## Moment Bar / Principle of Moment

i am Aiswarya i am doing experiment based on principle of moment aim is to find unknown mass apparatus Theory its based on the principle of moment in rotational equilibrium net torque is zero F1 produce anticlockwise rotation F2 produce clockwise rotation when they cancelled scale become horizontal we know equation of torque=r X F this separation is r1 , with direction outward to left this weight produce force F1 , downward due to F1 , using right hand rule , direction of torque is from the screen ( thumb ) this separation is r2 , with direction outward to right weight of stone produce force F2 , downward due to F2 , using right hand rule , direction of torque is into the screen ( thumb ) T1 is towards you and T2 is opposite at preset bar is horizontal since no weight is acting due to support that point of axis of rotation is center of mass gravity of scale ( at 50 cm ) if at 50cm it is not horizontal its better to add some extra weight on the inclined end may be tie using a thick wire or something to make balancing at 50 cm like this now scale balanced at 50 cm ( its easy for calculation and better result ) we can start experiment at present equilibrium is on 50 cm marking and mass m1 is at 30 cm marking so separation r1=50-30=20cm not marking on CG value of r1=20 cm we can fix m1=100 g weight holder (50 g ) + one extra mass ( 50 g )=100 g now m2 portion is hanging down so we have to reduce r2 ( move m2 close to center ) to aligning the scale horizontally we moved the string close to center now we can check whether it is balanced still not balanced , m2 portion is still hanging below so move little more close to center move little more now m2 portion inclined up so we have to increase r2 ( m2 , move away from center ) here r1 is kept constant always no need to move m1 its a simplest idea to balance easily , and reduce confusion now aligned horizontally position of m2 is 62 cm marking therefore r2=62 – 50=12 cm here m1=100 g at that time corresponding r2=12 cm now taking m1=150 g by adding one more slotted weight of 50 g now m2 portion is inclined up so r2 have to increase ( move away from CG ) now it become horizontal now m2 is at 67 cm marking r2=67 – 50=17 cm when m1=150 g , corresponding r2=17 cm now fix m1=200 g adding one more 50 g weight now try to balance now m2 is inclined up , so increase r2 m2 move away from CG now aligned horizontal position of m2 is at 73 cm marking r2=73 – 50=23 cm like this choose m1=250 , 300 and 350 find corresponding r 2 values now calculations calculate first observation substitute in the equation m1=100 , r1=20 & r2=12 we get m2=166.67 g similarly next set m1=150 , r1=20 and r2= 17 repeat like this for all observations after that find mean value of m2 should convert to Kg ( SI unit )

Pakka bro

Iniyum upload cheyy bro cvm

Parallogram law of vectors idu please

Ethra parasyam ittu verupikalle

Metre bridge vdo uplodiyoo,connections..

Nth kidu ayitt anu ella video yum bhayakara helpful anu sir ithellm👏

Prism idd

thanks sir

This video is very helpful

This video is very helpful

use full video

Nale model practical ah..helpful

Helpful video

center of gravity kadu pidikumbol known mass etherayaaa edaedathe? ?????

Refraction through a prism please idu

Thank u so much,innu practical exam ayirunnu itha chodichethu

Tnqs for this video

Mass of given scale video idumo

Niceee

Adipoli chechi super ayi parannu thannu 👌👌👌👌

Helpful video

She looks like priya prakash varrier

No hate just opinion 😊

സർ, ഞങ്ങളുടെ സ്കൂളിൽ ചെയുന്ന രീതി വേറെയാണ്…..

Known Mass ഒരേപോലെ നിർത്തി അത് 10, 20, 30….. എന്നിങ്ങനെ നീക്കുകയും അതിനൊത്ത് unknown mass align ചെയ്യുകയാണ്.

Tnx

Main Practical ആയത് കൊണ്ട് കാണാൻ വന്നവർ അടി ലൈക്Very help full. Thanku.. 😘

Sir njangal unknown mass 10 cm akale vachitt known mass aanu move cheyyippikkunne. Is it right?

Thank u sir very helpful thank you soo much

are we finding the mass or weight

Helpfull

Helpful

Njan ente recodil inganella chythethu😥R2 ella redngsinum samanu

Nale practical ullavar ondo

Priya varrierde poleyund ah kutty 😁

Karthave minnichekkane…😂😁

Inn public an😢

Priya prakash warrior nte pole und

Thank u

Thank you for this video

Forward characteristics of a semi conductor diode onnu video cheyyuo

Tnx

Sir mass mathram kandal poore tao okke veeno athin

എല്ലാവരും dislike അടിക്കുകഇത് നമ്മുടെ Priya kuttoos അല്ലേCentre of gravity kaanumbol known mass full idano

Itz very helpful sir..she done it in a easy grasping way.👍

Ayyo naala exam aanu…myr…classil uzapp kanicheppo orthilla ithinte seriousness ithrem indaarnenn

help full…thank you….നാളെ practice ആണ്…😵✌

Valare vykthamai parannutharunnu super

Spherometer demo please

നന്ദിയുണ്ട് സുഹൃത്തേ നന്ദിയുണ്ട് 🙏🙏

Tnk u .. nale main practical exama moment barokke ithra simple ayirino

U are

exam saver😅it's very helpful to me

Nalla parizha ayathukoddu mathram

Sir plz add coefficient of friction also

TEACHER STRICT AANENNUPARanjathukonde VANNATHA

Adipoli

josh124 adi lik

Physics. Practical examin tharunna quastionpaper inganeyaano undavuka allenkil nammal thanne kallikal varachu cheyyano please reply(model pareeksha eyuthiyittilla mathramalla neeyanu pareeksha oru idea polumilla athukondanu chodikkunnathu )

thank you….. use full

Gud

Language problem

Focal length of liquid lens please

Nyx presentatiom

നാളെയാണ് നാളെയാണ് നാളെയാണ്😥sir njan kannadi school la padikunne innale aayirunnu phy practical public. sir ningalude enne sahayichu athupole njangalki vanna sirum adipoli ayirunnu

innu main exm.😏

Thanks for this

Thanks.eth kundapol exam azhuthan confedence feelaythu

Yanik nale main pratical aa

Helpfull video thank u sir .

Thank you…tomarrow is my practical exam…

The video is very helpfull..😘

Nala polekum

Ennu padetham nala exam result anthanoo antho

Thanks for this video

wish me luck ,tommorrow i have practical exam

Nala helpfull aanuto ee channel polichuu 👍👍👍

So help full video

Tnx , this helped me lot

ഒരു ആടാർ help ,

Well explained

can make a video in English so that it would be more clear

ശരിക്കുo പൊളിച്ചു