Rotation | Rolling On Inclined Plane IIT-JEE/NEET #neet #jeemains #rotation #rolling #neet #jee
In this live Lecture I'll discuss Rolling On Inclined Plane 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 Rolling On Inclined Plane 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 the last lecture we discussed about the exhalation of the body when it moves
00:28the body will roll, in this way at theta angle, if a body rolls on inclined plane, the center
00:37of mass exhalation is of the same value, g sin theta divided by 1 plus i upon m r square
00:48i.e. g sin theta divided by 1 plus k square upon r square, k is the radius of gyration
01:02this is the formula, in the next lecture we will discuss about questions, in the last
01:12lecture we will discuss about questions, the kinetic energy of rolling, there will be two
01:20types of kinetic energy, two types of motion are combined in rolling, translation and rotation,
01:29translation and rotation both will have kinetic energy, so half m vcm square plus half i omega
01:38square, this is translational kinetic energy and this is rotational kinetic energy, the content of all the lecture we are teaching
01:51the pdf will be uploaded on our telegram channel, students who want to download the pdf, can
02:01download the pdf, now we will do questions in this, the first question is, starting from rest at the same time, a coin and a ring rolls down an incline without slipping, which reaches the bottom of the incline first,
02:27see here there is one ring and another coin, these two will roll on the inclined plane without slipping, in rolling forward slipping, backward slipping and pure rolling, there are three cases,
02:43here there will be pure rolling, the body will roll without slipping on the inclined plane, so which body will reach the bottom first, we have to find that,
02:56see here we make the inclined plane, this is the inclined plane, here there are two bodies, we are making only one body, you understand, here when this body will roll,
03:17both the bodies will start from rest, their mass m and radius r we assume for both, when they will roll, their center of mass acceleration, just now we have told that the center of mass
03:38acceleration comes, the value of a is theta angle, so the value of acceleration is g sin theta divided by I upon m r square or we can write it as
04:00g sin theta divided by 1 plus g square upon m r square, now see here when it will roll, the one whose acceleration will be more will reach the bottom first, and the one whose acceleration will be less will reach the bottom later, because here we will release them from the same point, so the length will be same,
04:28now the one whose acceleration will be fast will reach the bottom first, so we have to check the acceleration here, now how will we check the acceleration, so for the ring we will check for ring,
04:42the value of moment of inertia I is m r square for cylinder axis, and for disc it is ring, and for disc the moment of inertia is 1 m r square by 2,
05:06so here if we put the value in this formula, so the acceleration if we put m r square here, so this value will be 1, and 1 plus 1 is 2, so this acceleration will be g sin theta by 2,
05:23and if we put m r square by 2, so the acceleration will be g sin theta divided by 1 plus m r square divided by m r square, so it will be g sin theta by 2,
05:43and for disc if we put, so this acceleration is for ring, and for disc we will find, so this value will be g sin theta divided by 1 plus m r square by 2 divided by m r square,
06:08so here this m r square will get cancelled, so here this m r square will get cancelled, so this value will be g sin theta divided by 1 plus 1 by 2,
06:25so here this value will be g sin theta divided by 1 plus 1 by 2, so here this value will be g sin theta divided by 1 plus 1 by 2,
06:55so here disc is more and ring is less, so disc will reach first and ring will reach later,
07:02the coin reach the bottom first,
07:32so disc is reaching first at bottom, and ring will reach later, so coin will reach first and ring will reach later, so there is no problem, so let's start the next question,
07:47next question is, starting from rest, a coin roll down on inclined plane without slipping, when it reach the bottom, which is greater its translational or rotational kinetic energy,
08:11here there is a coin, it is rolling down, when it reach the bottom, it will have translational or rotational kinetic energy,
08:26now which kinetic energy will be greater, translational or rotational, we have to find this,
08:42so for this also we will draw the diagram,
08:45here there is a coin, it will start from rest,
09:07when it reach the bottom, it will have two kinetic energy, one is translational kinetic energy and other is rotational kinetic energy,
09:16in translational kinetic energy, its center and mass velocity will be Vcm, and when it will rotate, it will have angular velocity omega,
09:29so we will write the kinetic energy of translation and rotation separately,
09:36kinetic energy of translation, translational kinetic energy will be half m Vcm square,
09:45and rotational kinetic energy will be half I omega square,
09:55when the coin is rolling, it will roll above the cylinder axis, so the value of moment of inertia will be m r square by 2,
10:08and value of omega will be Vcm upon r,
10:15this is half m r square by 2, and this is Vcm square upon r square,
10:34this r square will be cancelled, so this will be m Vcm square by 4k,
10:48now the value of translational kinetic energy is half m Vcm square at the bottom,
10:57and the value of rotational kinetic energy is m Vcm square by 4k,
11:02means the value of translational kinetic energy is more than rotational kinetic energy,
11:09so the answer will be correct, its translational kinetic energy is,
11:26translational kinetic energy is half m Vcm square by 4k,
11:36and the value of rotational kinetic energy is half m Vcm square by 4k,
11:46translational kinetic energy is greater,
11:57its answer will be A,
12:02next question,
12:07each solid cylinder is rolling without slipping,
12:13what fraction of its kinetic energy is rotational,
12:25here a solid cylinder is rolling without slipping,
12:32so what fraction of its kinetic energy is rotational,
12:43here we first make a solid cylinder,
12:47we make it roll,
12:54when it will roll,
12:59this is the surface, it will roll on this,
13:04its center of mass velocity will be Vcm,
13:09and its omega angular velocity will be Vcm upon r,
13:15we write radius r as mass,
13:19here we will write its translational kinetic energy and rotational kinetic energy both,
13:26kinetic energy translation,
13:32translational kinetic energy will be half m Vcm square,
13:39and rotational kinetic energy will be half I omega square,
13:49here I will be its center of mass,
13:59for cylinder, cylinder axis value will be mr square by 2,
14:05omega value will be Vcm upon r,
14:19so this will be 1 by 4 mr square into Vcm square upon r square,
14:26here r square will cancel with r square,
14:30so this kinetic energy will be rotational Vcm square into m by 4,
14:39here we will get total kinetic energy first,
14:48then for rotation we will get fraction value,
14:51so the value of total kinetic energy will be,
14:57total kinetic energy,
15:01this kinetic energy cancellation plus kinetic energy rotation will be equal,
15:09when we add rotational kinetic energy and translational kinetic energy,
15:16we will get total kinetic energy of rolling,
15:20so translational kinetic energy will be m into Vcm square by 2,
15:27and m Vcm square by 4,
15:31here LCM will come,
15:33so this will be 3 m Vcm square,
15:38this 2 plus 1 will be 3,
15:41this is total kinetic energy,
15:45we can write this as kinetic energy of rolling,
15:48now the fraction we have to find is,
15:51kinetic energy rotation divided by kinetic energy rolling,
15:57we have found the kinetic energy of rotation,
16:01m Vcm square by 4,
16:05ok,
16:08m Vcm square by 4,
16:12and this is 3 m Vcm square by 4,
16:17we will solve this,
16:19m Vcm square by 4 will cancel,
16:23this will cancel,
16:283 will be left here,
16:30so this value will be 1 by 3,
16:34so its answer will be correct,
16:37A option will be correct,
16:39ok,
16:41so this kinetic energy was based on rotational and translational kinetic energy,
16:59now in the same way,
17:01there is a slight change in the same question,
17:06starting from rest,
17:09next question is,
17:11starting from rest,
17:13a solid sphere rolled down and inclined without slipping,
17:20here solid sphere is taken,
17:23when it reaches the bottom,
17:26the value of its translational and rotational kinetic energy,
17:33here we have to find the value,
17:36what will be the value of translational and rotational kinetic energy,
17:56so we will tell this,
18:00first we will draw its diagram,
18:13this is a solid sphere,
18:30we will release it from edge height,
18:40it will start from rest,
18:44it will roll without slipping,
18:48when it reaches the bottom,
18:53here we have to find its kinetic energy,
18:57for translation and rotation,
19:00we take m mass radius r,
19:03ok,
19:15now it will roll,
19:17when it rolls,
19:19in pure rolling,
19:21the energy is conserved,
19:25mechanical energy initial is equal to mechanical energy,
19:29we will put the final term,
19:31and at the bottom,
19:33we will find its kinetic energy and potential energy separately,
19:40so we take its reference gravity,
19:44reference gravity,
19:53when we take its reference gravity,
19:56gravitational potential energy will be zero,
19:59so here we will put,
20:01by conservation of mechanical energy,
20:06this mechanical energy initial is equal to mechanical energy final,
20:15so this k initial plus u initial is equal to k final plus u final,
20:23now initial kinetic energy starts from rest here,
20:28this will be initial position,
20:30and here it is final,
20:34initial kinetic energy is zero,
20:36but we will write initial potential energy as mgh,
20:43final potential energy will be zero,
20:45but kinetic energy will not be zero,
20:48here this kinetic energy will be half m,
20:52dcm square plus half icm,
20:56omega square,
21:03so this will be mgh is equal to half m,
21:10dcm square plus half i,
21:14moment of inertia of solid sphere is 2 by 5 mr square,
21:24this is 2 by 5 mr square,
21:29and value of omega will be dcm upon r whole square,
21:41we will solve this,
21:43so this will be 2 by 2 pencil,
21:47so value will be mgh is equal to,
21:54here we will take 2 by 5 10 rcm,
21:59so this will be 5 mvcm square plus 2,
22:12here this r square will get cancelled,
22:15it will be mvcm square,
22:17so this value will be 7 mvcm square by 10,
22:29from here we can write value of mvcm square,
22:37we will write mvcm square as 10 mgh divided by 7,
23:00there is no problem,
23:02we can cancel m but we have not cancelled,
23:05because we have to find different kinetic energy at bottom,
23:09so kinetic energy of translation is kinetic energy of translation,
23:16kinetic energy of translation motion is half mvcm square,
23:21value of mvcm square will be divided by 2,
23:26so translation will be kinetic energy,
23:31half mvcm square,
23:34here we will divide by 2,
23:37so it will be 5 mgh divided by 7,
23:43and kinetic energy of rotation is,
23:52mvcm square by 5,
23:55from here we will less mgh,
23:59so it will be kinetic energy of rotation,
24:03kinetic energy of rotation will be 2 mgh divided by 7,
24:10from total energy we will less kinetic energy of translation,
24:15so it will come,
24:25Thank you.
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