Rotation Motion | Pure Rolling Concept, IIT-JEE/NEET #neet #jeemains #rotation #rolling #neet #jee
In this live Lecture I'll discuss Pure Rolling 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।
Pure rolling is motion of the round object without any slipping or skidding at the point of contact between two bodies. The rolling motion is a combination of translational motion and rotational motion. In pure rolling, the point of contact with the surface has zero velocity.
In metalworking, rolling is a metal forming process in which metal stock is passed through one or more pairs of rolls to reduce the thickness, to make the thickness uniform, and/or to impart a desired mechanical property. The concept is similar to the rolling of dough.
धातुकर्म में, ढ़लाई या रोलिंग धातुओं के रूपान्तरण की वह विधि है जिसमें धातु को दो बेलनाकार रोलरों के बीच से होकर गुजारा जाता है और दोनों रोलर धातु को दबाते हैं।
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #rollingjee #rollingclass11th #purerollingclass11th
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In this live Lecture I'll discuss Pure Rolling 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।
Pure rolling is motion of the round object without any slipping or skidding at the point of contact between two bodies. The rolling motion is a combination of translational motion and rotational motion. In pure rolling, the point of contact with the surface has zero velocity.
In metalworking, rolling is a metal forming process in which metal stock is passed through one or more pairs of rolls to reduce the thickness, to make the thickness uniform, and/or to impart a desired mechanical property. The concept is similar to the rolling of dough.
धातुकर्म में, ढ़लाई या रोलिंग धातुओं के रूपान्तरण की वह विधि है जिसमें धातु को दो बेलनाकार रोलरों के बीच से होकर गुजारा जाता है और दोनों रोलर धातु को दबाते हैं।
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #rollingjee #rollingclass11th #purerollingclass11th
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LearningTranscript
00:00you
00:30Hello students, today we are going to talk about the concept of pure rolling.
00:42We have already discussed the concept of pure rolling.
00:45We are going to discuss the questions on this in today's lecture.
00:49Let us explain the concept of pure rolling once again.
00:55The concept of pure rolling is a very good concept.
01:04We are taking a body here.
01:09This is the body.
01:16Now when it will roll,
01:24it is rolling on this surface.
01:39There are three motions here.
01:42This is pure translation motion.
01:48This is pure rotation motion.
01:55And this is pure rolling.
02:00What is pure translation motion?
02:04In pure translation motion, the velocity of all the points of the body will be same.
02:09We have considered this as V0.
02:11The velocity of all the points will be same.
02:18This is pure translation motion.
02:24The velocity of all the points of the body will be same.
02:32This is pure translation motion.
02:35Here the body is doing pure rolling.
02:37So the body will rotate with omega about the center of mass.
02:43At the top point, the velocity will be in the right direction, V0.
02:49And here in the left direction, V0.
02:53At this point, V0 will be upward.
02:56And this is downward.
02:58So from here, the relation will come.
03:01The angular velocity of omega will be equal to V0 upon r.
03:09Now when we combine these two motions, we get pure rolling.
03:14Now in pure rolling, the center of mass velocity is Vcm.
03:21This will be equal to V0.
03:23When we combine these two motions,
03:25this V0 will be the center of mass velocity.
03:28Here the center of mass rotation will be zero.
03:33At the top point, this is 2V0.
03:39This will be the velocity 2V0.
03:41And at the bottom, the velocity of this point is zero.
03:45Zero.
03:49In this direction, V0 will be upward.
03:51And in the right direction...
03:57Here we have to tell a relation.
04:00The angular velocity of omega is equal to V0 upon r.
04:04That is, V tangential upon r.
04:07But in the case of pure rolling,
04:10the center of mass velocity is equal to V0.
04:16From here, we can write that
04:18omega is equal to Vcm upon r in pure rolling.
04:26This is a very important relation.
04:29The angular velocity of omega is Vcm.
04:33In pure rolling, the center of mass velocity is
04:37equal to Vcm upon r.
04:42Okay?
04:43No problem?
04:46Okay.
04:48Now let's ask a question.
04:50By the way, I had added a question.
04:53It got lost somewhere.
05:08It will take some time.
05:33See, this is a question.
05:38We will solve this.
05:55A light thread suspended from the ceiling
05:59is wrapped around a cylinder of mass capital M, radius capital R.
06:05If the cylinder is released from rest,
06:08find the acceleration of its center C
06:12in the subsequent motion.
06:14See, this is its center.
06:18C for cylinder.
06:22Its mass capital M, radius capital R.
06:25For this cylinder,
06:28what acceleration do we have to find for the center of mass?
06:32How will we find it?
06:35Now, the thread is wrapped around the cylinder.
06:41When the thread is wrapped around the cylinder
06:45and when it is released like this,
06:47it will roll and move downwards.
06:51When it rolls, it will have two motions.
06:54One will be the translation motion
06:56and the other will be the rotation motion.
06:58See, the acceleration of its center and mass
07:03Let us assume this to be A.
07:11Now, we will draw its FBD.
07:14What will we draw?
07:15FBD, Free Body Diagram.
07:17For its translation and rotation,
07:20we will make a free body diagram for both the motions.
07:23See, it is very simple to make a free body diagram.
07:26For pure translation motion,
07:28we are going to make a free body diagram here.
07:33FBD of cylinder
07:41in pure translation motion.
07:48See, the question is very simple.
07:50A light thread is suspended from the ceiling
07:54and this thread is wrapped around it.
07:56Its mass capital M, radius capital R.
07:58When it is released,
08:00when it is released from rest,
08:02it will accelerate downwards.
08:04We are going to find the acceleration of its center and mass.
08:08First, we will draw its FBD in pure translation motion.
08:13See, this tension T will be applied on top.
08:19We will show the force applied.
08:21Mg will be applied downward.
08:23In translation motion, T will be applied upward.
08:26Acceleration of center and mass is E.
08:31From here, the equation will be
08:33Mg minus T is equal to Ma.
08:40Mg minus T is equal to Ma.
08:43This is the first equation.
08:45It is a very simple equation.
08:48Okay.
08:51Now, we will draw its FBD in pure rotation motion.
08:56FBD of cylinder in pure rotation motion.
09:16In pure rotation motion,
09:18when we draw its FBD free body diagram,
09:21we have to draw the whole body.
09:24We will draw its diagram here.
09:29This will be the cylinder.
09:31We will draw its diagram here.
09:49See,
09:51Mg will be applied downward.
09:54Tension will be applied on its left side.
09:58This is the diagram.
10:00Tension T is applied upward on its left side.
10:04Mg will be applied on center and mass.
10:07So, Mg will be applied and T will be applied.
10:10Apart from this, no force will be applied on it.
10:13This is its center and mass.
10:17Now, when it will rotate,
10:23its angular acceleration will be alpha.
10:27This alpha will be its angular acceleration.
10:31So, in this direction,
10:35this is tangential acceleration.
10:39KT
10:40Like this.
10:41If angular acceleration is there,
10:43tangential acceleration will also be there.
10:47Now, we will write its center of mass about torque.
10:51We will write its center of mass about torque.
10:57See, this is the center of mass.
11:00Mg is passing through the center of mass.
11:03So, tau Mg about center of mass will be 0.
11:09See, Mg's line of action is passing through the center of mass of this cylinder.
11:16So, its torque will be 0.
11:20Okay.
11:21Now, T is left.
11:22See, T is the center of mass.
11:24This is the perpendicular distance R.
11:27Torque is force into perpendicular distance on the line of action.
11:36See, force is the line of action of tension.
11:39So, this will be the perpendicular distance.
11:42Torque will be the center of mass of tension about T into R.
11:51See, we can find its direction from vector R cross F.
11:55It is very simple to find the direction of torque.
11:59The vector form of torque is vector R cross F.
12:03See, we will find the direction like this.
12:06Here, the direction will come inside.
12:10Okay.
12:13Now, torque, net torque will be equal to I alpha about center of mass.
12:19Tau net about center of mass will be I alpha.
12:24And the value of torque is T into R.
12:29Cylinder.
12:30The moment of inertia of cylinder is
12:45Mr square.
12:48Here, we are taking it as a hollow cylinder.
12:51Mr will be square.
12:53And the value of alpha is A upon R.
12:56The tangential acceleration is 80 is equal to R alpha.
13:02So, the value of alpha will be A upon R.
13:11Now, we will solve this.
13:13This R will cancel its power.
13:15This R will cancel its R.
13:20See, the equation from here
13:26I am not able to understand.
13:28Okay.
13:34See, it is very simple.
13:36Here, we are talking about torque.
13:39Here, I have taken a cylinder.
13:41What have I taken?
13:42A cylinder.
13:43So, the force applied on the cylinder
13:46See, this was the question.
13:49This is a light thread.
13:52And this is a cylinder.
13:54So, the force applied on the cylinder will be mg.
13:57Tension T will be applied upward.
13:59So, we have made this diagram here.
14:02mg is applied downward.
14:04And tension T is applied upward.
14:07So, the torque applied on the center will be zero.
14:12mg is applied downward.
14:14And the line of action of mg is passing through the center.
14:18So, the torque is force into perpendicular distance.
14:22We have told about rotation in the first lecture.
14:25The torque is force into perpendicular distance on the line of action.
14:31Here, the line of action of mg is passing through the center.
14:37So, the torque will be zero when the perpendicular distance is zero.
14:43This is the line of action of tension T.
14:46And the perpendicular distance on this will be r.
14:49So, the torque will be force into perpendicular distance r.
14:57Now, the torque of one is zero.
15:00So, the net torque will be tau net of I alpha.
15:05Tau is equal to I alpha.
15:09We have told this in Newton's second law.
15:12So, the law of pure rotation is Newton's second law.
15:19Now, we will put the value of tau net T into r.
15:24I moment of inertia of cylinder is m r square.
15:30Alpha will be A upon r.
15:33The value of A tangential is equal to r into alpha.
15:38So, the value of alpha will be equal to A upon r.
15:42Now, we will solve this.
15:44We will get T is equal to m.
15:46We hope that there will be no problem.
15:49Now, we will add these two questions.
15:52First and second.
15:58We will add both the questions.
16:00First and second.
16:02When we add first and second,
16:06This is the first equation.
16:08mg minus T plus T is equal to ma plus ma is equal to 2ma.
16:14T minus T will cancel.
16:19mg is equal to 2ma.
16:24m minus m will cancel.
16:28The value of A will be g by 2.
16:32The value of acceleration of center of mass will be g by 2.
16:43Now, we will discuss one more question.
16:49What is the next question in this?
16:55.
17:14.
17:25Now, this is the question.
17:27Let's read this.
17:29This is a good question.
17:31This is a pure rolling question.
17:33This is a very good question.
17:35Cylinder 1 of mass m1 and radius r1.
17:43This is first cylinder and this is second cylinder.
17:45r1 rotates about a fixed axle.
17:49One end of a thread is connected to a block m mass.
17:54and the other end is wrapped on the periphery of cylinder 2
18:01which is kept on sufficiently rough table so that it does not slip.
18:08Cylinder 1 has a rough groove so that the string does not slip on it.
18:18This is the question.
18:20Let's see what we have to find in this.
18:28Acceleration of center of mass of cylinder 2.
18:32We have to find the acceleration of center of mass of cylinder 2.
18:36and acceleration of the block.
18:39and angular acceleration of cylinder 1.
18:42We have to find these values.
18:45First of all, the question asked here.
18:51This is one cylinder.
18:53and this is second cylinder.
18:55Its mass is m1 and radius is r1.
18:57and second cylinder's mass is m2 and radius is r2.
19:01and this is small m.
19:03The block of mass is hanging from it.
19:05This thread is passing over this cylinder.
19:13and the groove of this cylinder will be rough.
19:17The rope will not slip over the cylinder.
19:23Just like pulley.
19:25We pass thread over pulley.
19:27It will not slip over here.
19:30The pulley will rotate here.
19:33and when there is friction here.
19:35and there is mass also.
19:37Then tension will be t1 and tension will be t2.
19:42There will be different tension.
19:49Let's solve this.
19:51To solve this, first of all.
19:56If cylinder is kept on rough surface.
19:59Then rolling friction will be applied here.
20:01What will be applied?
20:02Rolling friction.
20:03I have told you about rolling friction.
20:05We can assume the direction of rolling friction in any direction.
20:11How do we assume?
20:13By our own will.
20:15Now some people will think.
20:17How can we assume by our own will?
20:19How can someone run here by his own will?
20:22See, running by our own will means.
20:25I have told you about rolling friction in last lecture.
20:30Rolling friction is a conservative force.
20:34If we assume rolling friction in any direction.
20:38If the value of rolling friction is positive.
20:42Then the direction we have considered is correct.
20:45If the value of rolling friction is negative.
20:48Then the direction we have considered is incorrect.
20:53Direction is only reversed.
20:55Magnitude remains same.
20:58We have assumed rolling friction in backward direction.
21:02Now we will draw FVD of heavy body.
21:07First of all, we will draw FVD of small m.
21:12It is very important to understand this question.
21:17This cylinder is passing a row above 1st cylinder.
21:22Here small m block is hanging.
21:26Mass of this cylinder is m1 and radius is r1.
21:30Mass of second cylinder is m2 and radius is r2.
21:33This cylinder will roll and this cylinder will rotate.
21:39Out of its fixed axis.
21:43Here we have to find acceleration of center of mass cylinder 2.
21:48We have to find acceleration of block.
21:51We have to find angular acceleration for cylinder 1.
21:55When we draw FVD here.
21:59First of all, we will draw free body diagram of small m.
22:06FVD of small m.
22:15No problem.
22:24Here we have translation motion.
22:28From small m mass, mg will be downward.
22:31And here we have taken tension T1.
22:35And its acceleration will be T downward.
22:41We assume this acceleration.
22:45From here its equation will be mg minus T is equal to ma.
22:52Now we will draw FVD of cylinder 1.
22:57Pure rotation motion.
22:59Cylinder 1 will rotate.
23:03Out of its fixed axis.
23:06Here we will draw FVD.
23:08Free body diagram of cylinder 1.
23:16Its mass is m1 and radius is r1.
23:28This is cylinder 1.
23:42First of all, we will draw diagram of cylinder 1.
23:46Then we will draw FVD of cylinder 1.
23:53Here the force will be m1g.
23:57This will be downward.
24:01Here tension T1 will be downward.
24:05And here T2 will be in backward direction.
24:10So in right direction.
24:13And its radius is center of mass c1.
24:17From here its distance is r1.
24:24Here normal action will be applied.
24:26Here it is fixed hinge.
24:28Here normal action will be applied.
24:30We assume this as n1.
24:32So this force will be applied on this cylinder.
24:37This is cylinder 1.
24:39mg will be downward.
24:41T1 will be downward.
24:43And this is hinge.
24:45So normal action will be applied on this.
24:49We will take pure rotation motion.
24:52What kind of motion we will take?
24:54Pure rotation motion.
24:58Its angular acceleration will be alpha 1.
25:06This is tangential acceleration a.
25:08So this will be alpha 1.
25:12Mass m1.
25:14Radius r1.
25:16Now we will write the equation for this.
25:19Here we will consider torque as center of mass about.
25:23In pure rotation motion we consider torque.
25:27So the torque of center of mass about mg will be 0.
25:34Tau m1g about c1 will be 0.
25:40The line of action of m1g will pass through its center.
25:47So the torque of m1g about c1 will be 0.
25:53If line of action of normal action will pass through its center.
25:58Then the value of tau m1 about c1 will be 0.
26:04The torque of c1 will not be 0.
26:09Tau t1 about center of mass will be t1 into r1.
26:19And the direction of t1 will be vector r cross f inside.
26:28And the line of action of vector r cross f will be perpendicular distance r1.
26:37So this will be tau t2 about c1.
26:43This will be t2 into r1.
26:47The direction of this will be outside.
26:52If we write net torque then these two torques are 0.
26:56One is outside and one is inside.
26:58This is not rotating clockwise.
27:00Tau t1 will be big.
27:02So we will write net torque here.
27:07Tau net will be equal to I alpha.
27:12And this will be t1 r1 minus t2 r1.
27:17And value of I will be m r1 square.
27:29One minute.
27:37This will be value of moment of inertia m1 r1 square.
27:49Sorry mass is m1 and value of alpha will be A upon r1.
27:54See this is tangential acceleration of r alpha.
27:59Here value of At will be equal to r1 alpha.
28:03Value of alpha will be equal to At upon r1.
28:16Now we will solve this.
28:18Here r will be common.
28:20This will be t1 minus t2 into r1.
28:25And here its power will be cancelled.
28:28m1 r1 into A.
28:34See this r1 will also be cancelled.
28:36So from here this equation will come.
28:39t1 minus t2 is equal to m1 A.
28:45This will be second equation.
28:47Now we will draw FPD of second cylinder.
28:53When we will draw FPD of second cylinder.
28:56It will have two motions.
28:58Because it will roll.
29:00Here one will be translation motion.
29:03And second will be rotation motion.
29:07See here we will draw FPD.
29:13First we will draw FPD of translation motion of second cylinder.
29:18FPD of cylinder second.
29:25And we will draw this for pure translation motion.
29:33When we will draw FPD for pure translation motion.
29:37We will show the force.
29:42M2g will be downward.
29:45This will be normal action.
29:47And in this direction tension will be applied.
29:50And in backward direction we have considered friction.
29:54This much force will be applied.
29:56No more force will be applied.
29:58Now in its forward direction there will be acceleration.
30:02Who can tell how much acceleration will be there?
30:09We will consider this as A2.
30:11What will be its value?
30:14See here we have considered its acceleration as A downward.
30:19So in this direction we have shown A.
30:22But this will not be A.
30:24We can consider this as A2.
30:29We will take acceleration of its center of mass.
30:32What we have shown here as A.
30:34This will be its tangential acceleration.
30:38This A will be its tangential acceleration.
30:41So the acceleration of its center of mass.
30:44Its value will be half of the acceleration of its top.
30:48See I have just started.
30:50A little while ago I have told you about velocity.
30:55This.
30:56The velocity of center of mass is half of the velocity of top point.
31:04Same relation of acceleration.
31:08So here the acceleration of its center of mass will be Ay2.
31:14So according to this we will put the value here.
31:17The value of A2 will be Ay2.
31:21See this is in pure rolling.
31:23The acceleration of center of mass is half of the acceleration of top point.
31:29And the acceleration of surface contact at bottom is zero.
31:35This is the concept of pure rolling.
31:37Only if you know this, the question of rolling will be solved.
31:40Otherwise no.
31:44See here the value of M2 will be M2g.
31:48And here this tension is T2.
31:51So T2 minus Fr is equal to M2 A by 2.
31:59This will be the next equation, third.
32:02Now we will make FVD of second cylinder for pure rotation motion.
32:13FVD of cylinder 2.
32:24Pure rotation.
32:28Pure rotation motion.
32:33For pure rotation motion we will draw FVD.
32:39For pure rotation motion we will draw FVD.
32:50See here the tension at top point of cylinder is in right direction, T2.
32:58And here the friction we have considered is in backward direction.
33:06Now according to this if it is rolling then it will roll like this.
33:11Its tangential acceleration will be in this direction.
33:16And its value will be A by 2.
33:21This will be angular acceleration.
33:23We will write it as alpha 2.
33:26Its mass M2 is radius R2.
33:30And we will apply force on it.
33:32Its Mg will be M2g downward.
33:35And this will be normal action.
33:39We will write its center as C2.
33:41See here we will write torque outside its center.
33:46So its Mg will be zero.
33:51Because the line of action of M2g will pass through its center.
34:00This will be zero.
34:01Torque of M2g will also be zero.
34:07Because normal action will also pass through its center of mass.
34:11Its torque will also be zero.
34:15If we take out torque of T2 then its distance from center of mass is R2.
34:21So its torque will not be above zero of center of mass.
34:26Its value will be tau T2 about C2.
34:31T2 into R2.
34:33Its direction will be inside vector R cross F.
34:38Where will be the direction?
34:40Inside.
34:45Okay.
34:48What is okay?
34:50You are asking such a good question.
34:53Now when we see the torque of rolling friction then see
34:56FR and this perpendicular distance will be R2.
35:00So the torque of rolling friction will be force into perpendicular distance.
35:06Means FR into R2.
35:11Sorry we have written this earlier.
35:17This will be the torque of rolling friction.
35:20Tau FR.
35:25Its direction will also be inside.
35:29Let's find the direction.
35:31This is vector R cross F.
35:35So this will be vector R and this is force in this direction.
35:39So its direction will also be inside.
35:42Means net torque will be FR into R2.
35:46So here both the torque will be inside.
35:49Means both will be added.
35:51So net torque will be
35:55We will add both of them.
35:57I alpha 2.
36:00The value of tau net will be
36:02T2 into R2 plus FR into R2.
36:08And value of I is M2 R2 square.
36:13So value of alpha will be
36:16A tangential upon R2.
36:21And we have written A tangential.
36:24It is Ay2.
36:30So value of alpha 2 will be A upon 2R2.
36:36Let's put it here.
36:37See R2 will be common.
36:44M2 R2 square.
36:48And this is
36:51Value of AT is A upon 2R2.
36:58See we will solve this.
37:00This R2 and this R2 will get cancelled.
37:04So T2 plus FR will be equal to M2 A by 2.
37:13This will be the fourth equation.
37:15There are so many equations.
37:17We have to add them.
37:19We have to solve them.
37:27T2 is also here.
37:31Now see how we will solve this.
37:33First we will add third and fourth equation.
37:37We will add third and fourth.
37:40After that we will put the value.
37:43So third and fourth we will add first.
37:56Third and fourth equation we will add.
37:59So fourth equation is T2 minus FR.
38:05And fourth equation is T2 plus FR.
38:09And the value on right side is T2 plus FR.
38:13M2 is A by 2.
38:15So this is M2 A by 2 plus.
38:20If we see third equation.
38:22So this is T2 minus FR.
38:27And this is M2 A by 2.
38:29So we will write M2 A by 2.
38:35See from here we will cancel rolling friction.
38:392 T2 will be equal to M2 A by 2.
38:44So this will be M2 A by 2.
38:49So from here value of T2 will be M2 A.
38:54From this we will give fifth equation name.
39:00Now we will need this in future.
39:04Sorry see here T.
39:10Now see here this is our first equation.
39:15And this is second equation.
39:20And this is third equation.
39:24But we don't need third.
39:26The first equation we will square.
39:30And here this is second equation.
39:33And with this we will add fifth equation to cancel T2.
39:49Crystal clear.
39:50Aditya Kumar just joined now.
39:53So first, second and fifth.
40:09See this is very simple question.
40:11Let's add this.
40:14First equation was Mg minus T1.
40:19Second equation was T1 minus T2.
40:23We are writing left side value.
40:25And fifth equation is T2.
40:31Which is equal to right side value.
40:34Let's see.
40:35By the way I remember.
40:41See in first equation right side value is ma.
40:45And in second equation value is M1 A.
40:49And in fifth equation value is M2 A.
40:56Not M2 A, M2 A by 2.
41:00So we put this.
41:05This is small ma plus M1 A by 2.
41:14And this is M2 A by 2.
41:16In this M1 A by 2.
41:26M1 A, M1 A by 2.
41:30This is M1 A.
41:35See this will come.
41:37Now from here we will take A common.
41:39So this is small m plus M1 plus M2 by 2.
41:45See very good value is coming.
41:47And see here in left side T1, T1, T2, T2 will cancel.
41:52This Mg will remain.
41:54So value of A will come.
41:57Mg divided by small m plus M1 plus M2 by 2.
42:04Ok.
42:06Hello.
42:13Is this in syllabus?
42:15Very good.
42:16Yes it is in syllabus.
42:18We are teaching pure rolling.
42:20Pure rolling is in syllabus.
42:22And physics is the best topic of 11th.
42:35So see we have found value of A.
42:37Now the question which we have written.
42:39Question was A.
42:42Exhalation of block was A.
42:45Value of A has come.
42:46Means our B part of question is solved.
42:51Exhalation of the center of mass of cylinder 2.
42:56What is center of mass?
42:58Exhalation of cylinder 2 will be A by 2.
43:02Means the value of A which has come.
43:04In that we will divide by 2.
43:06We will do now.
43:07Angular exhalation of cylinder 1.
43:10We will find this.
43:12Angular exhalation.
43:15So see here first.
43:21Exhalation of A block has come.
43:25Now we will find exhalation of center of mass of cylinder 2.
43:33This is the answer of second part.
43:35We are doing first part.
43:37Exhalation.
43:45Of.
43:48Center of mass of cylinder 2.
43:56Exhalation of cylinder 2 will be.
43:59AC2.
44:01This will be A by 2.
44:04And value of A is mg divided by m plus m1 plus m by 2.
44:10In this we will divide by 2.
44:14So this will be mg divided by 2 in the bracket.
44:19Small m plus m1 plus m2 by 2.
44:26This will be the exhalation of center of mass of cylinder 2.
44:32It is also in neat.
44:34Pure rolling, pure rotation.
44:36Yes the level of question will be little less.
44:40But rotation or rolling will not be removed from neat.
44:46Ok.
44:50It is a very good concept.
44:52This is a good level question.
44:57But rolling and rotation is also in neat.
45:00Pay attention to rotation.
45:05Now we will write the angular exhalation of cylinder 1.
45:13We will make cylinder 1 little.
45:16We have already made it.
45:23This is cylinder 1.
45:28Value of alpha 1 is A upon r1.
45:32We will directly write here.
45:35Value of alpha 1 is A upon r1.
45:41This is angular exhalation of cylinder 1.
45:45A upon r1.
45:47We have already written this equation.
45:50We will put the value of A here.
45:54mg divided by r1.
45:57small m plus m1 plus m2 by 2.
46:02This is the value.
46:04value of A is mg divided by small m plus m1 plus m2 by 2.
46:11This is the angular exhalation.
46:14We will find it like this.
46:17Can you see that?
46:20Hello.
46:28I didn't do it in my coaching.
46:33Why wasn't rolling taught in your coaching?
46:38Rolling is in the syllabus of 11th.
46:42Rolling is taught in rotation chapter.
46:46Yes, that is a different thing.
46:49Some people believe that rolling is in 11th.
46:52Some people believe that rotation is the most difficult in 11th.
46:56That is a different case.
46:58Rolling is in the syllabus of both 8th and GE.
47:03Rolling friction is a concept like this.
47:10Aditya Kumar.
47:15We have discussed two questions in one hour.
47:19On pure rolling.
47:21You can understand from this.
47:23We will discuss the standard of the question.
47:26It takes some time.
47:28Rolling is a good concept.
47:30What we have done is easy.
47:32Such a big question is not asked.
47:36Read it.
47:37Rolling is easy.
47:39The way to do it.
47:41In rolling, we discuss two motions.
47:45One is pure translation motion.
47:47Second is pure rotation motion.
47:50In rolling, we have to solve two parts.
47:57Will there be any problem?
48:05We are going to discuss one more question.
48:08We have taken this question on torque.
48:11We will discuss more questions on rolling tomorrow.
48:14We will tell you the concept of it.
48:20This is the question.
48:25Three forces act at the vertex of square of side A.
48:32It is small a side.
48:34As shown in the figure.
48:36To find net torque about point O.
48:40About O point.
48:42We have to find net torque.
48:47Ok.
48:49We have been given three forces.
48:52This is a square.
48:54Three forces are acting on it.
48:56About O point.
48:58We will find net torque.
49:00We will start the next lecture tomorrow.
49:04We will discuss one more question on torque.
49:07It has taken a lot of time.
49:10We have been given three forces.
49:13We call them F1.
49:16F2.
49:18F3.
49:23We will find the value of these three forces.
49:26If we see this.
49:29This angle will be 45 degree.
49:32This is a square.
49:35This angle will be 45 degree.
49:39This angle will be 45 degree.
49:41No problem.
49:43When we do the component.
49:46In this direction.
49:48Root 2 F.
49:50Cos 45.
49:52In this direction.
49:54Root 2 F.
49:56Sin 45.
49:59If we put the value.
50:01Root 2 F.
50:03Cos 45.
50:051 by root 2.
50:07Sin 45.
50:091 by root 2.
50:11Root 2 F.
50:13Cos 45.
50:15In this direction.
50:17F.
50:19In this direction.
50:21F.
50:23In this direction.
50:25F1.
50:27Torque.
50:29Tau F1.
50:31About O.
50:33Torque.
50:35Force into.
50:37Perpendicular distance.
50:39F.
50:41Into.
50:43Perpendicular distance.
50:45Perpendicular distance.
50:47Total distance.
50:49A.
50:51A.
50:53Under root.
50:55A square.
50:57A square.
50:59A root 2.
51:01Right angle triangle.
51:03Total distance.
51:05A by root 2.
51:07Total distance.
51:09A by root 2.
51:11Divide by 2.
51:13Half distance.
51:15Into.
51:17A by root 2.
51:19Torque.
51:21Vector R.
51:23Cross F.
51:25Direction inside.
51:29Direction inside.
51:31Second.
51:33F2.
51:35Torque.
51:37Force into.
51:39Perpendicular distance.
51:41A by root 2.
51:43Tau F2.
51:45About O.
51:47F.
51:49Into.
51:51A by root 2.
51:53Torque.
51:55Direction inside.
51:57Vector R.
51:59Cross vector F.
52:01In this direction.
52:03Vector R.
52:05F direction.
52:07Horizontal direction.
52:09Torque direction.
52:11Inside.
52:15Equation first.
52:17Second.
52:19For root 2 F.
52:21Torque.
52:23Two components.
52:25F and F.
52:27Line of action.
52:29In this direction.
52:31F component.
52:33Line of action.
52:35O center.
52:37Torque 0.
52:39Torque F.
52:43F dash.
52:45Torque.
52:47Force into.
52:49Perpendicular distance.
52:51Tau F3.
52:53About O.
52:55F2.
52:57A by root 2.
52:59Direction.
53:01Vector R.
53:03Cross F.
53:05Vector R.
53:07About O.
53:09Force direction.
53:11Torque direction.
53:13Outside.
53:15Sir.
53:17Why A by root 2.
53:19Ok.
53:21One minute.
53:23No.
53:25No.
53:27No.
53:29No.
53:31No.
53:33No.
53:35No.
53:37No.
53:39No.
53:41No.
53:43No.
53:45No.
53:47No.
53:49No.
53:51Ok.
53:53Now we make the same diagram.
53:55We make the same.
53:59We have to
54:03find this distance.
54:07This side is A.
54:09This is A.
54:11When we find the total length.
54:13This is 90 degree angle.
54:15In square.
54:17Ok.
54:19Under root.
54:21A square plus A square.
54:23So A root 2.
54:25No problem.
54:27Now this is the center.
54:29O.
54:31We have to find this distance.
54:35This distance.
54:39We consider this as A point.
54:41So this distance will be from O to A.
54:43Total of this distance.
54:45This is AB.
54:47AB by 2.
54:49Means.
54:51A root 2 by 2.
54:53We will write A root 2.
54:552.
54:57Root 2 into root 2.
54:59This root 2 is cancelled.
55:01So this distance is.
55:03A by root 2.
55:05Ok.
55:07Let's find this.
55:09We have found this value.
55:11When we find net torque.
55:13After O point.
55:15One torque is inside.
55:17And one torque is outside.
55:19So we will add these three equations.
55:21First.
55:23Second.
55:25Third.
55:27So this tau net.
55:29About O will come.
55:31This F.
55:33Into A by root 2.
55:35Plus F.
55:37Into A by root 2.
55:39And this will be minus.
55:41F A by root 2.
55:43See here.
55:45Two torque are inside.
55:47And one torque is outside.
55:49So the inside values.
55:51Will be added.
55:53And outside we have subtracted.
55:57This will be cancelled.
56:01So this value will come.
56:03F A by root 2.
56:05So this torque.
56:07Native out will come.
56:09And its direction will be inside.
56:13Ok.
56:15See here.
56:17In this way.
56:19We will find torque value.
56:21So this was the question on torque.
56:23No one will have any doubt.
56:31Ok.
56:33We will complete today's lecture here.
56:35And.
56:37In the next lecture.
56:39In the next lecture.
56:41On energy.
56:43Means in this.
56:45Conservation of angular momentum.
56:47Is the concept.
56:49And the question is on the concept of energy.
56:51So that concept.
56:53We will discuss in the next lecture.
56:55Ok.
57:05See here.
57:07When net is not 0.
57:09When net torque is not 0.
57:11Then there will be no equilibrium.
57:15See this is the diagram.
57:17Now here.
57:19The equilibrium is.
57:21One is translational equilibrium.
57:23Second is rotational equilibrium.
57:25So in translational equilibrium.
57:27See here.
57:29Two forces are in the right direction.
57:31And one is downward.
57:33And one is upward.
57:35These two will cancel.
57:37But these two are not in the right direction.
57:39Fnet will be equal to 2F.
57:41See here.
57:43I will show you.
57:47See this force is like this.
57:51Here at the top point.
57:53F is in this direction.
57:55And F is upward.
57:57And here.
57:59There are two components.
58:01This F here and this F here.
58:03See here.
58:05Translational equilibrium is not there.
58:07This F will cancel this F.
58:09So here.
58:11The net force will be.
58:132F.
58:15This body is not in the translational equilibrium.
58:19And for rotational equilibrium.
58:21We have found the center of mass.
58:23About its torque.
58:25This is the value of torque.
58:27This is the net torque.
58:29When torque net is not 0.
58:31So rotational equilibrium.
58:33Sorry.
58:35Rotational equilibrium will also not be there.
58:37Ok.
58:39So here.
58:41This body is not in the rotational equilibrium.
58:43And also not in the translational equilibrium.
58:45Ok.
58:47See now.
58:49We will complete today's lecture here.
58:51In the next lecture.
58:53We will discuss the topic of Fnet.
58:57And this.
58:59We will upload the PDF.
59:01On our telegram channel.
59:03SRV Physics Quota.
59:05You can download the PDF.