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08:30hello hello hello
08:37hello
08:46hello
08:56angular momentum of system
09:00plus l3o plus lno
09:10angular momentum
09:13by the way, let's repeat it again, when we are not able to hear the sound from the beginning
09:21okay, now you can hear the sound clearly, let's explain it again
09:28angular momentum, here we have written angular momentum for a single particle
09:36angular momentum for point mass is
09:41vector r cross p, vector r cross m into v
09:47vector r cross v will be m v r sin theta, no problem
09:56here we have found angular momentum for the whole system
10:02here we have taken n particles, m1, m2, m3, mn
10:07they all have different velocities, v1, v2, v3, vn
10:12now when we find the angular momentum for O point
10:17the angular momentum of the first particle will be m1 vector r1 cross v1
10:25the angular momentum of the second particle will be m2 vector r2 cross v2
10:31the angular momentum of the third particle will be m3 vector r3 cross v3
10:38the angular momentum of the nth particle will be mn vector rn cross vn
10:44when we will sum all these, the angular momentum of the whole system will be
10:50angular momentum, no problem
10:55you must have understood till here
10:58now we will find the parts of this
11:01it is very important, now we will find this only in pure rotation of rigid body
11:08then we will find the angular momentum of combined motion
11:12translation plus rotation
11:15for a rigid body in pure rotation motion
11:22when a rigid body is in pure rotation motion
11:26then what is the angular momentum in that condition
11:30here we take a general shape of the body
11:36like this I have taken this body
11:39and we will find the angular momentum of this body
11:46we will take reference of this O point
11:49reference point means the point about which the body will rotate
11:57this body is rotating about this O point with the angular velocity of omega
12:04after this we will find the angular momentum
12:07the angular momentum is in pure rotation of I omega
12:17this is the angular momentum of pure rotation motion
12:22the torque tau is equal to I alpha
12:28and we will write tau as dL by dt
12:33and we will write alpha as d omega by dt
12:37from here this relation comes
12:40dL by dt is equal to d by dt I omega
12:49after comparing this we will write the formula L angular momentum as I omega
12:57here we will write moment of inertia about O point
13:07so this formula is of pure rotation motion
13:11angular momentum about O will be of I omega
13:16I moment of inertia will be about O point
13:19and about O point the body which is rotating with angular velocity is omega
13:26so this angular momentum will be of pure rotation motion
13:31for pure translation motion m vector r cross v is there
13:38ok
13:42after this for a rigid body in general combined motion
13:48here the body will be of general combined motion
13:53means translation and rotation will be there for that body
13:57so for that body we will find angular momentum
14:02this is the general shape of the body
14:14here the velocity of center of mass of this body is vcm
14:20and this body will rotate about its center of mass with angular velocity of omega
14:32so here we will find angular momentum about this point
14:45we will write angular momentum about O point
14:49this vector is rcm
14:51here I have taken rigid body
14:56this is in general combined motion
14:59translation motion is also there and rotation motion is also there
15:04center of mass about omega is rotating with angular velocity
15:08and vcm is translation velocity
15:16this will be its line of action
15:24this will be line of action for vcm
15:27this will be line of action for vcm
15:30here we will write angular momentum about O point
15:36vector l about O
15:39this center of mass about omega is rotating with angular velocity of omega
15:43so this is icm into omega
15:47this center of mass about will be its angular momentum
15:51and its center of mass will become point mass
15:56for O point
15:58and point mass velocity and angular momentum is m vector rcm
16:17cross vector vcm
16:19this formula will come
16:21for general combined motion
16:23we will find angular momentum about center of mass of first body
16:30this will be icm into omega
16:35this angular momentum will come about center of mass
16:39icm into omega
16:42and this is translation motion of center of mass
16:47this m vector rcm cross vcm
16:51this will be angular momentum because of translation motion
16:54and this is rotation motion
16:56so we will add both angular momentum in general combined motion
17:09now we will do question on this
17:11this is question
17:14a solid cylinder mass m radius r
17:19rolls on rough surface
17:22edge zone
17:24to find angular momentum about point P and Q
17:30for this body this is solid cylinder
17:33whose mass is m and radius is r
17:36this is rolling on rough horizontal surface
17:40when it will roll, translation and rotation will be both motions
17:45to find angular momentum about point P and Q
17:51first of all we will find angular momentum about point P
17:57first of all we will find angular momentum about point P
18:10this is pure rolling
18:14in rolling, angular velocity of omega is equal to vcm upon r
18:24about P angular momentum will be
18:30icm into omega plus m vector rcm cross vc
18:47to find angular momentum about point P
18:50first of all we will write icm
18:52value of center of mass about moment of inertia
18:57value of solid cylinder about moment of inertia
19:03this is mr square by 2
19:07value of omega will be vcm upon r
19:11means v knot upon r
19:15we will find direction of this
19:17vector r cross v
19:19direction will be inside
19:22we will show this with cross
19:26if we write x and y axis like this
19:30we can write minus k cap
19:40now for translation motion
19:45we have written i omega
19:48this will also be inside
19:51value of vector rcm will be r
19:54and value of vcm is v knot
20:01this will also be direction of minus k cap
20:05we will solve this
20:08this r will cancel with r
20:11so mr v knot by 2
20:14and mr v knot
20:16if we add this
20:18this value will be
20:203 mr v knot by 2
20:24direction will be minus k cap
20:27this angular momentum will come about P point
20:33we hope there will be no problem in this question
20:37we have found angular momentum about P point
20:42now we will find about Q point
20:49let us make a diagram
20:51this is body
21:03this is Q point at 2 hour distance
21:06and this is its velocity
21:09for this we will find angular momentum about Q point
21:15LQ will be
21:19Icm into omega plus m vector rcm cross vcm
21:28this is the formula
21:30we will put value
21:33Icm will be
21:36value of moment of inertia about center of mass
21:40mr square by 2
21:44value of moment of inertia about solid cylinder
21:48mr square by 2
21:51and this is pi omega
21:53direction of omega is inside
21:55value of omega is v knot upon r
21:59and this direction will be minus k cap
22:02from x and y axis
22:06minus z axis will be inside
22:12now we will put value of m vector rcm
22:17from here this vector r value will be
22:20equal to 2r
22:222r into
22:24value of vcm is v knot
22:26and this direction will be
22:28about Q point
22:32vector r cross v will be outside
22:39vector r cross v
22:42direction will be outside
22:50power of r will be cancelled
22:53this will be m v knot r by 2
22:57minus k cap
23:00plus
23:01sorry we will write outside as k cap
23:07and this will be 2m v knot r
23:11in k cap
23:13we will solve this
23:152m v knot r
23:17from k cap
23:19m v knot r by 2
23:21let's do this
23:22this value will come
23:303m v knot r by 2
23:34k cap
23:38ok
23:45next question
23:46next question
23:51this is also on general combined motion
23:54we have to find angular momentum
23:56a solid sphere of mass capital m radius r
24:00rolls on rough horizontal surface
24:04as shown
24:06to find angular momentum about point P and Q
24:10here also P and Q
24:13to find angular momentum about these two points
24:19here we have to find angular momentum of combined motion
24:24so we will write angular momentum of both
24:28for general combined motion
24:32formula of angular momentum
24:35for
24:38general combined
24:43motion
24:50so here
24:52angular momentum about P will be
24:55I cm into omega
24:58plus m vector r cm
25:02cross v cm
25:04this will be
25:06center of mass velocity is v knot
25:12we will put value
25:13I cm
25:14here it is solid sphere
25:16center of mass of solid sphere
25:20i.e. diameter axis
25:22value of moment of inertia
25:25is 2 by 5 m r square
25:322 by 5 m r square
25:35and omega will be
25:37v cm
25:39in pure rolling value of omega is
25:42v cm upon r
25:45i.e. v knot upon r
25:50and here we take x and y axis
25:54here it is
25:56it will roll like this
25:58so due to omega
26:00this direction will come inside
26:03minus k cap
26:06plus m
26:09about P point
26:11vector r cm
26:13cross v cm
26:15vector r will be equal to r
26:18and value of v cm is v knot
26:23direction will be minus k cap
26:26we will solve this
26:29power will be cancelled by r
26:33this will come
26:352 m v knot r by 5 minus k cap
26:402 m v knot r by 5
26:44this direction is minus k cap
26:48and m v knot r
26:51k v direction is minus k cap
26:54m v knot r
26:59when we add these two
27:02angular momentum will come about P point
27:11we will take 5 l cm
27:135 plus 2
27:157 m v knot r
27:17this direction will come
27:19minus k cap
27:24so like this we have found angular momentum
27:27about P point
27:29ok
27:32now we will write angular momentum about Q point
27:43Q point is the top point
27:49we will make it here
27:52Q is the top point
27:55Q is the top point
27:59center of mass velocity v knot
28:02this is Q point
28:04this will be omega
28:06this is surface
28:09after Q point we will write
28:12I cm into omega
28:14plus
28:16m vector r cm cross vector v cm
28:22after center of mass
28:25the value of moment of inertia will be
28:282 by 5
28:37m r square
28:40and value of omega will be
28:43P knot by r
28:47this direction will come inside
28:50minus k cap
28:55and here about Q point
28:58vector r cross v
29:01m into r v knot
29:04this direction will come k cap
29:07m vector r cross v
29:10this direction will come
29:13positive will come in k cap
29:16here we will solve both of them
29:21if its power is cancelled
29:24this value will be 2 by 5
29:27m v knot r minus k cap
29:322 by 5
29:38m v knot r minus k cap
29:43plus m v knot r
29:46k cap
29:49here we will take 5 lcm
29:52so this 5 m v knot r
29:56minus 2 m v knot r
30:02and this direction will come k cap
30:09from here in this question
30:12value of angular momentum
30:15when we find about Q point
30:18this value will come
30:213 m v knot r by 5
30:24direction will come in k cap
30:30about Q point angular momentum will come
30:34ok
30:40now see here
30:43after angular momentum
30:46we will tell conservation of angular momentum
30:49conservation
30:52of
30:55angular momentum
31:03first of all
31:06when we apply conservation of angular momentum
31:09we should know that
31:12when we apply conservation of angular momentum
31:15see for conservation of angular momentum
31:18the condition is
31:21any body any system
31:24except one axis
31:27if torque is 0
31:30tau external
31:33value of torque is 0
31:36at any point torque will be 0
31:39at that point net external torque
31:42will be 0
31:45so about
31:48d l by d t
31:51this will also be 0
31:54torque is d l by d t
31:57torque is 0
32:00d l by d t value will also be 0
32:03at that point
32:06when d l by d t is 0
32:09in that condition
32:12possibility is
32:15value of angular momentum
32:18will be constant
32:21and when it is constant
32:24see here
32:27we take a general body
32:30we have taken this body
32:33and on this body
32:36we have taken some point O
32:39about this point
32:42if torque is 0
32:45whatever torque is applied on this
32:48if net external torque value is 0
32:51then d l by d t will also be 0
32:54tau is equal to d l by d t
32:57value of d l by d t will also be 0
33:00so angular momentum will be constant
33:03and hence conserved
33:06as we have told in linear momentum
33:09if net is equal to 0
33:12then d p upon d t value
33:15is 0
33:18this is constant
33:21and hence conserved
33:27if net force is linear
33:30in translation motion
33:33if net force is 0
33:36then d p by d t value will be 0
33:39d p by d t value will be 0
33:42so p will be constant and conserved
33:45similarly here
33:48if torque is 0 at any point
33:51in rotation
33:54then d l by d t value will be 0
33:57and value of angular momentum
34:00will be constant and conserved
34:08now see here we take a diagram
34:15we have a disk
34:18and let's say
34:21we have a disk
34:24and let's say
34:27we have a disk
34:30and let's say
34:33we have a disk
34:36and let's say
34:39the axis of rotation of this disk
34:42This is the axis of rotation for this disk.
34:52Omega.
34:55This disk will rotate around this axis.
34:58So here, around this axis, this is the disk.
35:03Apart from this axis, if it has a torque zero, then in that condition, we will apply conservation of angular momentum.
35:14Like here, we have placed one person on top of this disk.
35:33Now this is the first limit of this disk.
35:48This is initial.
35:51This is final.
36:04Now see, here in this case, we apply conservation of angular momentum.
36:10This man is standing on top of this disk.
36:13So why will his Mg fall down?
36:16Why will his normal action fall up?
36:18Initially, his hands are spread like this.
36:21And after that, he has passed his hands.
36:25Let's make it a little better.
36:28Like this.
36:30See, his hands were spread like this.
36:33He passed his hands from here.
36:35So here, the force applied on the disk, the Mg will fall downward.
36:40Normal action will fall upward.
36:42See here.
36:45This Mg will fall downward.
36:52And normal action will fall upward.
36:55So apart from this axis of rotation, if we see the torque, the net torque will be zero.
37:03Tau, we write the axis of rotation as OO dash.
37:08And its value will be zero.
37:11Because both the forces, Mg and normal are passing from the same axis.
37:17So the perpendicular distance of the line of action will be zero.
37:21And when the torque is zero, then the value of dl by dt will also be zero.
37:31When the torque is zero, then the angular momentum of the axis of rotation will be constant.
37:39And hence, it will be conserved.
37:45The angular momentum of this axis of rotation will be constant and conserved.
37:52Okay.
37:55And this is also called the pure concept.
38:02What is it called?
38:04Pure concept.
38:07The diagram that we have taken here is the pure concept for the conservation of angular momentum.
38:16Now what is the pure concept?
38:19Here, the value of the torque of this axis is actually zero.
38:26Now in the concept of collision, the value of angular momentum is not zero.
38:35Still, we apply the conservation of angular momentum.
38:40Here, the value of the torque of this axis of rotation is actually zero.
38:46Hence, we apply the conservation of angular momentum.
38:56Now let us ask a question on this.
39:16Here, the mass of this disk is m and the radius is r.
39:37And here, this man is standing on it.
39:42He is holding two dumbbells in his hands.
39:49The mass of the dumbbell is m and the distance is r.
39:56This is the disk.
39:58Here, the angular velocity of this is omega knot.
40:12When he will bring his hands closer, the angular velocity will be find.
40:22This is omega find.
40:26Here, we have to find the angular velocity of the dumbbell.
40:35Now, we have to find the angular velocity of the dumbbell.
40:55Now, let us solve this question.
41:10Here, the value of the torque of this axis of rotation is mg.
41:18Hence, the value of torque of this axis of rotation is zero.
41:26So, the value of tau about axis of rotation is zero.
41:34Now, when the value of torque about axis of rotation is zero, then the value of dL by dt is also zero.
41:42angular momentum initial by conservation of angular momentum initial by conservation of
42:08angular momentum
42:19l initial o or l final
43:08final
43:22omega
43:41moment of inertia
43:53square by 2 plus 2 m r square
43:58capital R divided by m r square by 2 k
44:04angular velocity final
44:27m mass length l
44:54perpendicular axis
44:56about
44:58initial
45:00omega
45:02ball
45:04sphere
45:06mass
45:08m
45:10point
45:12mass
45:14axis
45:16of
45:18rotation
45:20angular
45:22velocity
45:25velocity
45:27angle
45:29velocity
45:31angular
45:33velocity
45:35angle
45:37now
45:39angular
45:41valocity
45:43h
45:45and
46:03body
46:18if final
46:23omega
46:28omega
46:33omega
46:38omega
46:43omega
46:48omega
46:53omega
46:58omega
47:03omega
47:08omega
47:13omega
47:18omega
47:23omega
47:28omega
47:33omega
47:38omega
47:43omega
47:48omega
47:53omega
47:58omega
48:03omega
48:08omega
48:13omega
48:18omega
48:23omega
48:28omega
48:33omega
48:38omega
48:43omega
48:48omega
48:53omega
48:58omega
49:03omega
49:08omega
49:13omega
49:18omega
49:23omega
49:28omega
49:33omega
49:38omega
49:43omega
49:48omega
49:53omega
49:58omega
50:03omega
50:08omega
50:13omega
50:18omega
50:23omega
50:28omega
50:33omega
50:38omega
50:43omega
50:48omega
50:53omega
50:58omega
51:03omega
51:08omega
51:13omega
51:18omega
51:23omega
51:28omega
51:33omega
51:38omega
51:43omega
51:48omega
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