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
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