Class 11 Physics | Rigid Body Dynamics | #37 Analysis of Rolling with Sliding | For JEE & NEET

[@asimpradhan3388]

Sir why you take initial linear momentum to be zero in case of vrw ???

[@hermionegranger8833]

In this case, sir told that the body was spinned with some ang. vel. and gently placed on the ground, no translational vel. given to the body..hence linear initial momentum is zero...

[@0oooo041]

At 4:31 why initial angular momentum is taken 0 If it have linear velocity it must have angular velocity So must have angular momentum

[@adeelsabir904]

Initially the motion was sliding only, no rotation was there

[@sumitsalunke395]

00:02 now we’ll discuss how to analyze, the motion of rolling, with sliding. we’ve already 00:10 studied in the previous article, that whenever there exist rolling with sliding, kinetic 00:18 friction acts on body. and kinetic friction acts, that means, v is not equal to r omega. 00:30 where v is the translational speed and omega is the angular speed about center. and 2 cases 00:37 are framed for this situation, 1 is v more than r omega and other v less than r omega. 00:44 for the 2 situations, we can have some specified methods to analyze, different kind of numerical 00:53 problems. let’s discuss the situation, when no external force is acting on the body. so 01:02 we talk about, when f external is equal to zero. and in our first case, when v is more 01:13 than r omega. let’s see how we analyze the situation, and let’s handle a simple problem, 01:25 in a specified manner. like this is a rough surface with coefficient of friction mu, and, 01:33 a body is placed on this rough surface. say if this is a round shaped body having a moment 01:41 of inertia i. and suddenly a translation speed v-not is applied to it, without any rotational 01:50 speed. so this is the situation v is more than r omega because initial translational 01:56 velocity is v-not, and no angular speed is given, so v is more than r omega. in this 02:03 situation, the point of contact will move translationally in forward direction, so it’ll 02:08 experience the friction in backward direction. and this kinetic friction we can write as, 02:14 mu m g, because the normal contact force will be balancing the weight of the body. and, 02:21 due to this friction, this translational speed will decrease because it is opposing, and 02:29 the torque provided by this friction will impart some angular speed, which will increase 02:35 due to the torque. and say at t equal to zero, we’ve placed the body and imparted an impulse, 02:42 due to which it started moving at speed v-not. and say we analyze the motion after time t, 02:51 one we can state the body is having some decreased linear speed v-f, and some increased angular 02:58 speed omega f, such that v-f is equal to r omega f. so as no external force is present, 03:06 and if v-f is r omega f, after this point onwards, we can write the friction is equal 03:12 to zero. upto this point the body was experiencing the friction mu m-g. so if we wish to find, 03:21 the time after which the body will start pure rolling, because now onwards we can say the 03:27 motion will be pure rolling. say we’re required to find the time, translational and angular 03:34 speed, one the body will start pure rolling, or the distance travelled by the body in this 03:40 period. so all such things we can very easily analyze by writing, impulse momentum equations. 03:47 if we write impulse equation, for motion of the body. we can say that, initial momentum 03:56 of body was, m v-not. the friction acted for a time t, in opposite direction, so i can 04:03 write minus mu m g t, then impulse imparted on the body in opposite direction, which is 04:11 product of force into time. and this must be equal to m-v final. so this is our equation 04:19 1. for rotation motion if i talk, this is for translational motion, for rotational motion, 04:27 obviously as no angular momentum was their, initial angular momentum we write as zero. 04:34 and the torque applied due to friction will be mu m g r, if r is the radius of this body 04:40 . and this torque is also acting for the period, zero to t. so we can add the angular impulse 04:49 which is mu m g r into t, which is equal to the final angular momentum of body about the 04:55 center, which can be written as, i omega f. and omega f as it is pure rolling , we can 05:02 write as, v-f by r, which is equation 2. now from these 2 equations, if we evaluate, we 05:12 substitute the value of mu m g t here like, this r i can cancel and square it. by just 05:17 taking this radius term on the other side of equality . and if we add the 2 equations, 05:24 this mu m g t cancells out, it is m v-not is equal to, m plus i by r square, into v-f. 05:36 so in this situation we can see, the final speed with which, the body is translating 05:43 at the time of pure rolling, will start, can be written as, m v-not by, m plus, i by r 05:51 square. this is the way how we find out the final speed, if i substitute final speed over 05:58 here, i’ll get the time, when the pure rolling will start. similarly if we wish to calculate, 06:05 the displacement of body, for the time it was in rolling with sliding. so this can also 06:13 be easily calculated by speed equations, as the acceleration of body we can write as, 06:20 minus mu g, because the friction is acting opposite to velocity, and acceleration will 06:26 by friction by mass that is minus mu g. so the time t, in the beginning of which the 06:34 speed of body was v-not and, if it has travelled a distance s, in same direction which is the 06:41 displacement covered by the body during its motion of rolling with sliding. this s can 06:47 be easily calculated as, v-not t minus, half mu g t square. and as we’ve already calculated 06:57 the value of t, we can get the displacement also. this is the way to easily analyze all 07:03 the parameters related to this motion. now, we similarly analyze our second case 07:13 analytically, which was the situation when v is less than r omega. where we can state 07:22 on a rough surface, we just take up a situation, which can be handled mathematically. for a 07:30 rough surface with friction coefficient mu, we’ve placed a body, in such a way that 07:39 initially it was spinned at an angular speed omega not, and it was dropped gently on the 07:46 surface. and was not imparted any, translational velocity, along the surface. so due to this 07:55 rotational motion we can say that the point of contact at the bottom will be, sliding 08:00 in backward direction with a speed r omega not due to which the friction will act on 08:05 it in forward direction. and again if the coefficient is mu, the friction can be written 08:11 as, mu m g because in this situation the normal contact force will be balancing the weight 08:16 of the body. now due to this friction, obviously the body will gain some translational speed. 08:25 and the torque due to friction we can say is in the direction opposite to the angular 08:30 speed . so it’ll decrease the angular speed to omega. say this was the time t equal to 08:36 zero when the body was spinned and placed on the rough surface. and if we talk about 08:43 the parameters of motion, that is final speed v-f, angular speed omega f and the time t, 08:51 when pure rolling will start. in this situation we can state, if finally v-f is equal to r 09:03 omega f, again we can write impulse equation as, we can state, for translational motion 09:14 , say initial momentum was zero. so whatever or whichever momentum it gains, is due to 09:23 the force of friction, which can be written as mu m g t. and this is equal to m v final. 09:29 this is equation 1. and for rotational motion initially it was having some angular momentum 09:36 i omega not, and the torque of friction is opposing it for the time t. so we can write, 09:43 minus mu m g r t is equal to, i omega final, which can be written as, v-f by r, as now 09:52 after time t pure rolling will start, and friction becomes zero, as no external force 09:57 is present. here in these equations, this r if i cancel out, this’ll be divided and 10:05 this’ll be r square. now if we add the 2 equations, we can see this term will cancel 10:11 out and it’ll be, i omega not by r, is equal to, m plus i by r square into v-f. you can 10:21 see we’re getting the final speed of body, when pure rolling will start. which is, i 10:29 omega not, by r, into m plus, i by r square. this is the expression we’re getting for 10:38 various kind of objects very easily we can analyze this. and this v-f, if we substitute 10:44 it here, we get the value of time, for rolling with sliding motion after which pure rolling 10:51 starts. and if i know the time certainly i can calculate the displacement, covered by 10:58 the body during the time it was sliding along with rolling, s as, initial speed is zero, 11:05 here we can write it as only, half a t square. and acceleration of body we know is, which 11:13 is caused only due to friction it can be written as, plus mu g. so this can be written as half 11:19 mu g t square. by substituting the value of time from here we’ll get the displacement 11:25 . any other parameter whichever are required in various kind of problems, we can evaluate 11:30 by using various speed equations, if you know the time and final speed and, angular speed 11:37 using these expressions.

[@kushagrawal7204]

bro you written whole lecture !!!

[@namishbaranwal3522]

Arre bhai poora subtitle de diye

[@timepass4783]

Transcript copy karke cool banne ke koshish kar rahe🤦‍♂️

[@notrps]

Just Downloaded The transcript and Pasted it here😂

[@vinashsharma330]

Great sir...also such an old video yet so useful

[@dhruvbhardwaj1580]

sir this was very helpful for me thank you sir

[@alphaiitd1]

Great explanation sir

[@paragdadhich6469]

Thanks sir

[@alexs6284]

very helpful ...

[@namanbiyani4u]

Sir can we use conservation of angular momentum about point of in rolling every time?

[@physicsgalaxyworld]

Not necessary... it depends on the given situation in the question... if there is some external torque present in the situation about the bottom point then it cannot be applied...

[@deepambanerjee5594]

Naman Biyani ask Binay Kumar kasera

[@vijayabhaskarareddy1820]

Sir you have taken initial angular velocity as w. But in equation 2 you have taken as 0. I did not get Sir

[@architsharma9404]

Sir, in case 1 when you are writing the impulse momentum equation for rotating body, why is the initial angular momentum taken to be zero? shouldn't it be w(initial)*I(moment of inertia) ?

[@kartiknarayansahoo3260]

Because the body is given a linear velocity only at that instant Hence no angular velocity

[@aravindgopal3555]

@@kartiknarayansahoo3260 thanks

[@prathitprasad]

So when a body pure rolls there is not at all any friction acting on it??

[@yashanchaliya7234]

As it is sliding so can we conserve angular momentum anywhere or we can do only at ground

[@physicsgalaxyworld]

bottom point...

[@jagadeshkrishna6229]

Sir,why T (torque)=dI/dt is not applicable in non inertial frames ?

[@kalpeshmalde1588]

External torque present instead use impulse equation

[@knightradar7552]

6:38 why didn't you take initial angular momentum in impulse momentum equation sir

[@janakpalit6197]

Because initial angular velocity is 0

[@vijayabhaskarareddy1820]

@@janakpalit6197 No. He have taken As Wo. But while application taken as zero

[@janakpalit6197]

@@vijayabhaskarareddy1820 yeah so it's zero. That's what I told

[@ZEELANIPATAN]

when the above blocks are arranged in such a manner, that each block projects out by same distance.......... system of particles and rotational motion problem sir, pls solve this problem

[@vaibhavmourya65]

Then that means that here friction is not dissipative force here but friction force implies it does not have its own energy (so no transfer of energy) so how can it give torque.?

[@codeine5642]

4:31 why have we considered that initial angular momentum as 0? Isn't it that initial angular momentum= 'Moment of inertia' times 'angular velocity'

[@divyansh5592]

because initial angular velocity here is taken zero

[@surajkumarbehera]

sir Are the topics instantaneous axis of rotation and toppling ( that am frequently finding in comments ) important for my AIIMS exam or I can skip them...........

[@physicsgalaxyworld]

Not exactly needed... but if you cover thats good...

[@huzaifaabedeen7119]

Selection hua????

[@subhrojitdas9541]

@@huzaifaabedeen7119 tujhe kya tu padhai kar

[@pramodvc2958]

Sir the body wont slip again in the due course of time

[@arpangupta3775]

sir change in momentum =final velocity -initial velocity ...but in eq 1 it is coming final +initial velocity....

[@physicsgalaxyworld]

+Arpan Gupta We are using the same difference... it depends upon the direction of impulse whether it increases or decreases the momentum...

[@liitkgp6570]

Sir I have one doubt if any external force is applied by cue to billard ball then during writing of impulse momentum equation we will write mv1 + impulse by friction or impulse by ext force = mv2

[@sujeetparth2893]

Sir here how can i understand in impulse equation,when will be positive or when it will be negative???

[@venkatbhardwaj9593]

if angular velocity increases then +ve if angular velocity decreases then -ve

[@neharikasinha7827]

sir in the above situation why are we not considering RVo in the equation as initial momentum

[@physicsgalaxyworld]

Pl mention the time instant concerned to your query... as it is already taken...

[@HarshvardhanNigam11]

at 7:49 sir clearly mentions that the body was not imparted any translational velocity initially it was just spun . hence we didnt consider MVo as intial linear momentum

[@saicharanreddy1053]

friction is internal force then how it is in impulse equations

[@physicsgalaxyworld]

for the rolling body it is an external force...

[@amankapoor7665]

@@physicsgalaxyworld how?

[@jangir.10]

@@amankapoor7665 on the body only the friction is acting as ext. force but for the body-surface system it is action reaction pair.

[@prabhatkumarverma200]

Sir I want to study IAOR... Instantly...Pl tell from where I can study it...Because I found on Google for booster classes but was not able to find yesterday...Pl help ....My tests are approaching ...

[@physicsgalaxyworld]

This we have covered in booster classes of dynamics in detail... you can search for these or wait for the schedule on PG LIVE mobile app...

[@ashutoshmanorkar5556]

why didn't you cover toppling?

[@physicsgalaxyworld]

Toppling is already covered in booster classes....

[@hermionegranger8833]

@@physicsgalaxyworld Sir, please upload the previous booster classes..or start again for jee 2021..it's a humble request sir..

[@anshumanngarg6814]

sir would the friction instataneously drop to zero at the point rolling starts or would it decrease continuously till rolling . thnx

[@physicsgalaxyworld]

In case of pure rolling in absence of external forces it is always zero...

[@physicsgalaxyworld]

In presence of external force friction increases with external force but rolling starts instantly as external force is applied...

[@alexs6284]

sir i think this translational impulse equation is nothing but first equation of motion ....right??

[@sumeetjain2973]

Saumya Singh I

[@sumeetjain2973]

Yes that’s right

[@lovvyparhar393]

Sir you said that there is no slipping tendency at the point of contact during pure rolling with no external force as the relative velocity is 0 at that point. So when a force is applied, a static friction comes into play to save the pure rolling motion as the applied force wanted to accelerate it translationally beyond ωR and the friction causes some increase in angular velocity and regulates translational velocity to keep it equals to ωR. Sir is it correct, i need help as it is a doubt in my mind. Please reply.

[@physicsgalaxyworld]

Yes... by external force there is a tendency of slipping at the bottom point and it may be in forward or backward direction depending upon the point of application of force...

[@ashijain123]

Sir, the force which has an action-reaction pair in a system, that force is considered to be as an internal force of that system. You said that for rolling body, friction is an external force......then for a block moving on a rough horizontal surface, friction will again be considered as an external force and for many other cases where the reaction of friction is to the ground [questions (in which friction was one of the force) of system of particles where w/e theorem is applied]. If yes, then whats so unique that you specified like FOR ROLLING BODIES, it is an external force (in the comment below)?? Bit confused now....plz clear this query

[@physicsgalaxyworld]

There is nothing specific about it... I replied to it in reply of a query... read the query and then read my reply...

[@gurushiksha634]

Sir please tell me... Can we apply work energy theorem at 06:10 to calculate displacement

[@codeguy21]

Accelerated pure rolling?

[@antoniomantovani3147]

Not possible

[@rohitsanjay1]

why is it -1/2 (mu)gt^2 and not just -1/2gt^2 ? (at 6:55)

[@physicsgalaxyworld]

because due to sliding the acceleration is µg...

[@rohitsanjay1]

Yes sir I got to know thank you. You have removed all my apprehensions towards mechanics and I want to thank you for that!! :D

[@arjunmohammed7865]

Very goodly done

[@zeeshanfazili9081]

Sir which among these reduce double or triple bond

[@nitish0075]

Are u mad...?

[@namanbiyani4u]

point of contact with ground**

[@physicsgalaxyworld]

What exactly you wish to ask... pl specify...

[@nitish0075]

XD

[@kushtaneja7737]

topic Toppling is not hr...

[@physicsgalaxyworld]

+kush taneja What you wish to ask... pl clarify...

[@kushtaneja7737]

+Physics Galaxy Toppling is also a topic in RORATIONAL DYNAMICS..I couldn't find it