Exercises — Torque τ = r × F — definition, physical meaning

Kya pata hai: perpendicular ka matlab $\theta =90^{\circ }$, aur $\sin 90^{\circ }=1$. Why $\sin$: push ka sirf sideways wala hissa bolt ko spin karta hai (figure mein mint arrow); $90^{\circ }$ par saari force sideways hoti hai, toh maximum possible twist milti hai. $\tau =rF\sin \theta =0.25\times 20\times 1=5\text{N⋅m}.$ Answer: $\tau =5\text{N⋅m}$.

Kya pata hai: "spanner ke saath bolt ki taraf" ka matlab hai $\vec{F}$ $\vec{r}$ ke same direction (ya opposite direction) mein point karta hai, isliye $\theta =0^{\circ }$ (ya $180^{\circ }$). Why zero: pivot ki taraf aim ki gayi push kuch bhi circulate nahi kar sakti — jaise door ko uske hinges mein push karo. $\vec{r}$ aur $\vec{F}$ ka parallelogram picture karo: jab wo line up ho jaate hain toh woh flat ho jaata hai, area $=0$. $\tau =0.25\times 20\times \sin 0^{\circ }=0.25\times 20\times 0=0\text{N⋅m}.$ Answer: $\tau =0\text{N⋅m}$.

Unit: newton-metre, $\text{N⋅m}$. Type: torque ek vector hai — iska size bhi hota hai aur rotation axis / sense bhi (right-hand rule). Yeh dimensions joule ke saath share karta hai lekin energy nahi hai.

Step 1 — tool chuno: hume $r$, $F$, aur unke beech ka angle pata hai, isliye $\tau =rF\sin \theta$ direct route hai. Step 2 — why $\sin 30^{\circ }$: force ka sirf perpendicular slice, $F\sin \theta$, twist karta hai (figure mein mint arrow). $\sin 30^{\circ }=0.5$, isliye exactly aadhi force useful hai. $\tau =0.40\times 60\times 0.5=12\text{N⋅m}.$ Answer: $\tau =12\text{N⋅m}$.

Step 1 — tool chuno: vectors components mein diye gaye hain aur $xy$-plane mein hain, isliye 2D shortcut fastest hai: ${\tau }_z=x{F}_y-y{F}_x$. Step 2 — plug in: yahan $x=0,y=2,F_x=3,F_y=4$. ${\tau }_z=(0)(4)-(2)(3)=-6.$ Step 3 — sign padho (upar se convention): negative ${\tau }_z$ ka matlab clockwise hai, yaani torque vector page ke andar point karta hai ($-\hat{k}$). Answer: $\vec{\tau }=-6\hat{k}\text{N⋅m}$ (clockwise).

Why angle nahi chahiye: $rF\sin \theta$ ke do readings hain — (full distance) × (sideways force) aur (perpendicular distance) × (full force) — figure ko group karne ke do tarike hain. Agar kisi ne already perpendicular distance $d_{\perp }=r\sin \theta$ (butter arrow) de di, toh $\sin \theta$ us mein baked in hai — isliye simply poori force se multiply karo. $\tau =d_{\perp }F=0.15\times 80=12\text{N⋅m}.$ Answer: $\tau =12\text{N⋅m}$ (magnitude; sign depend karta hai is force ke body ko kis direction mein spin karne par).

Step 1: A ka torque: ${\tau }_A=0.90\times 50\times \sin 90^{\circ }=45\text{N⋅m}$. Step 2: B ka torque: ${\tau }_B=0.90\times 50\times \sin 50^{\circ }$. $\sin 50^{\circ }\approx 0.766$ ke saath, ${\tau }_B\approx 34.47\text{N⋅m}$. Step 3 — compare: extra fraction $=\frac{{\tau }_A-{\tau }_B}{{\tau }_B}=\frac{45-34.47}{34.47}\approx 0.305$. Answer: A, B se lagbhag $30.5\%$ zyada torque produce karta hai, sirf better angle par push karne se.

(a) Component route: ${\tau }_z=x{F}_y-y{F}_x=(1)(-2)-(3)(5)=-2-15=-17$. Isliye $\vec{\tau }=-17\hat{k}\text{N⋅m}$. Upar se sign convention ke hisaab se, negative ${\tau }_z$ = clockwise. (b) Magnitude cross-check.

• $r=\sqrt{1^2+3^2}=\sqrt{10}$.
• $F=\sqrt{5^2+(-2{)}^2}=\sqrt{29}$.
• Dot product se $\cos \theta$ milta hai: $\vec{r}\cdot \vec{F}=(1)(5)+(3)(-2)=5-6=-1$, isliye $\cos \theta =\frac{-1}{\sqrt{10}\sqrt{29}}$.
• Phir $\sin \theta =\sqrt{1-{\cos }^2\theta }$, aur $rF\sin \theta =\sqrt{10}\sqrt{29}\sin \theta$.

Numerically $\sqrt{290}\approx 17.03$, $\cos \theta \approx -0.0587$, $\sin \theta \approx 0.9983$, isliye $rF\sin \theta \approx 17.03\times 0.9983\approx 17.0$. ✓ Answer: $|\vec{\tau }|=17\text{N⋅m}$, component method se match karta hai. (Magnitude formula sign kho deta hai; component formula use rakhta hai — isliye hume pata hai yeh clockwise hai.)

Set-up: $\vec{r}$ rod ke saath point karta hai (length $1.2\text{m}$, horizontal se $60^{\circ }$ upar). $\vec{F}$ seedha neeche point karta hai. Horizontal se $60^{\circ }$ upar tilted arrow aur neeche wale arrow ke beech angle hai $\theta =90^{\circ }+60^{\circ }=150^{\circ }$. Why lever arm yahan neat hai: pivot se weight ki vertical line tak perpendicular distance (figure mein butter arrow) horizontal reach hai, $d_{\perp }=r\cos \varphi =1.2\cos 60^{\circ }=0.6\text{m}$. $\tau =d_{\perp }F=0.6\times 40=24\text{N⋅m}.$ $\sin \theta$ se cross-check: $\sin 150^{\circ }=0.5$, isliye $\tau =1.2\times 40\times 0.5=24\text{N⋅m}$. ✓ Answer: $\tau =24\text{N⋅m}$ (weight rod ko clockwise neeche spin karta hai, isliye signed quantity ke roop mein $\tau =-24\text{N⋅m}$).

Why hum sirf add kar sakte hain: fixed axis ke baare mein torque ek signed number hai; net twist individual signed twists ka sum hota hai. Upar se convention use karte hue (anticlockwise $=+$):

• $0.5\text{m}$ par downward force: yeh beam ko clockwise rotate karta — negative. Magnitude $=0.5\times 30=15\text{N⋅m}$. Contribution $=-15$.
• $2.0\text{m}$ par upward force: yeh beam ko anticlockwise rotate karta — positive. Magnitude $=2.0\times 50=100\text{N⋅m}$. Contribution $=+100$.
• (Dono forces horizontal beam ke perpendicular hain, isliye poori problem mein $\sin \theta =1$.) ${\tau }_{\text{net}}=+100-15=+85\text{N⋅m}.$ Answer: ${\tau }_{\text{net}}=85\text{N⋅m}$, anticlockwise.

(a) Torque — why lever arm exactly $R$ hai: rope rim ko tangentially chhohti hai, yaani circle ke edge ke saath. Centre se rim ke us point tak position vector $\vec{r}$ ek radius hai, jo hamesha tangent se right angle par milta hai. Isliye $\theta =90^{\circ }$, $\sin \theta =1$, aur centre se rope ki line of action tak perpendicular distance full radius $R$ ke barabar hai (butter arrow radius ke saath hi rehta hai). Ataah: $\tau =RF\sin 90^{\circ }=RF=0.10\times 8=0.8\text{N⋅m}.$ (b) Angular acceleration — new tool: jaise $F=ma$ force ko linear acceleration se jodta hai, Newton's Second Law for Rotation torque ko Moment of Inertia ke through angular acceleration se jodta hai: $\tau =I\alpha ⟹\alpha =\frac{\tau }{I}=\frac{0.8}{0.02}=40{\text{rad/s}}^2.$ Answer: $\tau =0.8\text{N⋅m}$, $\alpha =40{\text{rad/s}}^2$.

Why yeh steps: angular velocity steady rate $\alpha$ se rest se badhti hai, isliye $\omega =\alpha t$. Phir Angular Momentum hai $L=I\omega$ — linear momentum $p=mv$ ka rotational cousin.

• $\omega =\alpha t=40\times 3=120\text{rad/s}$.
• $L=I\omega =0.02\times 120=2.4{\text{kg⋅m}}^2/\text{s}$. Answer: $L=2.4{\text{kg⋅m}}^2/\text{s}$.

Why foot ke baare mein torque: foot par ground ka normal force aur uski friction dono act karti hain — wahan torques lete hue woh do unknowns zero lever arm ki wajah se vanish ho jaate hain, ek equation mein ek unknown bachta hai. Yeh standard Equilibrium of Rigid Bodies trick hai. Weight ka torque (clockwise, top ko neeche slide karne ki tendency):

• Weight midpoint par act karta hai, ladder ke saath $2\text{m}$ upar (Center of Mass & Gravity).
• Foot se uska horizontal lever arm $=2\cos 60^{\circ }=1.0\text{m}$.
• ${\tau }_W=W\times 1.0=200\text{N⋅m}$ (clockwise → hamari convention mein $-$). Wall force ka torque (anticlockwise):
• Wall top par horizontally push karti hai, height $=L\sin 60^{\circ }=4\sin 60^{\circ }=3.464\text{m}$.
• Foot se uska vertical lever arm woh height hai: ${\tau }_N=N\times 3.464$ (anticlockwise → $+$). Balance (net torque $=0$): $N\times 3.464-200=0⟹N=\frac{200}{3.464}\approx 57.7\text{N}.$ Answer: $N\approx 57.7\text{N}$.

Why hinge ke baare mein torque: hinge ki reaction force wahan zero lever arm ke saath act karti hai, isliye woh drop out ho jaati hai — sirf $W$ aur $T$ bachte hain. Weight ka torque (clockwise → $-$): weight free end par neeche pull karta hai, $1.5\text{m}$ bahar, horizontal rod ke perpendicular ($\sin 90^{\circ }=1$): ${\tau }_W=1.5\times 120=180\text{N⋅m (clockwise)}.$ Wire ka torque (anticlockwise → $+$): wire free end par rod se $40^{\circ }$ par pull karti hai, isliye sirf perpendicular slice $T\sin 40^{\circ }$ rod ko twist karta hai; uska lever arm poori rod length $1.5\text{m}$ hai: ${\tau }_T=1.5\times T\sin 40^{\circ }\text{(anticlockwise)}.$ Balance (hinge ke baare mein net torque $=0$): $1.5T\sin 40^{\circ }-180=0⟹1.5T\sin 40^{\circ }=180.$ $T$ ke liye solve karo: $T=\frac{180}{1.5\sin 40^{\circ }}=\frac{120}{\sin 40^{\circ }}\approx \frac{120}{0.6428}\approx 186.7\text{N}.$ Answer: $T\approx 186.7\text{N}$.

Step 1 — angular acceleration: $\alpha =\tau /I=4/0.5=8{\text{rad/s}}^2$ (Newton's Second Law for Rotation se). Step 2 — angle turned: $v^2=2as$ ke rotational analogue use karte hue, yaani ${\omega }^2=2\alpha \vartheta$: $\vartheta =\frac{{\omega }^2}{2\alpha }=\frac{20^2}{2\times 8}=\frac{400}{16}=25\text{rad}.$ Step 3 — rotational work: Work-Energy Theorem (Rotational) constant torque ke liye work $=\tau \vartheta$ deta hai: $W=\tau \vartheta =4\times 25=100\text{J}.$ Step 4 — kinetic energy se check: rotational KE $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.5\times 20^2=0.25\times 400=100\text{J}$. ✓ Answer: $\vartheta =25\text{rad}$, $W=100\text{J}$, gained KE ke barabar.

Components se directly torque kaunsa formula deta hai? ::: ${\tau }_z=x{F}_y-y{F}_x$ (2D, signed). Magnitudes aur angle se torque kaunsa formula deta hai? ::: $\tau =rF\sin \theta$. Equilibrium problems mein hinge/pivot ke baare mein torque kyun lete hain? ::: Wahan reaction force ka zero lever arm hota hai aur woh drop out ho jaata hai, fewer unknowns bachte hain. Torque ko angular acceleration se link karo? ::: $\tau =I\alpha$. Constant torque ka rotational work? ::: $W=\tau \vartheta$, jahan $\vartheta$ radians mein hai.