Rotation | Angular Momentum | IIT-JEE/NEET #neet #jeemains #rotation #angularmomentum #neet #jee
Rotation | Angular Momentum | IIT-JEE/NEET #neet #jeemains #rotation #angularmomentum #neet #jee
In this live Lecture I'll discuss Angular Momentum in pure translation motion, pure rotation motion and in combined motion in Rotation motion for JEE Mains and NEET.
Rotation or rotational motion is the circular movement of an object around a central line, known as axis of rotation. A plane figure can rotate in either a clockwise or counterclockwise sense around a perpendicular axis intersecting anywhere inside or outside the figure at a center of rotation.
घूर्णन या घूर्ण गति एक केन्द्रीय रेखा के परितः किसी वस्तु की वृत्तीय गति है। एक समतल आकृति एक लम्बवत् अक्ष के परितः दक्षिणावर्त या वामावर्त दिशा में घूम सकती है, जो घूर्णन के केन्द्र पर आकृति के अन्दर या बाहर कहीं भी प्रतिच्छेद करती है।
Angular momentum is defined as: The property of any rotating object given by moment of inertia times angular velocity. It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object.
In physics, angular momentum is the rotational analog of linear momentum. It is an important physical quantity because it is a conserved quantity – the total angular momentum of a closed system remains constant. Angular momentum has both a direction and a magnitude, and both are conserved.
भौतिक विज्ञान में कोणीय संवेग, संवेग आघूर्ण या घूर्णी संवेग किसी वस्तु के द्रव्यमान, आकृति और वेग को ध्यान में रखते हुए इसके घूर्णन का मान का मापन है। यह एक सदिश राशि है जो किसी विशेष अक्ष के सापेक्ष जड़त्वाघूर्ण व कोणीय वेग के गुणनफल के बराबर होती है।
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #angularmomentum #angularmomentumclass11th
rotation motion, rotation motion class 11th, rotation motion class 11, rotation motion jee mains, rotation motion neet, rotation motion IIT JEE, rotation motion JEE Main, moment of inertia, rotation, JEE Mains, JEE Main, JEE Mains 2025, NEET, NEET 2025, Physics, Physics JEE Mains, class 11th, 11th, AK Sir, class 11th rotation motion, angular momentum, angular momentum class11th, angular momentum class11, angular momentum jee mains, angular momentum jee main, angular momentum IIT JEE,
In this live Lecture I'll discuss Angular Momentum in pure translation motion, pure rotation motion and in combined motion in Rotation motion for JEE Mains and NEET.
Rotation or rotational motion is the circular movement of an object around a central line, known as axis of rotation. A plane figure can rotate in either a clockwise or counterclockwise sense around a perpendicular axis intersecting anywhere inside or outside the figure at a center of rotation.
घूर्णन या घूर्ण गति एक केन्द्रीय रेखा के परितः किसी वस्तु की वृत्तीय गति है। एक समतल आकृति एक लम्बवत् अक्ष के परितः दक्षिणावर्त या वामावर्त दिशा में घूम सकती है, जो घूर्णन के केन्द्र पर आकृति के अन्दर या बाहर कहीं भी प्रतिच्छेद करती है।
Angular momentum is defined as: The property of any rotating object given by moment of inertia times angular velocity. It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object.
In physics, angular momentum is the rotational analog of linear momentum. It is an important physical quantity because it is a conserved quantity – the total angular momentum of a closed system remains constant. Angular momentum has both a direction and a magnitude, and both are conserved.
भौतिक विज्ञान में कोणीय संवेग, संवेग आघूर्ण या घूर्णी संवेग किसी वस्तु के द्रव्यमान, आकृति और वेग को ध्यान में रखते हुए इसके घूर्णन का मान का मापन है। यह एक सदिश राशि है जो किसी विशेष अक्ष के सापेक्ष जड़त्वाघूर्ण व कोणीय वेग के गुणनफल के बराबर होती है।
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #angularmomentum #angularmomentumclass11th
rotation motion, rotation motion class 11th, rotation motion class 11, rotation motion jee mains, rotation motion neet, rotation motion IIT JEE, rotation motion JEE Main, moment of inertia, rotation, JEE Mains, JEE Main, JEE Mains 2025, NEET, NEET 2025, Physics, Physics JEE Mains, class 11th, 11th, AK Sir, class 11th rotation motion, angular momentum, angular momentum class11th, angular momentum class11, angular momentum jee mains, angular momentum jee main, angular momentum IIT JEE,
Transcript
00:00you
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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