1.5.18 · D5 · HinglishRotational Mechanics
Question bank — Equilibrium of rigid bodies — translational + rotational
1.5.18 · D5· Physics › Rotational Mechanics › Equilibrium of rigid bodies — translational + rotational
Do words jinhe hum throughout use karte hain, ek baar define kiye gaye hain taaki kuch bhi use se pehle build na ho:
Traps se pehle, teen figures har wo symbol build karti hain jo is page par use hote hain. Neeche kuch bhi use nahi hota jab tak yahan draw na ho.
True or false — justify
A body with is automatically in equilibrium.
False — ek extended body phir bhi spin kar sakti hai. Ek couple (do equal, opposite, offset forces) deta hai lekin ek nonzero net twist , isliye tumhe bhi chahiye; upar couple figure aur Couple and Moment of a Couple dekho.
"In equilibrium" ka matlab hai "at rest".
False — iska matlab hai zero acceleration, dono linear aur angular. Ek body jo constant velocity par glide kar rahi ho ya constant rate par spin kar rahi ho wo equilibrium mein hai; static equilibrium zyada strict case hai at rest rehne aur waise hi rehne ka.
Agar koi body momentarily at rest hai, toh wo equilibrium mein honi chahiye.
False — apni throw ke top par ball instantaneously at rest hai lekin nonzero net force (gravity) hai, isliye wo accelerate ho rahi hai, equilibrium mein nahi. Ek instant par rest ≠ balanced forces.
Jab ek baar ho, toh net torque same rehta hai chahe tum pivot ke roop mein koi bhi point choose karo.
True — pivot-shift term hai , aur ke saath yeh vanish ho jaata hai (upar inline sketch dekho). Isliye tum sabse clever axis choose kar sakte ho.
Agar forces balance nahi karte, toh net torque phir bhi pivot par depend nahi karta.
False — sirf tab jab hota hai torque axis-independent hota hai. Unbalanced forces ke saath bachi hui torque ko genuinely point to point change karti hai.
Ek single nonzero force jo ek free rigid body par act kare, usse kabhi equilibrium mein nahi chhod sakti.
False — ek force ka matlab hai , toh translation immediately break ho jaata hai, chahe yeh kahi bhi act kare.
Ek uniform body ka weight iska geometric centre par act karta hua treat kiya ja sakta hai.
True uniform bodies mein uniform gravity mein — centre of gravity wahan Centre of Mass ke saath coincide karta hai. Non-uniform field mein ye differ kar sakte hain; dekho Centre of Gravity vs Centre of Mass.
Ek couple plane mein har point ke baare mein same torque produce karta hai.
True — kyunki ek couple ka construction se hi hai, iski turning effect us axis se independent hai jis par tum ise measure karo (same reason pivot upar free hai).
Ek axis ke baare mein torque balance rotational equilibrium guarantee karne ke liye kaafi hai.
True sirf tab agar force balance already hold karta hai — phir ek point ke baare mein ise sabhi points ke baare mein force karta hai. Force balance ke bina, ek axis kaafi nahi hai.
Torques add karte waqt, ek force jo seedha pivot ki taraf point karti hai spin mein contribute karti hai.
False — ek force jiska line of action pivot se guzarta hai uska zero moment arm hai (, kyunki ), isliye yeh koi torque contribute nahi karta. Isliye smart axis choice support reactions ko erase kar deti hai.
Spot the error
Ek student likhta hai pivot tak ki full distance use karte hue.
Error — torque hai , perpendicular distance use karte hue, jahan aur ke beech ka angle hai (figure s01). Sirf ka component jo ke perpendicular hai body ko twist karta hai; dekho Torque.
Ek student ladder ka weight top rung par rakhta hai "kyunki wahan yeh lean karta hai".
Error — ek uniform ladder ke liye weight iske Centre of Mass par, yaani uske midpoint par act karta hai, moment arm deta hai, nahi (wahi midpoint exactly produce karta hai).
Seesaw balance karne ke liye ek student do weights ko equal set karta hai.
Error — torque balance do products weight×distance ko equal set karta hai, weights ko nahi. Ek light child jo door baitha ho ek heavy child ko jo paas baitha ho balance kar sakta hai.
Ek student kehta hai "pivot ka normal force torques balance karne mein madad karta hai, isliye ise include karo."
Error — agar tum torques pivot ke baare mein lete ho, toh iska reaction zero moment arm hai aur completely drop ho jaata hai. Ise ek torque term ke roop mein include karna ek unknown ko bina kisi kaam ke waste karta hai.
Ek student conclude karta hai ki constant rate par spin karta flywheel "equilibrium mein nahi hai kyunki yeh rotate ho raha hai".
Error — constant angular velocity ka matlab hai , isliye ; yeh hai rotational equilibrium mein. Motion theek hai; sirf acceleration equilibrium break karta hai.
Ek student ek force ke liye clockwise ko positive choose karta hai aur doosre ke liye counter-clockwise ko positive, usi equation mein.
Error — sign convention ek equation ke sabhi torques mein consistent honi chahiye. Ek sense fix karo (maano counter-clockwise ) aur use throughout stick karo.
Ek frictionless wall ke against ladder ke liye, ek student wall ke along ek friction force add karta hai.
Error — ek frictionless wall sirf ek normal (perpendicular) force exert karta hai. Friction rough floor par hota hai, smooth wall par nahi.
Why questions
Ek extended body ko ek doosri condition kyun chahiye jo ek point particle ko nahi chahiye?
Kyunki ek point ka koi size nahi hota, isliye "kahan" force act karti hai meaningless hai aur ise sirf push kiya ja sakta hai, kabhi twist nahi. Ek extended body mein geometry hoti hai, isliye offset forces ise spin kar sakte hain even jab wo pushes ke roop mein cancel ho jaate hain.
Problems mein pivot ko support par choose karna kyun itna useful hai?
Support reaction us point se through act karta hai, ise zero moment arm deta hai, isliye yeh torque equation se vanish ho jaata hai. Tum wo quantity solve karte ho jo tum chahte ho bina us reaction ko introduce kiye jo tum nahi chahte.
Steeper ladder ko less floor friction kyun chahiye?
Foot ke baare mein torque wall push deta hai — weight ke ladder ke midpoint par act karne se aati hai (arm ) versus wall ka full-height arm . Friction ko isse match karna hoga (), aur shrink hota hai jab , isliye steeper ladder ko less floor friction chahiye; dekho Static Friction.
Do forces zero mein sum ho sakte hain phir bhi body ko equilibrium mein nahi chhod sakte — kyun?
Agar wo alag lines par act karte hain toh wo ek couple form karte hain: force cancel hoti hai lekin har ek twist same rotational sense mein contribute karta hai, isliye torque mein add ho jaata hai (figure s02). Equilibrium demand karta hai ki dono sums vanish hon.
"Turning" describe karne ke liye force ki jagah torque kyun use karte hain?
Force Centre of Mass ki linear motion ko change karta hai; wo quantity jo rotation ko change karti hai wo torque hai, jo har force ko uske moment arm se weight karta hai — dekho Newton's Laws of Motion aur iska rotational analogue .
Do particles ke beech internal force equilibrium equations mein kabhi kyun nahi appear karta?
Newton's third law ke anusaar internal forces equal-opposite pairs mein aate hain jo force sum aur torque sum dono mein cancel ho jaate hain, sirf external forces balance karne ke liye bacha kar.
Edge cases
Agar sabhi external forces ek single common point se guzarti hain, toh kya torque balance automatic hai?
Us common point ke baare mein haan — har moment arm zero hai. Lekin tumhe phir bhi force balance alag se check karna hoga; concurrent forces phir bhi cancel karne mein fail ho sakti hain.
Ek body par exactly do external forces hain aur wo equilibrium mein hai — unke baare mein kya sach hona chahiye?
Wo magnitude mein equal, direction mein opposite, aur same line of action share karte hue hone chahiye. Same line couple ko khatam karta hai; opposite-aur-equal net force ko khatam karta hai.
Zero net torque teen non-parallel forces ke saath — unke lines of action kahan milne chahiye?
Wo concurrent hone chahiye (ek common point se guzarne chahiye), ya ek special parallel set hona chahiye; warna ek leftover couple appear hota hai aur rotational balance fail ho jaata hai.
Frictionless floor ke saath-saath frictionless wall par ek ladder — kya yeh kabhi static equilibrium mein ho sakti hai?
Tab wall ki push ko oppose karne ke liye koi horizontal force nahi ho sakti, isliye impossible hai jab tak wall push zero na ho, jiske liye ladder vertical chahiye. Practically yeh slip ho jaati hai; Static Friction floor par essential hai.
Jab ladder angle (almost flat), required friction ka kya hota hai?
Minimum coefficient of static friction, (sliding rokne wala sabse chhota ratio ), ke saath without bound grow karta hai, isliye koi real surface ise hold nahi kar sakta — ek flat leaning ladder ko zaroor slip hona chahiye.
Agar gravity switch off ho jaaye, kya ek floating couple phir bhi equilibrium fail karega?
Haan — couple ki twist do applied forces se aati hai, weight se nahi. Koi gravity nahi toh body phir bhi angularly accelerate karegi, kyunki regardless of jahan gravity act kare.
Kya ek body translational equilibrium mein ho sakti hai lekin rotational mein nahi, ya vice versa?
Dono ke liye haan — ek couple force balance deta hai bina torque balance ke; ek single off-centre force dono mein se kuch nahi deta, aur cleverly arranged forces ek point ke baare mein torque balance kar sakte hain jabki phir bhi slide karte hain. Do conditions independent hain.