00:00in the last class
00:10in the moment of inertia
00:15in the moment of inertia point mass
00:27for
00:30point mass
00:35mr square
00:37as a moment of inertia
00:39that depends on
00:54distribution
00:59of mass
01:03and access of rotation
01:10and
01:24depends on
01:30access of rotation
01:31the moment of inertia
01:37point mass
01:38mr square
01:40point mass
01:42and
01:44other body
01:49of mass
01:54and access of rotation
01:57of mass
01:57and
01:58clear
01:59clear
02:04moment
02:08of inertia
02:10of
02:18rod
02:20uniform
02:21rod
02:27mass
02:28capital
02:29length
02:30capital
02:48length
02:59length
03:00moment
03:01of inertia
03:02point
03:03and
03:04carrying it
03:05and
03:06and
03:07and
03:08and
03:09and
03:10and
03:11and
03:12and
03:13and
03:14and
03:17and
03:18and
03:19and
03:20and
03:21and
03:22and
03:23and
03:24and
03:25all
03:26and
03:26for
03:27and
03:28and
03:29of road, we find about one end, one end of the road, we find the right thing.
03:49So, if you find the road, you can find the moment of inertia.
03:57Point mass
04:17linear mass density
04:22linear mass
04:26and speed
04:29lambda is equal to
04:34mass upon length
04:35dm
04:38value
04:38lambda
04:41into dx
04:43equal
04:43m upon
04:48l into dx
04:49now
04:50moment of inertia
04:54di
04:56dm into x square
05:01dm
05:06value
05:07m upon l
05:09x square
05:12dx
05:13x
05:13x
05:15x
05:17x
05:19x
05:20x
05:21x
05:22x
05:23x
05:24x
05:24x
05:25x
05:26x
05:27x
05:28x
05:29x
05:30x
05:31x
05:32x
05:33x
05:34x
05:35x
05:36x
05:37x
05:38x
05:39x
05:40x
05:41x
05:42x
05:43x
05:44x
05:45x
05:46x
05:47x
05:48x
05:49x
05:50x
05:51x
05:52This will be x3 by 3.
06:09One end of moment of inertia will be mL square by 3.
06:17One end of moment of inertia will be mL square by 3.
06:47One end of moment of inertia will be mL square by 3.
06:53One end of moment ofult August match.
06:55We will mL square before a cosmic rake.
06:59Two end of moment of inertia will be mL square by 3.
07:04How upon fire withinSHOPs 0.
07:08Unt cuántip 잘�月ns Direct.
07:103.
11:11e i
11:16m into x square
11:22over
11:41so here we can write m upon l x square dx
11:50here di i find moment of inertia integration of di
12:00k ho ga limit kya laghe ghi
12:11bolo
12:14minus l y 2 plus l y 2
12:28कि यह एधर x negative लेंगे इधर x को फोच्टिव लेंगे तो यह
12:37minus l y 2 इधर भी तो रोड ना क्या दि रोड का निकालना है
12:44अलू पुरी रोड का निकालना है न तो इधर जो रोड है
12:48minus l y 2 इधर लिमेट लगेगी इधर प्लस l y 2
12:58m upon l, this is x cube by 3, minus ly2 plus ly2.
13:08यह x की जगे पे, ly2 का cube करेंगे,
13:31और यह 3 होगा,
13:42call करिए,
13:49l cube by 8 plus l cube by 8 divided by 3,
13:552 l cube by 8 हो जाएगा,
14:04और यह आपका 2 l cube by 8 into 324 के होगा,
14:10एक l cancel हो जाएगा,
14:13यह moment of inertia आजाएगा,
14:17ml square by 12 होगा,
14:21ठीक है, देखोगे,
14:26यह माँ राएगे,
14:28मतम आएगा जटड़ा राएगा,
14:34की जाएगा जाएगा,
14:37भीखन दीरे,
14:47की जाएगा,
14:51यह जाएगा,
14:55you
15:06you
15:08you
15:09you
15:14you
15:25Can I, let's see.
15:55Can I, let's see.
16:25Can I, let's see.
16:55Can I, let's see.
17:25Can I, let's see.
17:55Can I, let's see.
18:25Can I, let's see.
18:55Can I, let's see.
19:25Can I, let's see.
19:55Can I, let's see.
19:57Can I, let's see.
20:03Can I, let's see.
20:05Can I, let's see.
20:07Can I, let's see.
20:09Can I, let's see.
20:11Can I, let's see.
20:13Can I, let's see.
20:15Can I, let's see.
20:17Can I, let's see.
20:19Can I, let's see.
20:29Can I, let's see.
20:31m r square upon two pi d theta integration theta limit zero set two pi limit apply
20:42कर दीजिए कि यह मुमेंट आफ इनर्स्टी आजाएंगा
21:03m r square
21:04टीक है
21:18वो दिकाद है
21:23लेख लिया होगा नए निए लेसे लेकिये
21:34you
21:48oh
21:53oh
21:59oh
23:04To find the moment of inertia of semi-circular ring.
23:22A uniform ring, semi-circular.
23:34It's a mass capital M and radius capital R.
23:58This is a mass capital R.
24:06This is a mass capital R.
24:08This is a mass capital R.
24:16This is a mass capital R.
24:18So find the area.
24:20This is a mass capital R.
24:28This is a mass capital R.
24:30This is a mass capital R.
24:38This is a mass capital R.
24:40This is a mass capital R.
24:48This is a mass capital R.
24:50This is a mass capital R.
29:03Because,
29:10you
29:11you
29:12you
29:13you
29:14you
29:15you
29:16you
29:17you
29:18you
29:19disk
29:20find
29:21you
29:22so
29:23this
29:24is
29:25this
29:26is
29:27uniform
29:28disk
29:29okay
29:30this
29:31moment of
29:34inertia
29:35this
29:36is
29:42a
29:43a
29:44a
29:45a
29:46a
29:47a
29:48a
29:49a
29:50a
29:51a
29:53a
29:54a
29:56a
30:05a
30:06a
30:07a
30:08m
30:09a
30:10a
30:13e
30:14a
30:16a
30:17a
30:18g
30:19a
30:20e
30:20e
30:22a
30:24a
30:25a
30:26a
30:31a
30:31a
30:35surface mass density sigma mass upon area pi r square
30:48now this is dm find for the sigma surface mass density find for the d a find
30:57for the d a can be a ring length 2 pi r
31:11and this is d a value to pi r
31:19ndm
31:23ndm
31:28ndm
31:32ndm
31:32ndm
31:37ndm
31:39ndm
31:41ndm
31:49So, this dm of value is 2m upon r square small r dr.
32:05Now, this is the dm into small r square.
32:14So, this is the dm into r square, dm into r square, dm into r square, dm is 2m upon r square small r dr into r square.
32:38So, this is the dm.
32:44Now, this is the dm of s.
32:47So, this is the dm of the dm.
32:53Now, this dm of s.
32:58So this will be done here.
33:252m upon r square
33:28R into R square R cube r cube r cube r cube dr
33:38limit kya lagayenge
33:40Dekhye yahaanpa jho ring file cut hogey
33:45Sabhse chhoti ring
33:48Khaan cut hogey center peyani center peyati nahi hogey
33:54R equal to zero hogey cut hogey ring nahi manegy
33:57Sabhse last ring
34:03R equal to capital R
34:05Sabhse chhoti ring R tending to zero pey maan saktay ho
34:09Jab radius zero hogey
34:11Sabhse capital R pey
34:13Limit jho hogey woh khaan se khaan tuk lagey
34:15Zero se capital R
34:19Or yye ho jayega integration R ki power 4 upon 4
34:312m upon R square
34:45R ki power 4 upon 4
34:49Call kariye kithna a gaya
34:51M R square by 2
35:01Disk ke liye jom moment of inertia
35:05Ata hai
35:07Excess of rotation kave out
35:09Parish cylindrical excess kave out
35:11Ata hai
35:13M R square by 2
35:15Vf find karte hai
35:17Ata hai
35:19Ata hai
35:21Ata hai
35:23Ata hai
35:25Ata hai
35:27Ata hai
35:29Ata hai
35:31As a
35:39stronger
35:41Ata hai
35:45Ata hai
35:49Ata hai
35:51Ata hai
35:53Ata hai
35:55A moment of inertia of uniform rectangular
36:25I am late
36:38I am rectangular plate
36:40moment of inertia find
36:42is
36:44key
36:46one side
36:48is
36:50about
36:52about
36:55c
36:56ु
37:02कि इसकी यह लेंत है
37:04और यह लेंत है
37:09कि अधिर देती है
37:10है
37:11काईसे पायन दिए यहां पर
37:12रैक्टैंगुल लेंत है
37:18एम एक आया है ना आल बी
37:21will be so how can we find here rectangular plate is the mass capital
37:27you
37:31yeah surface mass density you have to find me
37:37okay
37:43yeah
37:44mass density sigma mass upon area
37:57we have a part of the key i am a car
38:12and you have a master
38:17and you have to say
38:27so dm mass find
38:30karenge dm ki value
38:32d a pahela likh lietay
38:34d a hooga b into
38:36d x k
38:37d a hooga
38:39sigma into d a k
38:44m upon
38:47l b
38:47or yye ho jayega
38:49b into d x
38:59m upon
39:07l into d x
39:09so yye d a ho
39:11kuchh gala likh kya
39:13haan
39:15yaan dm
39:17lithna thar
39:19so yye dm
39:25a gaya
39:26kiske baad
39:27d i find
39:28karenge
39:28as a dm
39:37nai d i nikale
39:38nikale
39:40nikale
39:40nikale
39:40nikale
39:42d i dm
39:46into x
39:46sqr kye hoga
39:48korekat
39:50korekat
39:50korekat
39:55korekat
39:56korekat
39:57korekat
39:57korekat
39:58dm ki value
40:00m upon
40:02m upon
40:02l into
40:03d x
40:05and
40:17yyaan
40:18integration
40:19of d a
40:20m upon
40:23l
40:24x
40:26sqr dx
40:27limit
40:280
40:29sik
40:29l
40:30lagayeng
41:00So, kitna a gaya moment of inertia?
41:13ML square by 3 aya.
41:19Kitna aya?
41:20ML square by 3.
41:30Kitna aya?
41:42Kitna aya?
41:43Kitna aya?
41:54or find another no they can
42:00moment of inertia
42:06of
42:09follow cylinder
42:18about
42:23yes, uniform
42:25uniform
42:27more
42:34cylindrical
42:49yeah,
42:53you
42:59yes
43:17oh
43:19and this radius is R
43:23mass capital M and length capital M
43:33How do we find it?
43:41So hollow cylinder is so hollow cylinder
43:44in which we cut ring.
43:46What are we cut?
43:47Ring.
43:51This ring is cut.
44:12DM mass is this.
44:14And this is radius is r.
44:23So this is moment of inertia.
44:26Find.
44:28Find.
44:30Hollow cylinder.
44:32Surface mass density.
44:35Sigma.
44:37Mass upon.
44:38Is that area?
44:392 pi r into l.
44:52DM find.
44:54DM ke liye.
44:59DA find.
45:01DA ki value hooghi.
45:032 pi r into.
45:05Chikna sa it ki.
45:07DX.
45:16So DM a jaega.
45:22Sigma into DA.
45:25Sigma kitna?
45:26M upon.
45:282 pi.
45:30R.
45:31L.
45:32R.
45:33L.
45:34और यह.
45:35DA है.
45:362 pi r.
45:37DX.
45:382 pi r.
45:392 pi r.
45:402 pi r.
45:41Hensil.
45:42हो जाएगा.
45:43DM की value.
45:44M upon.
45:45L into.
45:46DX.
45:47के होगे.
45:48अब इस DM के लिए.
45:49DI find करेंगे.
45:51करिये.
45:52DI.
45:53O o o ndes.
45:54अब यह देखी.
45:55ring करके है.
45:56तो ring की cylindrical axis के बाउट,
45:58moment of inertia कितना होता है.
46:02MR square.
46:03तो DMR square हो जाएगा.
46:05Radius R है.
46:07तो यह हो जाएगा.
46:09DMR square.
46:10dmr square
46:13so this is a d i
46:21a d i a g a
46:22moment of inertia
46:24i find
46:25i
46:26i
46:40So,
46:42this is
46:44m
46:46r
46:48square upon l
46:50constant common
46:52dx
46:54integration
46:56limit
46:58length
47:00cylinder
47:02m r
47:04square upon l
47:06dx
47:08zero to l apply
47:10m r square upon l
47:12into l
47:14ho jayega
47:38i
47:40i
47:42i
47:44i
47:46i
47:48i
47:50i
47:52i
47:54i
47:56i
47:58i
48:00i
48:02i
48:04i
48:06i
48:08i
48:10i
48:12i
48:14i
48:16i
48:18i
48:20i
48:22i
48:24i
48:26i
48:28i
48:30i
48:32i
48:34i
48:36i
48:38i
48:40i
48:42i
48:44i
48:46i
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48:58i
49:04i
49:06i
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