Rotational Mechanics | Rod Rotating With Disc | Similar to Irodov Problem 1.270 | Two Methods

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[@infinitevoid5141]

Hello sir,i have a doubt at 7.17,why did you took distance between Fcp and pivot 2l/3

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[@AyushKumarmaurya-r4b]

Why we take point of application of centrifugal force at 2l/3 distance

[@PhysicswithAkashGoyal]

You can confirm it by finding the point of application of centrifugal force ... For any linearly varying force which starts from zero will have point of application at L/3 .. just like pressure force

[@AyushKumarmaurya-r4b]

@PhysicswithAkashGoyal oh, thanks sir

[@Dr1z.]

Sir, in the first method why isn't the centrifugal force passing through COM? If it passes from L/3 distance in the first method, why doesn't it pass from the same point in the second method?

[@sohansingh2022]

Beautiful problem with explaination 🧡

[@preetamraj4955]

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[@sanghmitrarath6831]

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[@vaibhav5839]

Sir can we find angular acceleration using alpha=wdw/dtheta,btw nice explaination Sir

[@anuragmishra9778]

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[@bhaveshkumar2198]

Sir why can't we use mv(centre of mass) cross r in calculating angular momentum 🤔🙏

[@PhysicswithAkashGoyal]

Angular momentum calculation doesn't follow such rules. For every purpose you can't assume mass to be concentrated at CM. With this logic in this case angular momentum about CM should be zero, But it's not.

[@bhaveshkumar2198]

@@PhysicswithAkashGoyal sir after thinking over it I found a new method for the calculation of angular momentum which is as follows Since moment of inertia is a tensor quantity hence we can take its components.so moment of inertia of rod about the vertical axis will be ML^2/3sintheta(since angle is 90-theta with the vertical) and angular momentum is simply ML^2/3sin theta×w(omega) which matches with the result

[@bhaveshkumar2198]

@@PhysicswithAkashGoyal yes sir it was a silly question. forgive me 🙏🙏

[@PhysicswithAkashGoyal]

@@bhaveshkumar2198 Not at all silly. ☺️

[@PhysicswithAkashGoyal]

Yes, it can be done like this also.

[@nikhilchauhan2097]

Op sir back in form

[@nishchalsrivastava4444]

Thankyou sir 💯

[@meetnoob6059]

Sir isme ek vertical force bhi hai rod par, top end par jo axis ke sath juda hai, usko kyu count nahi kiya?

[@mrmajestical]

Rod isn't kept in contact with axis, the centrifugal force is keeping the rod in a slanted position

[@physicsformulas8403]

sir centrifugal force is a pseudo force. shouldn't we consider its point of application to be centre of mass of the rod. please answer sir.

[@PhysicswithAkashGoyal]

Point of application will be at CM only if force is distributed uniformly... Even mg should be applied at centre of gravity as value of g is different at different points... But if the body is not very tall then there is no appreciable change in g and it is treated as uniform. In case of translational acceleration of frame the pseudo force is uniformly distributed so net pseudo force can be applied at CM. In case of centrifugal force it's non uniform in general... Distance of points varies from axis of rotation

[@saksham_ks1670]

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[@anuragmishra9778]

sir ji 2l/3 wala mujhe yaad th but aap bta skte h iska explanation

[@PhysicswithAkashGoyal]

You can prove it by equating torque calculated using integration and then torque due to net centrifugal force. Assume some distance of point of application of centrifugal force to write it's Torque

[@deltazone1133]

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[@ashmit3675]

u can watch ns sir

[@deltazone1133]

Okk

[@aryansudan2239]

done thank you sir

[@sambhavmishra5423]

Happy Holi Akash sir

[@PhysicswithAkashGoyal]

Happy Holi to you too

[@Loseralways0]

thank you sir

[@musicloverr547]

Op 😍

[@vallarisharma7391]

Sir,why can't angular momentum be conserved when centrifugal force is also considered in fbd of the body ?

[@PhysicswithAkashGoyal]

About which point will you conserve it

[@K_PRIYAM]

🙂