this is horizontal circular motion
07:59so in horizontal
08:01circular motion
08:03we will call it horizontal circular motion
08:07you don't know
08:09what is horizontal
08:11it will be done
08:13it will be written
08:15horizontal circular motion
08:17clear?
08:19if you don't know
08:21then take a garland
08:23and remember Mahadev
08:27done?
08:29can we proceed?
08:31this formula
08:33will be useful later
08:35remember
08:37one more
08:39very good concept in this
08:41this is horizontal circular motion
08:43on rough surface
08:45what kind of surface?
08:47we have taken smooth horizontal table
08:49what kind of table?
08:51smooth horizontal table
08:53now how is the surface?
08:55rough
08:57friction
09:03see
09:05the car
09:07we run on
09:09horizontal circular path
09:15for that path
09:17you have to
09:19this is not a circular path
09:21it is straight
09:23from where we are studying
09:25our school ground
09:27a circle was made
09:31to make race
09:33which was made in olympics
09:35you can see
09:39circular path
09:41yes
09:47see here
09:49we have taken horizontal circular path
09:51here
09:55we will make Milka singh run
09:59no no
10:05run Milka run
10:07ok
10:09he is running
10:11this is v velocity
10:15mass m
10:17when he will run
10:19he is running on circular path
10:21to run on circular path
10:23do we need centripetal force?
10:25do we need?
10:27clear?
10:29ok
10:31now who will give centripetal force?
10:35now Milka is running
10:37Milka singh
10:39we have made Milka singh run
10:41we will not tie him with rope
10:43tie him with rope
10:45you will get centripetal acceleration
10:47run
10:49will it happen?
10:51no it will not happen
10:53where will you get centripetal acceleration?
10:55centripetal force
10:57let's make FBD
10:59here
11:01horizontal rough surface
11:03friction of rough surface
11:07gives centripetal acceleration
11:09run
11:15when he will run
11:17static friction will not be there
11:19we have studied in NLM
11:21kinetic will not be there
11:23we will make FBD
11:27when we will make FBD
11:29mg will be downward
11:31normal reaction will be upward
11:33and centripetal force
11:35here is Fs
11:37will be Fs
11:41and its acceleration will be
11:43this
11:45if there is any problem
11:47ask me
11:49centripetal or centrifugal
11:51see now
11:53the force which is applied towards center
11:55is centripetal force
11:57what kind of force is it?
11:59centripetal
12:01the force which is applied towards center is centripetal
12:03away from the center
12:05for centrifugal
12:07you have to wait
12:09you have to wait
12:11see here
12:13this force is applied
12:15mg will be downward
12:17normal reaction will be upward
12:19for circular motion
12:21force which is applied towards center is centripetal
12:23here friction
12:25will be applied towards center
12:27now here
12:29equation will come
12:31mg will be equal to
12:33and
12:35Fs
12:37will be of ma
12:39not mu s
12:41be patient
12:43we will make you meet mu s very soon
12:47no
12:49you have to wait
12:51Fs is the value of limiting
12:53mu s into n
12:55Fs value is not mu s into n
12:57let me tell you
12:59remember
13:01see here
13:03Fs is equal to
13:05here
13:07what happened
13:09Fs will be
13:11mu s into
13:13if you have any doubt ask me
13:15you
13:17don't have
13:19ok
13:23ok
13:31no
13:33no
13:35no
13:37see here
13:39Fs will be centripetal
13:41force
13:43will be
13:45now the one who is running
13:47if he will run fast
13:49he will need more centripetal force
13:51more velocity
13:53more centripetal force
13:55if he has to run fast
13:57he will need more centripetal force
13:59how much maximum centripetal force
14:01can we do
14:03Fs limiting
14:05how much can we do
14:07when
14:09Fs
14:11static friction
14:13Fs limiting
14:15in that condition
14:17the one who is running
14:19the velocity
14:21value of v will be
14:23v max
14:25now here
14:27Fs limiting
14:29replace
14:31ok
14:33replace
14:41condition
14:43Fs
14:45Fs
14:47Fs limiting
14:49maximum centripetal
14:51acceleration
14:53Fs limiting
14:55value of Fs limiting
14:57Fs into n
15:01and
15:03value of n
15:05mg
15:09and we can write
15:11v square upon r
15:17see here
15:19cancel m
15:21multiply r
15:25and
15:27mutual
15:29r
15:31g
15:33is equal to
15:35v max square
15:37from here
15:39v max maximum velocity
15:41coming
15:43under root
15:45mu s into
15:47r g
15:49this is maximum velocity
15:51yes
15:55tell me
15:59any doubt
16:01no
16:03understood
16:05see here
16:07vmax is the maximum velocity
16:09from which
16:11it is running in horizontal circle
16:25it is nothing like that
16:31how come v square upon r
16:35centipetal acceleration is
16:37a centipetal
16:39v square by r
16:41remember
16:43see the value of centipetal acceleration
16:55is v square by r
16:57or r omega square
16:59we write
17:01so put it like this
17:03ok
17:07now see
17:09horizontal circular motion
17:11we have told you on smooth table
17:13and after that
17:15rough horizontal surface
17:17ok
17:19now what has come
17:21we can
17:23rotate it with a thread
17:25like this
17:29no no no horizontal
17:35why
17:37this is horizontal
17:43how is this
17:45horizontal
17:47we do not call it vertical
17:49we call it vertical
17:53what we call
17:55this is vertical
17:57now we are telling horizontal
18:01ok see
18:03now here
18:05to understand horizontal circular motion
18:07here
18:09from the thread
18:11we have connected
18:13the body of mass
18:15when we connect it
18:17see how the equation will come
18:19force will be applied
18:21we will show it
18:23what will be applied on thread
18:25tension
18:27t
18:29now there will be two components of t
18:31this is theta angle
18:33in this direction
18:35t cos theta
18:37and in this direction
18:39t sin theta
18:41this horizontal circle
18:43is its center
18:45its length
18:47we have taken as l
18:49and its radius as r
18:51if you have any problem
18:53you can ask
18:55very good
18:59we make fbd
19:01of this
19:03fbd of m
19:05how fbd will be
19:07mg will be downward
19:11mg will be
19:13downward
19:15any other force
19:17through thread
19:19we have made component
19:21t cos theta
19:23and here t sin theta
19:25and center
19:27centripetal acceleration
19:29center
19:31clear
19:33without
19:35balance the equation
19:37here we have made
19:39tension component
19:41we can cut it
19:43because
19:45its component is done
19:47rest of the work is left
19:49t cos theta
19:51will be equal to mg
19:53and t sin theta
19:55t sin theta
19:57will be written as
19:59m into ac
20:01and t cos theta
20:03will be written as mg
20:05clear
20:07this will be first equation
20:09and this will be second equation
20:11in first equation
20:13divide by second equation
20:15you will get the result
20:17divide it
20:19clear
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