Non-inertial reference frames — pseudo forces: Question bank

1.2.13 · D5 · Physics › Newton's Laws & Dynamics › Non-inertial reference frames — pseudo forces

Shuru karne se pehle, is page par use hue har symbol ko plain words mein samjhate hain taaki kuch bhi mystery na rahe:

Definition Symbols ki cast

$a_{0}$ (padho "a-naught") = us frame ki acceleration jo tum mein baithe ho (bus, lift, car), jaise koi zameen par khade observer ne measure kiya ho (ek inertial observer).
$m$ = us object ki mass jise hum dekh rahe hain.
$g$ = gravity ke wajah se acceleration ( $\approx 9.8 m/s^{2}$ ), seedha neeche ki taraf; object ka real weight $m g$ hai, neeche ki taraf kheenchta hua.
$N$ = normal force: wo upar ki taraf push jo koi floor ya scale jo bhi uus par rakha hai use deta hai. Ek weighing scale literally $N$ dikhata hai — yahi tumhara "apparent weight" hai.
$T$ = tension: wo pull jo ek taut string apni hi length ke saath karti hai, bob se pivot ki taraf. Ek hanging pendulum bob ke liye, $T$ string ke saath upar ki taraf point karta hai.
$F_{pseudo} = - m a_{0}$ = woh fake force jo hum add karte hain taaki $F = m a$ accelerating frame ke andar bhi sahi rahe. Minus sign ka matlab hai "ise $a_{0}$ ki opposite direction mein point karo".

Teen scenarios jinhe ye traps baar baar revisit karte hain — bus, lift, tilted pendulum — yahan free-body diagrams ke roop mein draw kiye gaye hain taaki tum unke baare mein reason karne se pehle har ek arrow dekh sako:

Figure — Non-inertial reference frames — pseudo forces

Pendulum panel mein notice karo kaise tension $T$ do parts mein split hoti hai: ek upar wala part $T cos θ$ jo gravity se ladta hai, aur ek sideways part $T sin θ$ jo pseudo force se ladta hai. Har direction ko balance karne se milta hai $T cos θ = m g$ aur $T sin θ = m a_{0}$ ; divide karne se $T$ khatam ho jaata hai aur bacha rehta hai $tan θ = a_{0} / g$ . Arrows ka woh right-triangle hi "force triangle" hai jiska neeche zikr kiya gaya hai.

True or false — justify

Har item: true/false decide karo, phir reason do. Akela "true" yahan zero score karta hai.

A pseudo force non-inertial frame mein har mass par act karta hai, chahe chhota ho ya bada.

True — kyunki

F_{pseudo} = - m a_{0}

m

ke saath scale karta hai, isliye har object ek feel karta hai; ek bhaari object proportionally bada fake force feel karta hai, aur yahi wajah hai ki sab objects same fake acceleration

- a_{0}

share karte hain.

Ek pseudo force kisi object par work kar sakta hai.

True — iske paas ek magnitude aur direction hai, isliye frame mein kisi bhi displacement ke saath ye

F_{pseudo} \cdot d

karta hai; "fictitious" ka matlab koi source na hona hai, energetically inert hona nahi.

Pseudo force Newton's third law ka usi tarah paalana karta hai jaise koi bhi doosra force karta hai.

False — iske paas push karne wala koi physical agent nahi hai, isliye push back karne ke liye koi doosra body nahi hai; action–reaction pairs ke liye do real bodies chahiye, aur yahan sirf ek hi hai.

Ek inertial frame mein tum pseudo force add kar sakte ho jab tak tum consistent raho.

False — ek inertial frame mein

a_{0} = 0

hota hai, isliye

F_{pseudo} = 0

; iska "add" karna matlab zero add karna hai, aur agar tum chupke se ek nonzero add karte ho toh double-count ho jaata hai aur galat answer milta hai.

Agar do frames dono accelerate karte hain lekin same

a_{0}

se, toh ek mein observer ko doosre mein motion describe karne ke liye pseudo force chahiye.

False — unke beech relative acceleration zero hai, isliye ek doosre ke relative se wo inertial hain; same

a_{0}

share karne wale frames ke beech koi pseudo force ki zaroorat nahi.

Tum jis direction mein push feel karte ho wahi pseudo force ki direction hoti hai.

True (is case mein) — pseudo force

- m a_{0}

frame ki acceleration ki opposite point karta hai, aur yahi woh taraf hai jis taraf tumhari inertia tumhe "throw" karti hai; feeling aur fake force genuinely align karte hain.

Agar tum switch karo ki

a_{0}

ko kis inertial frame se measure karte ho toh pseudo force change ho jaata hai.

False — sab inertial frames acceleration par agree karte hain (wo sirf constant velocity se alag hote hain, dekho Galilean relativity), isliye

a_{0}

aur hence

F_{pseudo}

same rehta hai chahe tum kisi bhi inertial frame ko reference karo.

Spot the error

Har statement mein ek flaw hai. Flaw ka naam batao aur vector reasoning walk through karo jo ise fix karta hai.

"Bus aage accelerate karti hai, isliye passenger par pseudo force aage ki taraf point karta hai."

Galat direction. Rule ko vector ki tarah likhte hain:

F_{pseudo} = - m a_{0}

. Yahan

a_{0}

aage point karta hai (ise

+ x

kaho), isliye

- m a_{0}

mein minus sign

x

-component mein carry hota hai, use

- x

mein flip karta hai — peeche ki taraf. Step by step: forward

a_{0}

→

m

se multiply karo (abhi bhi forward) →

- 1

se multiply karo (ab backward). Reversal hi minus sign hai, isliye fake force peeche point karta hai, jo passenger ke seat mein pheke jaane se match karta hai.

"Kyunki smooth bus floor par ball par pseudo force hai, ground frame mein bhi use accelerate karna chahiye."

Frame mix-up — pseudo force sirf bus frame mein exist karta hai; ground frame mein koi horizontal real force nahi hai, isliye ground observer sahi taur par dekhta hai ki ball rest mein rahe.

"Upar accelerate karte lift mein, apparent weight

N = m (g - a_{0})

hai kyunki lift gravity mein add karta hai."

Sign error. Neeche ko negative set karo. Frame upar accelerate karta hai (

a_{0} = + a_{0}

), isliye

F_{pseudo} = - m a_{0}

neeche point karta hai (

- m a_{0}

), pehle se neeche ki taraf ke weight

- m g

ke upar stack ho ke. Upar ki taraf ka normal

N

ab dono ko balance karna chahiye:

N - m g - m a_{0} = 0

, jisse milta hai

N = m (g + a_{0})

. Minus-version tab aata hai jab

a_{0}

khud neeche point karta hai aur pseudo force upar flip hota hai.

"Car mein pendulum ke liye, force triangle se

tan θ = g / a_{0}

Ratio inverted hai. Force triangle mein (figure ka pendulum panel dekho), horizontal leg

T sin θ

hai jo horizontal pseudo force

m a_{0}

ko balance karta hai, aur vertical leg

T cos θ

hai jo gravity

m g

ko balance karta hai. Isliye

T sin θ = m a_{0}

aur

T cos θ = m g

. Pehle ko doosre se divide karne par

T

cancel ho jaata hai:

tan θ = \frac{m a _{0}}{m g} = \frac{a _{0}}{g}

. Claimed

g / a_{0}

triangle ki legs ko swap karta hai.

"Pseudo force ek real force hai jiska source unknown hai jo humne abhi tak find nahi kiya."

Category error — by construction iska koi source nahi hota; ye ek bookkeeping term hai jo accelerating frame choose karne se paida hoti hai, aur ye usi waqt gayab ho jaata hai jab tum inertial frame mein wapas aate ho.

"Kyunki centrifugal force ek pseudo force hai, ye ground frame mein appear hota hai jab car turn karti hai."

Frame error — centrifugal force sirf rotating (car) frame mein appear hota hai; ground frame mein car turn karta hai kyunki ek real inward force (friction/road) hai, bina kisi outward force ke bilkul bhi.

Why questions

Mechanism ke saath jawab do, restatement se nahi.

Pseudo force ka koi reaction partner kyun nahi hota jab har real force ka hota hai?

Kyunki reaction partner ek interaction mein "doosra body" hota hai, lekin pseudo force observer ke accelerating frame se arise hoti hai, na ki kisi body se — koi doosra body nahi hai isliye push back karne ke liye kuch bhi nahi hai.

F_{pseudo}

mass ke saath scale kyun karta hai, unlike, say, ek fixed applied push?

F = m a

ko rearrange karne mein

- m a_{0}

term se aata hai; ise

m

ke saath scale karna hi chahiye taaki

m

se divide karne par har object ko same fake acceleration

- a_{0}

mile, jo frame ka observer actually dekhta hai usse match karte hue.

Pendulum-in-a-car problem ko "accelerometer" kyun kaha jaata hai?

Steady tilt

tan θ = a_{0} / g

satisfy karta hai, isliye

θ

measure karna directly

a_{0}

reveal karta hai; pseudo force literally string ko frame ki acceleration ke proportion mein tilt karta hai.

Pseudo force add karna ek passenger ko

F = m a

(dekho Newton's Second Law) use karte rehne kyun deta hai?

Inertial frame mein true law se shuru karke

- m a_{0}

term ko force side mein move karna "

m a_{frame}

" ko wapas "(fake wale samait sab forces)" mein turn kar deta hai, jaani-pehchaani bookkeeping restore karte hue.

Ground observer ko same motion explain karne ke liye kabhi bhi pseudo force ki zaroorat kyun nahi padti?

Unke liye

a_{0} = 0

hai, isliye correction zero hai; real forces akele hi motion sahi predict kar deti hain, kyunki inertial frame wahi jagah hai jahan Newton's laws bina kisi chhed-chhad ke hold karti hain.

Pseudo force upar accelerate karte lift mein "neeche" kyun point karta hai lekin forward bus mein "peeche"?

Dono cases mein ye simply

- a_{0}

hai: lift upar accelerate karti hai isliye fake force neeche hai; bus aage accelerate karti hai isliye fake force peeche hai. Same rule, alag

a_{0}

Edge cases

Har scenario ko uski boundary tak push karo.

Pseudo force kya hota hai jab frame ki acceleration exactly zero ho (

a_{0} = 0

Ye exactly zero hota hai — frame tab inertial hai aur Newton's laws ko koi correction nahi chahiye; ye smooth limit hai jo ordinary mechanics recover karta hai.

Free fall mein ek lift mein (

a_{0} = g

downward), scale kya read karega aur kyun?

N = m (g - a_{0}) = 0

— downward pseudo force

m g

falling frame mein real gravity ko exactly cancel kar deta hai, isliye kuch bhi scale par press nahi karta; ye weightlessness hai (dekho Apparent weight & normal force).

Agar ek lift gravity se tez neeche accelerate kare (

a_{0} > g

), toh apparent weight ka kya hoga?

N = m (g - a_{0})

negative ho jaata hai, jo ek upar push karne wale scale ke liye impossible hai — iska matlab hai ki insaan floor se utha jaata hai aur use ceiling hold karni padegi; object lift ke relative upar uthta hai.

Accelerating-car pendulum ke liye,

θ

kya karta hai jab

a_{0} \to 0

aur jab

a_{0} \to \infty

Jab

a_{0} \to 0

tan θ \to 0

isliye string vertically hang hoti hai (koi fake tilt nahi); jab

a_{0} \to \infty

tan θ \to \infty

isliye

θ \to 9 0^{\circ}

, string horizontal ki taraf swing karti hai.

Kya pseudo-force rule

F_{pseudo} = - m a_{0}

ek rotating frame ke liye complete hai?

Nahi — rotating frame ke liye acceleration ek single constant

a_{0}

nahi hoti; extra velocity- aur position-dependent terms appear hote hain, yani centrifugal aur Coriolis pseudo forces.

Smooth bus floor par ball ka kya hota hai agar bus accelerate kare, phir constant velocity hold kare?

Accelerate karte waqt, pseudo force ball ko peeche push karta hai; jis instant

a_{0} = 0

(constant velocity), pseudo force gayab ho jaata hai aur ball simply apni nayi velocity rakhti hai — bus frame mein aur koi slide nahi.

Agar bus floor par friction ho, toh kya pseudo force gayab ho jaata hai?

Nahi — pseudo force abhi bhi

- m a_{0}

ke equal hai; friction ek alag real force hai jo ball ko jagah par rok sakti hai, isliye bus frame mein friction aur pseudo force balance karke zero net acceleration de sakte hain.

$- m a_{0}$ se centrifugal aur Coriolis tak

Ek-line rule $F_{pseudo} = - m a_{0}$ straight-line special case hai. Jab frame rotate karta hai, uski acceleration ab ek fixed vector nahi rehti, aur usi idea se do naye pseudo-force terms nikal aate hain. Neeche ka figure bridge sketch karta hai:

Straight-line $a_{0}$ wala frame → plain $- m a_{0}$ (yahi page).
Ek rotating frame → rest mein objects ek outward centrifugal push feel karte hain (axis se position par depend karta hai).
Rotating frame ke andar move karte objects ek sideways Coriolis push feel karte hain (velocity par depend karta hai).

Har baar same philosophy: frame jo bhi extra acceleration inject kare, us se $m$ times subtract karo aur result ko pseudo force kaho.

Connections

Newton's Second Law — fake force kyun add kiya jaata hai ye bhoolne ka trap.
Newton's Third Law — baar baar aane wala "no reaction pair" trap.
Centrifugal force — rotating frame mein pseudo force (rule extend hota hai).
Coriolis force — velocity-dependent pseudo force, $- m a_{0}$ se aage.
Apparent weight & normal force — lift edge cases yahan rehte hain.
Galilean relativity — kyun sab inertial observers $a_{0}$ par agree karte hain.