Exercises — Weightlessness — true (free fall) vs apparent

Jo hum jaante hain: rest matlab $a=0$, isliye $N=mg$. Step — mass nikalo. $N=mg$ se: $m=N/g=500/10=50\text{kg}$. Acceleration: $a=0$ (scale na move kar raha hai na accelerate). ==$m=50\text{kg}$, $a=0$.== Scale true weight tabhi dikhata hai jab $a=0$ — yahi baseline hai jis se baad ki har problem alag hoti hai.

Trap-buster: "neeche move karna" ek velocity hai, aur velocity kabhi $N=m(g+a)$ mein nahi aati — sirf acceleration aati hai. Step — $a$ nikalo. Constant velocity ⇒ $a=0$. Step — law apply karo. $N=m(g+a)=50(10+0)=500\text{N}$. ==$N=500\text{N}$ (normal weight — person ko kuch bhi unusual feel nahi hota).==

$N=m(g+a)=0$ set karo. Kyunki $m\ne 0$, humein chahiye $g+a=0$, yani $a=-g$. ==Answer: (c) $a=-g$.== Baaki check karo: (a) $N=m(2g)=2mg$ (double weight); (b) $N=mg$; (d) $N=m(g-2g)=-mg$ — negative $N$ floor ke liye impossible hai, matlab person floor chhod kar ceiling se press karne lagega.

Step — $a$ ka sign. Upar accelerate ⇒ $a=+30$. Step — apply karo. $N=70(10+30)=70\times 40=2800\text{N}$. Step — g-force. True weight $mg=700\text{N}$. Ratio $=2800/700=4$. ==$N=2800\text{N}$, yani 4 g — candidate apna weight chaar guna feel karta hai.==

Step — $a$ ki direction. Upar ja raha hai aur slow ho raha hai ⇒ acceleration neeche ki taraf hai ⇒ $a=-4$. Step — apply karo. $N=50(10-4)=50\times 6=300\text{N}$. ==$N=300\text{N}$ — true $500\text{N}$ se halka.== Dhyaan do: neeche-accelerate karte lift jaisa hi "halka feel" hota hai, kyunki upar ki rise ko slow karna aur neeche ki drop ko fast karna dono same downward acceleration hain.

Step — local $g$ use karo. Law hai $N=m(g_{\text{local}}+a)$ jahan $g_{\text{local}}=1.6$. Step — sign. Neeche accelerate ⇒ $a=-1.6$. Step — apply karo. $N=50(1.6-1.6)=0\text{N}$. ==$N=0$ — haan, weightless.== Weightless condition hamesha $a=-g_{\text{local}}$ hoti hai, chahe tum kisi bhi planet par ho. Moon par free fall bilkul Earth par free fall jaisi lagti hai — zero support force dono jagah.

Step — law ulta karo. $N=m(g+a)\Rightarrow a=\frac{N}{m}-g=\frac{480}{60}-10=8-10=-2{\text{m/s}}^2$. Direction: negative ⇒ acceleration neeche ki taraf, magnitude $2{\text{m/s}}^2$. Trick sub-question: Nahi — tum nahi bata sakte ki woh upar ja raha hai ya neeche. Scale sirf $a$ batata hai, velocity nahi. Ya toh neeche jate hue accelerate kar raha hai, ya upar jate hue decelerate — dono mein $a=-2$ aata hai. ==$a=2{\text{m/s}}^2$ neeche ki taraf; motion ki direction undetermined hai.==

(a) Free fall ⇒ $a=-g=-10$. $N=50(10-10)=0\text{N}$. Weightless feeling. (b) Ye Free Fall Kinematics hai: $v=u+at$ jahan $u=0$, $a=g=10$ (speed badhti hai, magnitude lo): $v=0+10(2.0)=20\text{m/s}$. (c) Nahi. True weight $mg=50\times 10=500\text{N}$ poore time — gravity puri tarah on hai; isliye hi elevator $g$ par accelerate karta hai. Sirf support force $N$ zero hai. ==(a) $0\text{N}$; (b) $20\text{m/s}$; (c) true weight $=500\text{N}$, zero nahi.==

Centripetal kyun? Rotating Earth par khada rehna gentle Circular Motion & Centripetal Acceleration hai: person din mein ek baar circle karta hai, isliye ek chhota net downward acceleration $a_c$ zaroori hai. Woh net force gravity se aati hai jo normal force ko thoda outpull karti hai. Step — Newton's 2nd law, acceleration direction neeche. Net downward force $m{a}_c$ ke barabar honi chahiye: $mg-N=m{a}_c\Rightarrow N=m(g-a_c).$ Step — numbers. $N=50(9.8-0.034)=50\times 9.766=488.3\text{N}$. Compare: true weight $mg=490\text{N}$. Toh scale lagbhag $1.7\text{N}$ kam dikhata hai. ==$N\approx 488.3\text{N}$, yani $\approx 1.7\text{N}$ true weight se neeche.== Ye wahi $N=m(g+a)$ hai jahan $a=-a_c$ (neeche). Earth ka rotation ek mild, permanent "elevator jo gently neeche accelerate kar raha hai" jaisa hai.

(a) $g=g_0{\left(\frac{R}{R+h}\right)}^2=9.8{\left(\frac{6.37\times 10^6}{6.77\times 10^6}\right)}^2=9.8(0.9409{)}^2\approx 9.8\times 0.8853\approx 8.68{\text{m/s}}^2.$ (b) Astronaut aur station dono ka same centripetal acceleration $a=g$ hai (Earth ki taraf). Station ke frame mein floor ke liye kuch bacha hi nahi supply karne ko: $N=m(g-g)=0\text{N}.$ Ground frame mein zyada carefully: dono ko $ma=mg$ inward chahiye, jo poori tarah gravity se milta hai, toh floor kuch contribute nahi karta. (c) ==Woh float karte hain kyunki astronaut aur station saath mein $g\approx 8.7{\text{m/s}}^2$ par free-fall karte hain; equal acceleration ⇒ koi mutual push nahi ⇒ $N=0$. Gravity surface ka ~89% hai, zero nahi.==

(a) Yahan $g=0$ hai lekin floor astronaut ko $a=+9.8$ par accelerate karna chahta hai. $N=m(g+a)=70(0+9.8)=686\text{N}$. (b) Woh $686\text{N}$ ka normal push feel karta hai — exactly Earth-surface weight ($70\times 9.8=686$). Bina khidki ke, woh kasam khaata ki woh Earth par khada hai. (Yahi Non-inertial Frames & Pseudo-forces aur Einstein ke equivalence principle ki neenv hai.) (c) ISS: $g\approx 8.7$ lekin free-falling, isliye $N=0$ — gravity present, koi support nahi. Deep-space ship: $g=0$ lekin thrusting, isliye $N=686$ — koi gravity nahi, poora support. Scale "real gravity" aur "acceleration" mein fark nahi bata sakta — woh sirf $N$ report karta hai. ==(a) $686\text{N}$; (b) Earth par khade hone jaisa lagta hai; (c) opposite causes, aur scale — sirf $N$ report karta hua — unhe alag nahi kar sakta.==

(a) Weightless ⇒ $N=0\Rightarrow a=-g=-10{\text{m/s}}^2$ (plane ko $g$ par neeche accelerate karna padega, yani parabola ke top par ek thrown ball ki tarah arc karna). (b) $N=m(g+a)\Rightarrow a=\frac{N}{m}-g=\frac{1200}{60}-10=20-10=+10{\text{m/s}}^2$ (upar ki taraf). g-force $=N/(mg)=1200/600=2\text{g}$. (c) Nahi — true weight $mg=600\text{N}$ poore time. Sirf $N$ $0$ aur $1200\text{N}$ ke beech swing karta hai. ==(a) $a=-10{\text{m/s}}^2$; (b) $a=+10{\text{m/s}}^2$, $2\text{g}$; (c) true weight constant $600\text{N}$ par.==

Phase (i): $a=+2$. $N=50(10+2)=600\text{N}$ (heavy). Phase (ii): $a=0$. $N=50(10)=500\text{N}$ (normal). Phase (iii): upar ja raha hai aur ruk raha hai ⇒ acceleration neeche ⇒ $a=-2$. $N=50(10-2)=400\text{N}$ (light). Max speed: phase (i) ke end par reach hoti hai: $v=u+at=0+2(3)=6\text{m/s}$. (Phase (ii) mein constant rehti hai, phir $0$ tak decelerate: check karo $6-2(3)=0$ ✓, consistent.) ==(i) $600\text{N}$, (ii) $500\text{N}$, (iii) $400\text{N}$; max speed $6\text{m/s}$.==

(a) $a=-12$. $N=50(10-12)=50(-2)=-100\text{N}$. (b) Floor sirf upar push kar sakta hai ($N\ge 0$); woh neeche pull nahi kar sakta. Formula ka $-100\text{N}$ signal karta hai ki floor "khatam" ho gaya — person actually floor se lift off karta hai aur, agar ceiling ho, toh ussse press karta hai. Ceiling phir person par neeche $100\text{N}$ supply karta hai (toh ceiling ka contact force woh role play karta hai, magnitude $|N|=100\text{N}$ ke saath). Ceiling ke bina, person falling lift ke relative upar float karta hai. ==$N_{\text{floor}}=0$ (person floor chhod deta hai); ceiling contact hone par $100\text{N}$ neeche push karta hai.== Yahi poora picture hai: $a<-g$ "extra-weightless" nahi hai, yeh inverted support hai — ground woh cheez rehna band kar deta hai jo tumhe thaami hui thi.