Foundations — Equilibrium of rigid bodies — translational + rotational

1.5.18 · D1 · Physics › Rotational Mechanics › Equilibrium of rigid bodies — translational + rotational

Is page par assume kiya gaya hai ki tum kuch nahi jaante. Parent note the parent topic ka har arrow, har letter, har chhota symbol yahan ground up se build kiya jayega, ek aisi order mein jahan har idea sirf usse pehle wale par lean karta hai.

0. "Rigid body" kya hota hai?

Topic ko yeh word kyun chahiye? Kyunki ek rigid body ke liye, ek end par lagayi gayi ek force poore object ke baare mein ek twist ki tarah feel hoti hai. Agar object bend ho sakta, toh door wala end care nahi karta. Rigidity hi woh cheez hai jo "tum kahan push karte ho" ko matter karaati hai.

Figure dekho: rigid rod ke top par same sideways push poori rod ko uske base ke baare mein swing karaata hai. Woh "ek point ke baare mein swing" is page ki har cheez ka seed hai.

1. Ek point aur uski position — vector $\vec{r}$

Forces se pehle, humein yeh batana hoga ki cheezein kahan hain. Ek reference point chuno (ise origin ya pivot kaho). Us point se kisi jagah tak ka arrow position vector hai.

Picture: pivot par khado, apni ungli wahan point karo jahan ek force act karti hai — tumhara haath hi $\vec{r}$ hai. Uski length yeh hai ki woh point kitna door hai; uski direction yeh hai ki tum kis taraf point kar rahe ho.

Topic ko yeh kyun chahiye: torque poochti hai "force kitni door apply hoti hai, aur kis direction mein?" Tum yeh pivot se ek arrow ke bina answer nahi kar sakte. Woh arrow $\vec{r}$ hai.

2. Force — symbol $\vec{F}$

Picture: ek arrow object se start hota hua, jis taraf tum use shove karte ho us taraf point karta hua. Weight bhi ek force hai — ek arrow seedha neeche ki taraf point karta hua, object ke centre of mass se drawn.

Topic ko yeh kyun chahiye: Newton's laws kehti hain ki forces hi woh cheezein hain jo motion change karti hain. Body ko still rakhne ke liye, humein us par act karne wala har arrow jaanna chahiye.

3. Vectors add karna aur symbol $\sum$

Picture: har force arrow ko tip-to-tail rakho. Tum jahan pahunchte ho — shuruat se lekar bilkul end tak ka ek akela arrow — woh $\sum \vec{F}$ hai, net force.

4. Ek force ko $F_x$ aur $F_y$ mein split karna — components

Tedhe arrows add karna mushkil hai. Trick: har arrow ko do perpendicular walls par ek shadow do — horizontal $x$-axis aur vertical $y$-axis.

Picture: seedha neeche se light chamkao — floor-shadow $F_x$ hai. Side se light chamkao — wall-shadow $F_y$ hai. Dono shadows milke arrow ko rebuild karte hain (woh $\vec{F}$ ko slanted side banake ek right triangle form karte hain).

Topic ko yeh kyun chahiye: ek mushkil 2-D balance ki jagah, hume do aasaan 1-D balances milti hain: $\sum F_x=0\text{and}\sum F_y=0.$ "Kuch bhi use sideways nahi khiinchta" aur "kuch bhi use upar ya neeche nahi khiinchta."

5. Angle $\theta$, aur $\sin \theta$ kyun aata hai

Ab twisting ke liye key idea. Position arrow $\vec{r}$ (pivot se jahan push karte ho tak) aur force arrow $\vec{F}$ lo. Unke beech ka angle $\theta$ hai.

$\sin$ kyun aur kuch nahi? $\sin \theta$ exactly woh fraction hai jo across (perpendicular) point karta hai rather than along. Jab $\theta =90^{\circ }$ force fully sideways hai, $\sin 90^{\circ }=1$: maximum twist. Jab $\theta =0^{\circ }$ (force $\vec{r}$ ke along), $\sin 0^{\circ }=0$: koi twist nahi. $\sin \theta$ ka answer hai "is push ka kitna fraction actually body ko turn karta hai?"

6. Torque — symbol $\vec{\tau }$ aur moment arm $d$

Ab hum "twisting power" ka naam de sakte hain.

Picture: force arrow ko ek poori straight line mein extend karo (uski line of action). Pivot se us line par ek perpendicular drop karo. Us perpendicular ki length $d$ hai. Torque $=F\times d$ — "kitna hard $\times$ kitna off-line."

7. Cross product $\vec{r}\times \vec{F}$ aur twist ka sign

Ek twist ki bhi ek direction hoti hai: clockwise ya counter-clockwise. Hum yeh sab cross product mein pack karte hain.

Picture: apne right hand ki ungliyon ko us taraf curl karo jis taraf force body ko spin karne ki koshish karti hai; tumhara thumb $\vec{\tau }$ ke along point karta hai. Thumb page se bahar = positive.

8. Mass $M$, acceleration ${\vec{a}}_{cm}$, aur Newton ka engine

Newton's second law poori body ke liye: $\sum {\vec{F}}_{\text{ext}}=M{\vec{a}}_{cm}.$ "Ext" ka matlab hai external — body ke bahar se aane wali forces; uske apne atoms ke beech internal pushes pairs mein cancel ho jaati hain. ${\vec{a}}_{cm}=0$ set karo aur tumhe pehli equilibrium condition milti hai, $\sum \vec{F}=0$.

9. Moment of inertia $I$, angular acceleration $\vec{\alpha }$ — twisting Newton's law

Newton's law ka rotational twin: $\sum {\vec{\tau }}_{\text{ext}}=I\vec{\alpha }.$ $\vec{\alpha }=0$ set karo aur tumhe doosri condition milti hai, $\sum \vec{\tau }=0$. Tumhe static problem ke liye $I$ compute nahi karna — lekin tumhe yeh jaanna chahiye ki yeh exist karta hai, kyunki yahi woh cheez hai jo "$\sum \tau =0\Rightarrow$ no spin-up" ko true banati hai.

10. Weight, centre of mass, aur friction — real problems mein players

• Weight $W=Mg$: ek downward force arrow, centre of gravity se drawn (uniform gravity mein centre of mass ke same). Ek uniform body ke liye woh geometric middle hai.
• Normal force $N$: ek support (floor, wall, pivot) apni surface ke perpendicular push karta hai.
• Friction $f$: ek sideways grip force, $f\le \mu N$, jahan $\mu$ friction coefficient hai. Yahi woh hai jo ladder ko slide hone se rokta hai.

Yahan kyun: parent mein worked examples (seesaw, ladder, couple) poori tarah inhi arrows aur do balance conditions se bane hain.

Yeh sab topic ko kaise feed karta hai

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