Vertical circular motion — minimum speed conditions: Question bank

1.2.18 · D5 · Physics › Newton's Laws & Dynamics › Vertical circular motion — minimum speed conditions

Pehle yeh setup samjho (sawaalon se pehle padho)

Definition Sign convention — "inward is positive"

Kyunki circular motion ke liye ek net force chahiye jo centre ki taraf point kare, hum radial axis ko inward (centre ki taraf) positive choose karte hain. Tangential axis motion ki direction mein point karta hai (radius ke perpendicular); hum ise sirf tab use karte hain jab koi force ball ko speed up ya slow down kare.

Inward-positive kyun? Is axis par Newton's second law padhti hai $F_{net, inward} = m a_{centripetal} = \frac{m v ^{2}}{r}$ . Inward positive rakhne se required side $+ \frac{m v ^{2}}{r}$ hoti hai, isliye jo bhi force inward point kare wo $+$ sign ke saath aata hai aur jo outward point kare wo $-$ sign ke saath. Koi guessing nahi.

Ek baar apply karo, hamesha ke liye:

Top par, centre neeche hai, isliye "inward" = down. Gravity (down) $+ m g$ hai, tension (down, centre ki taraf) $+ T$ hai: $T + m g = \frac{m v ^{2}}{r}$ .
Bottom par, centre upar hai, isliye "inward" = up. Tension (up, centre ki taraf) $+ T$ hai, gravity (down, centre se door) $- m g$ hai: $T - m g = \frac{m v ^{2}}{r}$ .

Figure — Vertical circular motion — minimum speed conditions

Upar ke free-body diagrams poora game hain. Dhyaan do, koi "centripetal" arrow nahi draw kiya — sirf real forces ( $m g$ aur $T$ ). Quantity $\frac{m v ^{2}}{r}$ equation ke doosri taraf rehti hai as required inward net, kabhi apne arrow ke roop mein nahi.

$\frac{1}{2} m v_{bot}^{2} = \frac{1}{2} m v_{top}^{2} + m g (2 r)$

$2/ m$ se multiply karo: $v_{bot}^{2} = v_{top}^{2} + 4 g r$ . Yahi " $+ 4 g r$ " climbing tax hai jo neeche baar baar milega. (Energy conservation laagu hoti hai kyunki tension motion ke perpendicular hai aur zero work karta hai.)

True or false — justify karo

True or false: Loop ke top par tension aur gravity opposite directions mein point karte hain.

False. Top par circle ka centre ball ke neeche hota hai, isliye string neeche centre ki taraf kheenchti hai — gravity ke same direction mein. Inward-positive use karo (inward = down yahan), dono positive hain aur add hote hain:

T + m g = \frac{m v ^{2}}{r}

True or false: Centripetal force ek extra arrow hai jo free-body diagram par zaroori draw karna padta hai.

False. Free-body diagram par tum sirf wahi real forces draw karte ho jinhe tum naam de sako — yahan sirf gravity

m g

aur tension

T

(figure s01 dekho). Quantity

\frac{m v ^{2}}{r}

koi force nahi hai; ye required inward net hai jo in real forces ko add karke banana padta hai, isliye ye equation ke doosri taraf hota hai, kabhi apne arrow ke roop mein nahi.

True or false: Loop complete karne ki minimum speed

g r

hai.

Jaisa kaha gaya, False.

g r

sirf top par minimum speed hai. Poora loop complete karne ke liye bottom par

5 g r

se enter karna padta hai: top ko

g r

chahiye, aur climb

v^{2}

mein

4 g r

add karta hai, isliye

v_{bot}^{2} = g r + 4 g r = 5 g r

True or false: Ball ko circling rakhne ke liye string push ya pull dono de sakti hai.

False. String sirf pull kar sakti hai, isliye

T \geq 0

. Poora "

g r

" minimum isi one-sided constraint se aata hai: jis moment equation

T < 0

(ek push) maangti hai, string slack ho jaati hai.

True or false: Ek rigid rod par ball ke liye top par minimum speed phir bhi

g r

hai.

False. Rod push bhi kar sakti hai pull bhi, isliye

T

negative ho sakta hai; slack condition

T \geq 0

khatam ho jaati hai. Rod ki sirf yehi requirement hai

v_{top} \geq 0

, jo minimum top speed

0

deta hai.

True or false: Tension difference

T_{bot} - T_{top} = 6 m g

sirf minimum speed par hold karta hai.

False — ye kisi bhi speed par hold karta hai. Do setup equations se:

T_{bot} = \frac{m v _{bot}^{2}}{r} + m g

aur

T_{top} = \frac{m v _{top}^{2}}{r} - m g

. Subtract karo, do

\pm m g

terms combine hokar

+ 2 m g

dete hain, aur

\frac{m}{r} (v_{bot}^{2} - v_{top}^{2}) = \frac{m}{r} (4 g r) = 4 m g

climbing tax use karke. Total:

2 m g + 4 m g = 6 m g

, speed se independent.

True or false: Minimum-speed top par, ball string ke relative momentarily weightless feel karti hai.

True.

T = 0

ke saath, string kuch nahi exert karti; gravity akela ball ko curve karta hai, isliye use koi support force feel nahi hoti — curve ke saath free fall ki tarah. Isliye paani nahi girta.

True or false: Radius double karne se minimum top speed double ho jaati hai.

False.

v_{top,min} = g r

, isliye

r

double karne par speed

2

se multiply hoti hai,

2

se nahi. Speed radius ke square root ke saath scale karti hai.

Error dhundo

"Top par, net force

= T - m g = \frac{m v ^{2}}{r}

, isliye minimum speed par

T = 0

rakhne se

- m g = \frac{m v ^{2}}{r}

milta hai."

T

ka sign galat hai. Top par inward-positive ke saath (inward = down), tension neeche centre ki taraf point karta hai, isliye wo

+ T

ke roop mein aata hai,

- T

nahi:

T + m g = \frac{m v ^{2}}{r}

T = 0

rakhne par

m g = \frac{m v ^{2}}{r}

milta hai, isliye

v = g r

"Energy:

\frac{1}{2} m v_{bot}^{2} = \frac{1}{2} m v_{top}^{2} + m g r

, kyunki top, bottom se height

r

upar hai."

Radius

r

ke loop ki top, bottom se height

2 r

(ek poora diameter) upar hai,

r

nahi — figure s02 dekho. Sahi term

m g (2 r)

hai, jo

v_{bot}^{2} = 5 g r

deta hai,

3 g r

nahi.

"Minimum speed par ball top par zero speed rakhti hai, isliye wo bas barely wahan pahunchti hai."

Yeh rod case hai. String ke liye, ball top par abhi bhi

g r

par move karti hai — zero speed ka matlab hoga string bahut pehle slack ho gayi. "Bas pahunchna" (zero top speed) ke liye ek rod ya track chahiye jo push kar sake.

"Tension ball par kaam karta hai jab wo upar jaati hai, isliye hum energy conservation use nahi kar sakte."

Tension hamesha motion ke perpendicular hoti hai (ye radius ke saath point karti hai, ball tangentially move karti hai), isliye ye zero work karti hai. Yahi wajah hai ki energy conservation yahan valid hai.

"Bottom par,

T_{bot} = \frac{m v ^{2}}{r}

kyunki yahi centripetal force hai."

Gravity missing hai. Bottom par, inward = up: tension

+ T

hai, gravity

- m g

hai, isliye

T_{bot} - m g = \frac{m v ^{2}}{r}

, jo

T_{bot} = m g + \frac{m v ^{2}}{r}

deta hai — zyada, equal nahi.

"Roller-coaster loop par, top par seat rider ko upar push karti hai."

Track/seat car ke bahar hai, isliye top par wo sirf inward = downward push kar sakta hai (ek normal force

N \geq 0

centre ki taraf). Gravity ke saath milkar ye centripetal force supply karta hai; ye kabhi outward push nahi karta.

Why questions

Gravity top par "help" kyun karta hai lekin bottom par "fight" kyun karta hai?

Centripetal requirement hamesha inward point karti hai. Top par centre neeche hai, isliye gravity (down) inward point karta hai aur

+ m g

ke roop mein aata hai — ye help karta hai. Bottom par centre upar hai, isliye gravity (down) outward point karta hai,

- m g

ke roop mein aata hai — tension ko ise overcome karna padta hai.

Bottom, top se ek fixed energy amount se faster kyun honi chahiye?

Bottom se top tak chadhne par ball

2 r

height (figure s02) utha jaati hai, kinetic energy gravitational potential energy mein convert hoti hai. Yahi "tax"

m g (2 r)

hai; energy equation ko

\frac{m}{2}

se divide karne par pata chalta hai ki ye

v^{2}

mein

4 g r

add karta hai:

v_{bot}^{2} = v_{top}^{2} + 4 g r

Ulte bucket mein top par paani kyun nahi girata?

Minimum speed par ya usse zyada par, gravity zyada se zyada utna centripetal force provide karti hai jitna chahiye, isliye bucket ka bottom utna hi tezi se neeche accelerate karta hai jitna paani girta. Paani aur bucket saath girte hain — bucket kabhi door nahi jaata, isliye kuch nahi girta.

String ki minimum speed top par dhundne ke liye hum

\frac{m v ^{2}}{r} = 0

kyun nahi set kar sakte?

Logic order mein follow karo. (1) Physical limit string ka slack hona hai, yaani

T = 0

— zero required force nahi. (2)

T = 0

T + m g = \frac{m v ^{2}}{r}

mein daalo: left side abhi bhi

m g

hai, kyunki gravity gayab nahi hui. (3) Isliye circle abhi bhi

\frac{m v ^{2}}{r} = m g

demand karta hai, jo

v = g r

deta hai.

\frac{m v ^{2}}{r} = 0

set karna galat hoga kyunki iska matlab hoga gravity switch off ho gayi.

Minimum-speed condition ek string/track rule kyun hai, physics ka law nahi?

Ye poori tarah constraint

T \geq 0

se aata hai (strings aur outer tracks sirf inward pull/push kar sakti hain). Connector badlo — rod, tube, ya inner track — aur constraint badal jaata hai. Ye connector ke baare mein hai, motion ke nahi.

Jab ball top par slow hoti hai to required centripetal force kyun shrink karta hai?

\frac{m v ^{2}}{r}

v^{2}

par depend karta hai. Slower

v

→ smaller required inward force. Kyunki gravity

m g

fixed hai, tension ko difference poora karne ke liye drop karna padta hai; jab requirement

m g

se kam ho jaaye, tension negative jaani padegi — string ke liye impossible.

Edge cases

Agar top par ball

g r

se faster ho to kya hoga?

Circle

m g

se zyada demand karta hai, isliye string extra provide karti hai:

T = \frac{m v ^{2}}{r} - m g > 0

. String taut rehti hai — bilkul theek, bas tighter.

Exactly

v_{top} = g r

par kya hota hai?

Tension exactly zero hoti hai — string slack hone ki verge par hai lekin abhi bhi perfect circle trace karta hai, gravity akela centripetal force ke roop mein. Yeh "loops fine" aur "falls out" ke beech ki boundary hai.

Agar ball ek rigid rod par hai aur top par

g r

se slower move kare to kya hoga?

Bilkul allowed hai. Rod sirf outward push karta hai (

T < 0

) net inward force ko

\frac{m v ^{2}}{r}

tak reduce karne ke liye. Zero top speed tak rod use circle par rakhe rakhta hai.

Rod case ke liye minimum bottom speed kya hai, aur ye smaller kyun hai?

v_{bot,min} = 2 g r

(from

v_{bot}^{2} = 0 + 4 g r

), kyunki rod ko sirf ball ka top tak pahunchna chahiye (

v_{top} = 0

), sirf

4 g r

climbing tax pay karta hai — taut rehne ke liye extra

g r

nahi.

Ek smooth tube ke andar ball ke liye kya hota hai (dono taraf walls hain)?

Tube dono inward aur outward push kar sakti hai, isliye rod ki tarah ye slack condition hata deti hai: minimum top speed

0

hai, aur

v_{bot,min} = 2 g r

. Outer-track/string case se compare karo jahan sirf inward push exist karta hai.

Agar bottom par speed

2 g r

aur

5 g r

ke beech ho (string case), to string exactly kahan slack hoti hai?

Angle

θ

par jo horizontal se upar hai (centre se measure karke) jahan tension pehli baar zero hoti hai.

v_{θ}^{2} = v_{bot}^{2} - 2 g r (1 + sin θ)

lo (bottom se height

r (1 + sin θ)

hai). Radial equation

T + m g sin θ = \frac{m v _{θ}^{2}}{r}

mein

T = 0

set karne par slack condition milti hai

sin θ = \frac{v _{bot}^{2} - 2 g r}{3 g r}

. Example ke roop mein

v_{bot}^{2} = 4 g r

ke saath ye

sin θ = \frac{2}{3}

deta hai (horizontal se lagbhag

4 2^{\circ}

upar); us point ke aage ball circle chhod kar projectile ban jaati hai.

Horizontal side par (height

r

), kya ball ki weight ka koi hissa centripetal force ka part hai?

Nahi. Side par centre horizontal hai, isliye gravity (vertical) poori tarah tangential hai — ye ball ki speed change karta hai, direction nahi. Wahan sirf tension centripetal force provide karta hai:

T = \frac{m v ^{2}}{r}

Connections

Centripetal force and acceleration
Newton's Second Law — net force form
Conservation of mechanical energy
Tension and constraint forces
Normal force in circular tracks
Banked curves and horizontal circular motion
Free-body diagrams