Class 11 Physics | Rigid Body Dynamics | #24 Conservation of Angular Momentum| For JEE & NEET

PG Concept Video | Rigid Body Dynamics | Conservation of Angular by Ashish Arora
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[@physicsgalaxyworld]

When the bodies are attached, by conservation of anular momentum we can calcuate its final omega which comes out to be MW/(M+2m). Now you can calcuate the difference of initial and final rotational KEs of the system. Ans will be option (4).

[@yashpandey9550]

when the smaller block is dropped then , the friction and the normal reaction b/w the block and the disc are internal forces , so we will not account them while calculating torque but the force mg on the block is not an internal force and its torque mgr must be accounted for , which will result in angular momentum not being conserved , which is not happening here , why ????

[@physicsgalaxyworld]

+Yash Pandey Think carefully... torque due to mg (mgr) is acting in horizontal direction not along the rotation of body.... it is balanced by the force applied by axis of rotation on thickness of the disc which is keeping the disc horizontal...

[@yashpandey9550]

Ooh , so only those torques which are along the AOR (and if making some angle with AOR then their component along AOR) is taken while solving problem....Ya i got it that the torque is along the disc but can you please tell me , the reason why the torques which are along the bpdy aren't able tp contribute to its rotation , as i didnt find any videos with that sort of thing being discussed ! Thanks :D

[@yashpandey9550]

*bpdy = body ; *tp= to

[@physicsgalaxyworld]

+Yash Pandey As in given cases AOR is fixed about which only the body can rotate so only those torques are considered of which vector is along the axis rest of the torques are balanced by the opposite torque by AOR so not considered in any cases and we ignore these but you must remain careful for such cases as questions can be asked on these as well...

[@yashpandey9550]

But if the torque makes an angle alpha with the AOR then we will take its component along the AOR AND NEGLECT THE component perpendicular to the AOR ??

[@alphaiitd1]

Thanks sir

[@amankapoor7665]

how to decide which force is internal and which one is external?

[@jigyasa1093]

sir have you covered the topic of hinge forces?

[@jigyasa1093]

sir the mg will not change the rotation but it will tilt the disc right?

[@tejakoncha1816]

Sir what u telling is enough for jee

[@tpsicmin]

Thank You Sir😎

[@sumanyugarg4576]

sir ..i got a little bit confused..why we havent taken the mg(mass of the small block placed)..?i reach to an explanation that the normal forces and frictional forces are internal..Mg(weight of bigger block) is through centre and also the hinge forces through the centre..(due to these torque become zero..) then..mg is left..and no other force to cancel its torque..?

[@physicsgalaxyworld]

On block mg and N are the only forces acting in vertical direction so these will get cancelled out... Only force left on disc will be N, the normal reaction in vertical direction which will extert a torque but it will not be affecting the rotational motion of disc so we are not using it.... Torque due to this N will tend to tilt the disc which is balanced by a couple exerted at center by hinge...

[@samikolarkar7227]

Sir why angular momentum act on any body while rotating?

[@jatinbhatt7826]

Sir since angular momentum is a vector quantity. If its direction is changing with time and not the magnitude then can we say there must be torque ??

[@physicsgalaxyworld]

Yes. Thats correct...

[@jatinbhatt7826]

Physics Galaxy thanks sir

[@namaste6313]

@@jatinbhatt7826 Kb kaha par ho

[@jatinbhatt7826]

@@namaste6313 ??

[@namaste6313]

@@jatinbhatt7826 ab konsa college me ho??

[@CristianoRONALDO-nw8hr]

Centrifugal force will not act on dropping the block on the moving disc???

[@next_thing_to_god]

friction will balance it

[@sparshagarwal3494]

Sir is newton's third law also has something to do with torque

[@physicsgalaxyworld]

Not directly... its for force only...

[@sparshagarwal3494]

Physics Galaxy A, coil of radius R carries current I. Another concentric coil of radius r (r

[@physicsgalaxyworld]

Here energy is conserved... and simply by conservation of energy above que can be solved... there is no such law for conservation of force or torque...

[@Ashutosh._.Gupta_]

Why friction acts in both directions?

[@physicsgalaxyworld]

Every force acts in pairs... one is action and other is opposite reaction... but the two forces always act on different bodies... here also the two friction is acting on different bodies...

[@indusaini1523]

Sir is it due to the fact that mg to produce torque should have a axis perpendicular to the given axis..

[@smittavate5790]

i have the same doubt....please elaborate what you understood

[@indusaini1523]

Sir is this disc is not hinged at centre...??? Plz help sir..

[@physicsgalaxyworld]

Yes it is hinged...

[@himanshumarmat3858]

Respected Sir if we change the definition of our system, we simply changes the external or internal force/torque but practically the situation is same so, will the linear/angular momentum still be conserved if not that why ?

[@physicsgalaxyworld]

Didn't get you... what exactly you want to ask... pl clarify... in any case if external force is acting on a system, its reaction is outside the system and due to net non zero force conservation laws cannot be applied, similar state is true for torque also...

[@himanshumarmat3858]

+Physics Galaxy Sir, consider two blocks lying on a friction less surface moving towards each other we consider both the blocks as our system, they collide then momentum is conserved as no external force is there. But now we consider one of the box as our system and the other box collides with it now the situation is practically same but in this case according to definition of our system there is external force, the momentum should be conserved practically but not theoretically . Please help.

[@physicsgalaxyworld]

As this query is not linked with this video, it may distract viewers of this video so I suggest you to post any other query in INTERACT tab of our website www.physicsgalaxy.com under the topic of your query, that will not only get other people in discussion but thats a forum where we can discuss on any topic....

[@Prodbybah]

@@himanshumarmat3858 bhai where are you now

[@HarishKumar-yr9gv]

​@@himanshumarmat3858 bro even practicaly in the second system momentum will not be conserved

[@vanshmishra7119]

sir but the gravity is acting on the whole body right? even if disc and block is the system then i understand ki AOR will cancel the effect fo force but not torque then Why did you conserve? pls sir do Answer half of my posted questions are not answered tho i follow all your vid.s almost and you ans the foolish comments those calling the accent bad an all ...and not my doubts i humbly ask you to clear my doubts thnx.....................in short if you have taken the disc block system then what force is cancelling the torque by gravity?

[@abhisheksahu6725]

sir if external torque is acting but it's resultant torque is zero Will the angular momentum will remain conserve?

[@physicsgalaxyworld]

Yes... but only about the point for which net torque is zero...

[@darkflamev]

Thank you so much! :)

[@hermionegranger8833]

Where are you these days?

[@GopalSharma-nj2zk]

hi uncle

[@sohruti]

@@hermionegranger8833 omg 😂

[@edunitianmanab2836]

@@hermionegranger8833 If you are so clever, you do this

[@prabhatkumarverma200]

sir....while balancing the angular momentum about any axis.... should we take into consideration only those angular momentum which are along axis of rotation ????

[@mehulsoni3581]

Sir suppose that we have a hemispherical bowl and a ball is projected along the rim of the bowl with velocity u and after some time its velocity (say v) makes an angle theeta with the horizontal tangent of the bowl. Can we write vxcosine(theeta) = u ?

[@physicsgalaxyworld]

No... use conservation of angular momentum to calculate it...

[@mehulsoni3581]

Physics Galaxy Sir but why horizontal speed won't be constant

[@physicsgalaxyworld]

because radius of path is decreasing and the normal reaction on ball is exerting a horizontal force on it...

[@mehulsoni3581]

Physics Galaxy Sir the horizontal force will be radial while the horizontal speed will be tangential. Then how can the force decrease the speed ?

[@adityarawat4964]

It will case like banking of roads

[@ashu1314]

Sir why friction is acting in 2 directions

[@HarishKumar-yr9gv]

One is exerted by mass m and other from disc

[@repudirajasekhar3884]

Sir... once plz explain how torque is zero...

[@physicsgalaxyworld]

Supporting external force on system is acting on the axis of rotation only where the platform is hinged... so torque of this force about this axis will be zero...

[@himanshuacharya215]

Thanks sir from the bottom of my heart 🙏🙏sita ram

[@darkflamev]

Could you help me out with this? A ring of mass M and radius R is rotating about its axis with angular velocity W. Two identical bodies each of mass m are now attached gently at the two ends of a diameter of the ring. What will be the loss in KE due to this? Options are: 1) ((M+m)M(W)^2(R)^2)/(M+2m) 2) (Mm(W^)2(R)^2)/(M+m) 3) (m(M+2m)(W)^2(R)^2)/M 4) (Mm(W)^2(R)^2)/(M+2m)

[@halamadrid13uclbarcelonado66]

I will help u

[@awesomeguy6171]

@@halamadrid13uclbarcelonado66 Kaafi Jaldi aa gaye help karne

[@halamadrid13uclbarcelonado66]

@@awesomeguy6171 😂🤣😂🤣8yrs ago lol

[@sandeepgarapati7965]

sir,in the below question they are given 2 masses so then final angular velocity is MW/(M+4m).....is it right or wrong..sir..if then ans is changed

[@hermionegranger8833]

Where are you these days?

[@edunitianmanab2836]

@@hermionegranger8833 In other words, Did you make it to IIT

[@next_thing_to_god]

@@hermionegranger8833 why emma watson pfp

[@next_thing_to_god]

@@edunitianmanab2836 did u tho?? (just asking )

[@hermionegranger8833]

@@edunitianmanab2836 yeah I did

[@adityapathak518]

Sir please help me ye chapter thoda samaj a rha hn sir Kain book solve kare jisase strong ho jye

[@physicsgalaxyworld]

you can use our 'Physics Galaxy' series of books...

[@adityapathak518]

Physics Galaxy Thanks a lot sir

[@varunrajesh1729]

@@lifestough2546 chup

[@divitgoyal5901]

sir is instantaneous AOR not in jee mains syllabus ??

[@physicsgalaxyworld]

No.

[@physicsgalaxyworld]

Only in case of pure rolling of a body on a surface you can use this concept of IAOR to quickly solve MCQs.

[@surajkumarbehera]

Physics Galaxy sir what is instantaneous AOR??

[@physicsfaadoooo5473]

Respected sir, In the example you told that there is no Torque due to ext firce about the com. How and why?

[@physicsgalaxyworld]

because line of action of force passes through COM so about COM torque of this force will be zero... as torque is the product of distance of line of action from the point...

[@YogeshSharma-ys3hm]

@@physicsgalaxyworld sir here please specify which force are you talking about and how acis of rotation can apply any type of force on the body please correcct me

[@physicsgalaxyworld]

If the AOR is passing through COM then due to rotation of body COM remain at rest but if this is not the case then COM will revolve in a circular path which is possible only if body experiences an external force along radially inward direciton... this is the force AOR applies in such cases...

[@alexs6284]

really very helpfull :):)

[@tanishjeetsingh4350]

what exactly is this external torque

[@MsRashi21]

sir i undeestoo

[@vishalgupta895]

When I first watch the video I never understand but watching after a few days I understand all the concept why is it happen to me sir

[@harrykane1347]

happens to everybody

[@hermionegranger8833]

Yes, it happens with everyone...where are you guys these days?

[@repudirajasekhar3884]

Thank u sir