Rotation Motion | Energy In Pure Rolling, IIT-JEE/NEET #neet #jeemains #rotation #rolling #neet #jee
Rotation Motion | Energy In Pure Rolling, IIT-JEE/NEET #neet #jeemains #rotation #rolling #neet #jee
In this live Lecture I'll discuss Energy In 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।
Pure rolling is a combination of translation and rotation motion in which the point of contact comes to rest instantly as there is no sliding. The sum of the kinetic energies of translational motion and rotation gives the kinetic energy of such a rolling body.
Kinetic Energy of Rolling Motion
The kinetic energy in terms of the radius of gyration is given by K.E = (½)mv 2 [1 + k 2 /r 2 ], where k is the radius of gyration.
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #rollingjee #rollingclass11th #purerollingclass11th #energyinpurerolling
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In this live Lecture I'll discuss Energy In 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।
Pure rolling is a combination of translation and rotation motion in which the point of contact comes to rest instantly as there is no sliding. The sum of the kinetic energies of translational motion and rotation gives the kinetic energy of such a rolling body.
Kinetic Energy of Rolling Motion
The kinetic energy in terms of the radius of gyration is given by K.E = (½)mv 2 [1 + k 2 /r 2 ], where k is the radius of gyration.
#srbphysicskota #aksir #aksirphysics #science #physics #yt #rotational_motion #rotation #class11 #jeemain #iitjee #rollingjee #rollingclass11th #purerollingclass11th #energyinpurerolling
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LearningTranscript
00:00Hello students, in today's lecture, we will discuss the concept of pure rolling, which I have discussed in the last lecture.
00:20We will discuss some questions in today's lecture and we will discuss the concept of energy in pure rolling in today's lecture.
00:33So let's start, first we will discuss some questions on pure rolling, then we will discuss kinetic energy in pure rolling.
00:44So here we have taken the first question.
00:48Velocity of the center of smaller cylinder is V, there is no slipping anywhere.
01:03Velocity of the center of larger cylinder is V, there is no slipping anywhere.
01:28See, it is written here that there is no slipping anywhere.
01:42If there is no slipping, then it is pure rolling.
01:46In the case of pure rolling, the velocity of the bottom point is 0,
01:51and the velocity of the center is just double the velocity of the top point.
01:57The value of V is 2V.
02:01If there is a rod here, then the velocity of A point is 2V,
02:13then the velocity of B point will also be equal to 2V.
02:21Now from here we will find the velocity of the center of larger cylinder.
02:27This is Vc, V is center of mass for larger cylinder.
02:36So this will be VV by 2, the velocity of the top point is half the velocity of the center of mass in the case of pure rolling.
02:48And this will be 2V by 2,
02:55the velocity of the center of larger cylinder will be equal to V.
03:03So the correct answer for this is B.
03:08Now see the next question.
03:14T tangential force F acts at the top of a thin spherical cell of radius capital R.
03:25The acceleration of the cell if it rolls without slipping is.
03:31So here we have a thin spherical cell of radius capital R.
03:45And at the top point capital F force is acting,
03:50and there will be no slipping, it will roll.
03:54Here rolling friction will be F R.
04:01So we have to find the acceleration of this.
04:04In the case of pure rolling, we draw two FVDs.
04:11One FVD is of translation motion and the other FVD is of rotation motion.
04:18So first of all we will draw FVD of this cell for translation motion.
04:28FVD of cell.
04:32First we will take pure translation motion.
04:36Pure translation motion.
04:42In pure translation motion we will draw FVD.
04:49It will be Mg downward, normal reaction upward.
04:53F is in right direction.
04:55And we can consider rolling friction in any direction.
04:59We have taken backward.
05:01No problem.
05:02We will solve it.
05:05This will be the acceleration.
05:07This will be the acceleration of center and mass.
05:09We will take A from this.
05:15This will be the acceleration.
05:17M is the mass.
05:19So the equation will be
05:21N is equal to Mg
05:23F minus F R rolling friction
05:27This will be M into A.
05:31Now for this cylinder we will draw FVD of pure rotation motion.
05:41FVD of cylinder.
05:47Here we will take pure rotation motion.
05:51For pure rotation motion we will draw FVD.
06:03Here we will draw cell.
06:21It is a thin cell.
06:23We will show the force applied on it.
06:27This will be Mg at center.
06:31This will be normal reaction.
06:33And at top point this will be F force.
06:37And this friction will be applied on surface.
06:41This will be force.
06:43We will see the torque of center.
06:47Moment of inertia will be I.
06:49And this will be alpha angular acceleration.
06:55When we see the torque of normal reaction.
06:59The line of action of this is passing from center.
07:03The torque of normal reaction will be zero.
07:09We take force into perpendicular distance.
07:11From line of action.
07:13And here line of action of normal reaction is passing from center.
07:19So perpendicular distance will be zero.
07:21Now we will see the torque of Mg.
07:25The torque of center of mass will be zero.
07:29Because line of action of Mg is passing from center of mass.
07:35So its torque will be zero.
07:37When we see the torque of force F.
07:41Torque of F about center of mass.
07:45This will be F into R.
07:47It will be clockwise.
07:49And torque of rolling friction.
07:51Tau Fr about center of mass.
07:55This will be Fr into R.
07:59Its direction will be inside.
08:01We will find the direction like this.
08:03We will find the direction like this.
08:05Vector R will cross F.
08:07So direction will be inside.
08:09And for this force.
08:11Vector R will cross F.
08:13So its direction will be inside.
08:17For net torque.
08:19We will add these two torques.
08:21So net torque will be.
08:23Tau net about center of mass.
08:29This will be I alpha.
08:31This will be F into R.
08:33Thus Fr is rolling friction.
08:35Into R.
08:37I moment of inertia.
08:39Its value is.
08:41For cell.
08:43So for cell.
08:452 by 3 Mr square.
08:51And value of alpha is.
08:57A upon R.
09:01Tangential acceleration.
09:0380 is equal to R alpha.
09:05Alpha is equal to 80 upon R.
09:07It is center of mass upon R.
09:09It is center of mass upon R.
09:11It is center of mass upon R.
09:13It is center of mass upon R.
09:15This will be.
09:17We will solve this.
09:19Here F plus Fr.
09:21Here F plus Fr.
09:23We will take capital R common.
09:252 by 3 Mr square.
09:272 by 3 Mr square.
09:292 by 3 Mr square.
09:31Here R will cancel.
09:33Here R will cancel.
09:35Here value will be.
09:37F plus Fr.
09:39Is equal to 2 by 3 Ma.
09:41Is equal to 2 by 3 Ma.
09:43This is second equation.
09:47And this was first equation.
09:49F plus Fr is equal to Ma.
09:51F plus Fr is equal to Ma.
09:53Here, we will add
09:55both of these equations.
10:01First plus second.
10:03When we will add
10:05both of these equations.
10:07First equation is F minus Fr.
10:09First equation is F minus Fr.
10:11And the second equation is
10:13F plus Fr.
10:15And the second equation is
10:17F plus Fr.
10:19And this M into e
10:21m into a plus second equation i.e. 2 by 3 m a
10:30so if we add this then this rolling friction will get cancelled
10:35here if we take m a lcm then it will become 3 for m a common
10:431 plus 2 by 3 and here we will take 3 lcm
10:483 plus 2 will become 5 and this is 2f
10:54from here we have to find the value of excitation
10:58we will cross multiply, we will multiply 3 here
11:02it will become 6f and we will divide 5m
11:07so the excitation will be 6f by 5m
11:17so we had taken small m in the question
11:21in the answer we have taken capital
11:256f by 5m so this will be the answer
11:29whose t option will be correct
11:33no problem
11:39so see it was a very good question
11:41it is not very difficult, it is simple
11:44only in rolling two concepts are used
11:47pure translation and pure rotation motion
11:54next question see this is also on pure rolling
11:59a ring of radius capital R is rolling purely
12:03on the outer surface of a pipe of radius 4R
12:11at some instant the center of ring has a constant speed
12:16c then the acceleration of the point on the ring
12:21which is in contact with the surface of the pipe
12:25so here in the question we have given a pipe
12:29whose radius is 4R
12:33and on it there is a ring of radius R
12:37which is rolling on it
12:39and the velocity of the center of this ring is V
12:44its radius is 4R of the pipe and its radius is R
12:49so in the contact of this point
12:53we have to find the acceleration of this point
12:59so let's find it
13:01it is easy to find, it is not difficult
13:18so here we will
13:49so here we will write its center of mass
13:53i.e. its acceleration of this point
13:55so it will be
13:57vector ACM
14:04plus
14:06its acceleration of this point
14:08see here
14:10the body will rotate around it
14:12so its one acceleration will be downward
14:15it is moving with V velocity
14:19if there will be circular motion
14:21then it will be centripetal acceleration
14:23and if it will rotate
14:25then its centripetal acceleration will be upward
14:29it is rolling
14:31so here there will be two accelerations
14:43so here we will put its values
14:46if we write the value of acceleration ACM
14:50then its center of mass will be
14:52its value will be
14:54V square upon
14:56R
14:58it will be V square upon
15:00R
15:06and this distance from this center
15:08is 5R
15:12this direction will be
15:15downward
15:17see here it is doing pure rolling
15:19so when it will do pure rolling
15:21see we will make its diagram
15:23here we will make it separately
15:27this is the body
15:29this is the small ring
15:35let's make two
15:37it will be useful
15:39see this is
15:41pure rolling
15:43in pure rolling
15:45the velocity of this point is V
15:47so it will be 2V
15:49and here we take 0
15:51so actually
15:53we divide it in two parts
15:55it is translation
15:57so all points velocity will be
15:59V
16:01and in pure rotation
16:03its velocity will be like this
16:05this will be like this
16:07this is pure rotation
16:09this is pure
16:12rotation
16:16so the centripetal acceleration
16:18of pure rotation will be
16:20towards A
16:22which we have written as A
16:24so if we write the value of A
16:26upward
16:28it will be V square upon
16:30R
16:34and this direction will be
16:36upward
16:38if we want to calculate net acceleration
16:41one is upward and one is downward
16:43so from this V square
16:45upon R
16:47we will less V square upon
16:495R
16:55this will be
16:57V square upon R
16:59minus V square upon
17:015R and this direction will be upward
17:03here we will take
17:05V square upon R
17:07very common
17:101 minus 1 upon 5
17:144 upon
17:165 will be the value
17:184V square upon 5R
17:20this acceleration
17:22will be in upward direction
17:24ok
17:40now see here
17:42we will tell
17:44the rolling on inclined plane
17:46on which we will tell
17:48rolling on inclined plane
18:04rolling on
18:07inclined
18:09plane
18:11when a body
18:13will roll on inclined plane
18:15here
18:17how much acceleration
18:19will be there
18:21in rolling
18:23we will find
18:25how much friction will be there
18:27we can calculate
18:29here we will take pure rolling
18:31pure
18:35here we are taking
18:37a body
18:40any body can be there
18:48its mass
18:50radius
18:52is R
18:56when it will roll
18:58the force applied on it
19:00this rolling friction will be applied
19:02we will take it
19:04in backward direction
19:06or far
19:10mg will be applied on its center
19:12and mass
19:14so
19:16there will be two motions
19:18one is pure translation
19:20and other is pure rotation
19:22here first of all
19:24we will take translation motion
19:26this is its center and mass acceleration
19:28a
19:30here first of all
19:32we will take
19:34FVD of m
19:36in pure
19:39translation motion
19:41in
19:43pure
19:45translation motion
19:49when pure translation motion will be there
19:51here
19:53when we draw FVD
19:55its mg
19:57will be downward
19:59this is theta angle
20:01in this direction normal reaction will be there
20:03and in backward direction
20:05we have written rolling friction
20:08mg cos theta
20:10mg sin theta
20:12and this will be acceleration
20:16we have done mg component
20:18we will cancel it
20:20from here m is equal to
20:22mg cos theta
20:26and this mg sin theta
20:32what is written here
20:38if mg sin theta
20:40is big
20:42then mg sin theta
20:44minus
20:46Fr
20:48is equal to
20:50mak
20:54see this equation
20:56we have written first equation
20:58mg sin theta
21:00is down the plane
21:02force
21:04and rolling friction is up the plane
21:07and acceleration is down the plane
21:09mg sin theta
21:11minus Fr
21:13is equal to m into ak
21:15this equation will come
21:17now we will make FVD
21:19for pure rotation motion
21:27FVD
21:29of m
21:31here
21:33in pure
21:35rotation motion
21:45pure rotation motion
21:57here we make body
22:05mg
22:07will be on center and mass
22:09so it will be downward
22:11and normal reaction
22:13is in this direction
22:17and this is rolling friction
22:19Fr
22:21no other force
22:23is there
22:25now we take angular acceleration
22:27m mass
22:29radius r
22:35see
22:37from here
22:39we will write
22:41this is normal reaction
22:43so we will write
22:45center of mass about torque
22:47tau
22:49normal reaction
22:51torque will be zero
22:53because normal reaction is passing from center
22:55so its torque will be zero
22:57mg's torque
22:59center of mass about zero
23:05because mg will also
23:07pass from center of mass
23:09only rolling friction's torque
23:11will be
23:15Fr
23:17into r
23:19and its direction will be outside
23:29and we will write this
23:31tau is equal to i alpha
23:34Fr into r
23:36i alpha
23:38will be
23:40a upon r
23:46so from here
23:48Fr is equal to
23:50i upon
23:52r square
23:54into ak
24:04now see
24:06first and second equation
24:08we write second equation
24:10and add both
24:12here i value
24:14is put
24:16we are taking general body
24:18we don't know
24:20moment of inertia
24:22so add both
24:24mg sin theta
24:26minus Fr
24:28plus Fr
24:30is equal to
24:33ma plus
24:35i upon r square
24:37into a
24:41this will cancel
24:43mg sin theta
24:45is equal to
24:47ma
24:51plus
24:53i upon r square
24:57into a
25:03from here
25:05we will
25:07calculate
25:09value of
25:11acceleration
25:13e value will be
25:15mg sin theta
25:17divided by
25:19ma
25:23plus i upon
25:25r square
25:27from here we cancel
25:29m common
25:32mg sin theta
25:34upon
25:36this m common
25:38a plus i upon
25:40m r square
25:48this m
25:50will cancel
25:56this acceleration will come
25:58g sin theta
26:01one minute
26:05what is this?
26:09sorry common
26:15this will be one
26:17one plus
26:19i upon
26:21m r square
26:23this value will come
26:25this
26:27should not be
26:29now
26:37now here we
26:39talk about energy
26:41kinetic energy
26:45kinetic energy
26:47in pure rolling
26:53kinetic
26:55energy
26:59in pure rolling
27:05kinetic energy in pure rolling
27:07means
27:09translation motion
27:11kinetic energy
27:13and rotation motion
27:15k rolling
27:17will be
27:19k translation
27:21plus
27:23k rotation
27:26translation
27:28motion
27:30kinetic energy
27:32and rotation motion
27:34kinetic energy
27:36will be
27:38half m
27:40bcm square
27:42and half i
27:44omega square
27:46this is
27:48kinetic energy of rolling
27:52in today's lecture
27:55we are not going to tell anything
27:57today we are
27:59taking leave