Acceleration of rolling objects on inclines — comparison: Worked examples

1.5.15 · D3 · Physics › Rotational Mechanics › Acceleration of rolling objects on inclines — comparison

Yeh page parent topic ka drill ground hai. Parent ne humein ek master formula diya tha. Yahan hum us par har tarah ka input throw karte hain — steep angles, flat angles, weird shapes, sliding blocks, energy, time, friction — aur har ek ko ground up se solve karte hain.

Koi bhi symbol aane se pehle, yeh ek tool hai jo hum har jagah reuse karte hain. Yeh parent note mein earn kiya gaya tha; hum ise poore plain words mein restate karte hain taaki kuch bhi assumed na ho.

Formula Master result (parent se)

Ek round object jo without slipping roll kar raha hai (contact patch kabhi skid nahi karta) ek slope par jo angle $θ$ par tila hua hai, neeche ki taraf: $a = \frac{g s i n θ}{1 + β}, β = \frac{I}{m R ^{2}} .$

$a$ = object ka centre slope ke along kitni tezi se speed up karta hai (metres per second, har second).
$g \approx 9.8$ = free-fall acceleration.
$θ$ = slope ka tilt flat ground se measure kiya gaya.
$sin θ$ = gravity ka woh fraction jo slope ke along pull karta hai (flat ho toh 0, vertical ho toh 1).
$m$ = object ki mass; $R$ = uski radius.
$I$ = object ka moment of inertia — ek number jo measure karta hai ki spin karna kitna mushkil hai, yaani mass kitni rotational "stubbornness" deta hai (dekho Moment of Inertia). Bada $I$ = mass spin axis se zyada door hai.
$β$ = shape factor $I / (m R^{2})$ : ek pure number jo mass aur radius ko strip out karta hai taaki sirf shape bache. Chhota = axis ke paas bunched (spin karna aasaan), bada = axis se door (spin karna mushkil).

Shape table ko apne dimag mein pin karke rakho:

Object	$β$	one-word memory
Solid sphere	$2/5 = 0.40$	marble
Solid cylinder / disc	$1/2 = 0.50$	coin
Hollow sphere (shell)	$2/3 \approx 0.67$	tennis ball
Ring / hoop / pipe	$1$	bangle
Sliding block (no rolling)	$0$	ice cube

Scenario matrix

Is topic mein jo bhi question aa sakta hai woh in cells mein se kisi ek mein hoga. Neeche ke worked examples mein se har ek par (Cell A2) jaisi tag hogi taaki tum dekh sako ki coverage complete hai.

#	Cell class	Kya special hai	Covered by
A1	Standard angle ( $0 < θ < 90°$ )	seedha plug-in	Ex 1
A2	Do shapes compare karo	ranking / kaun jeeta	Ex 2
B1	$θ \to 0$ (flat) degenerate	$sin θ \to 0$ , $a \to 0$	Ex 3
B2	$θ \to 90°$ (vertical) limit	$sin θ \to 1$ , max $a$	Ex 3
C1	$β = 0$ (sliding block) degenerate	rolling term vanish ho jaata hai	Ex 4
C2	$β \to \infty$ (mass door) limit	$a \to 0$	Ex 4
D1	Energy / final speed	$v = 2 g h / (1 + β)$ use karta hai	Ex 5
D2	Descend karne ka time	$t = 2 L / a$ use karta hai	Ex 6
E1	Required friction (find $f$ )	kya slope kaafi rough hai?	Ex 7
E2	Slip threshold (exam twist)	jab rolling break ho jaati hai	Ex 8
F1	Real-world word problem	words → symbols translate karo	Ex 9
F2	Unknown shape (back-solve $β$ )	$a$ measure karo, shape nikalo	Ex 10

Worked examples

Ex 1 — Standard angle, single object (Cell A1)

Worked example Solid cylinder ek 20° slope par

Ek solid cylinder ko rest se ek $θ = 20°$ incline par release kiya jaata hai. Uska acceleration nikalo. $g = 9.8$ use karo.

Forecast: guess karo — kya $a$ , $g sin 20° \approx 3.35$ se zyada hoga ya kam? (Ek rolling body hamesha sliding wale se slower hoti hai, toh expect karo kam.)

$β$ identify karo. Solid cylinder $\Rightarrow β = \frac{1}{2}$ . Yeh step kyun? $β$ единственный shape-dependent input hai; pehle ise nikalo.
Formula likho. $a = \frac{g sin θ}{1 + β} = \frac{g sin 20°}{1 + \frac{1}{2}} = \frac{g sin 20°}{1.5}$ . Yeh step kyun? Direct substitution — kuch aur vary nahi karta.
$sin 20° = 0.342$ compute karo. Toh $g sin 20° = 9.8 \times 0.342 = 3.352$ . Yeh step kyun? $sin θ$ slope ke along gravity ka fraction hai; divide karne se pehle hume uska number chahiye.
Divide karo. $a = 3.352/1.5 = 2.23 m/s^{2}$ . Yeh step kyun? $1 + β$ se divide karna woh "spin tax" apply karta hai jo body ko slow karta hai.

Verify: $2.23 < 3.35$ ✔ (rolling, pure sliding se slower hai, jaise forecast tha). Units: $\frac{m/s ^{2}}{pure number} = m/s^{2}$ ✔.

Ex 2 — Shapes ke beech race (Cell A2)

Worked example Marble vs bangle ek hi 30° slope par

Ek solid sphere aur ek ring ek $30°$ incline ke top se saath start karte hain. Sphere ka acceleration ring se kitne factor zyada hai?

Forecast: ring apni mass ko radius $R$ par chhupa leti hai, toh woh markedly slower honi chahiye — guess karo ratio around $1.3$ – $1.5$ .

Sphere: $β = \frac{2}{5}$ , toh $a_{sph} = \frac{g sin 30°}{1 + 2/5} = \frac{g sin 30°}{7/5} = \frac{5}{7} g sin 30°$ . Yeh step kyun? Sphere ka shape factor plug karo.
Ring: $β = 1$ , toh $a_{ring} = \frac{g sin 30°}{2} = \frac{1}{2} g sin 30°$ . Yeh step kyun? Same slope, alag $β$ — sirf yehi change hai.
Ratio. $\frac{a _{sph}}{a _{ring}} = \frac{5/7}{1/2} = \frac{10}{7} = 1.4286$ . Yeh step kyun? $g sin 30°$ cancel ho jaata hai — race sirf $β$ par depend karta hai, toh answer slope-independent hai.

Verify: $10/7 \approx 1.43$ , forecast window ke andar ✔. Sphere $43%$ faster accelerate karta hai ⇒ race jeetta hai, parent ki ranking se match karta hai.

Ex 3 — Angle ke extremes (Cells B1 & B2)

Yahan hum $θ$ ko hi stress-test karte hain. Neeche ki figure is example ka geometric heart hai: yeh increasing tilt ke chaar slopes draw karti hai aur har ek par full weight (plum, hamesha seedha neeche) aur uska along-slope component $g sin θ$ (orange) dikhati hai. Dekho orange arrow $10°$ par ek stub se $85°$ par almost full weight tak kaise barta hai — woh barta hua arrow hi hamare formula mein driving push hai. Neeche ke steps 1 aur 2 simply us growth ke do endpoints read off karte hain.

Figure — Acceleration of rolling objects on inclines — comparison

Figure: along-slope drive $g sin θ$ (orange) flat ground par $0$ se vertical par full weight tak barta hai, jabki total weight (plum) kabhi nahi badalti.

Worked example Flat slope aur near-vertical slope (solid sphere)

Ek solid sphere ( $β = 2/5$ ) ke liye, (a) $θ = 0°$ aur (b) $θ = 90°$ par $a$ kya hai? $g = 9.8$ use karo.

Forecast: flat ground = kuch nahi rolega, toh $a = 0$ . Seedha neeche = sphere basically free-fall mein hai lekin spin se slow — expect karo $a = \frac{5}{7} g \approx 7$ .

Case $θ = 0°$ (Cell B1). $sin 0° = 0$ , toh $a = \frac{g \cdot 0}{7/5} = 0 m/s^{2}$ . Yeh step kyun? Flat ground par figure mein orange arrow zero length tak shrink ho jaata hai — gravity ka surface ke along koi component nahi hai, toh kuch bhi aage push nahi karta.
Case $θ = 90°$ (Cell B2). $sin 90° = 1$ , toh $a = \frac{g}{7/5} = \frac{5}{7} g = \frac{5}{7} (9.8) = 7.0 m/s^{2}$ . Yeh step kyun? Vertical par orange arrow full weight tak pahunch jaata hai (figure mein sabse lamba); $sin θ$ 1 par max out ho jaata hai. Yeh idealised hai — ek truly vertical surface friction dene ke liye press back nahi kar sakti, toh rolling actually yahan break ho jaayegi (yeh Cell E2 ka kaam hai). Hum ise mathematical limit maante hain.

Verify: $a (0°) = 0$ ✔ (sensible — flat ground). $a (90°) = 7.0 < 9.8 = g$ ✔ (pure free-fall se abhi bhi slower kyunki energy spin mein jaati hai). Dono extremes ek solid sphere ke har real answer ko $0 \leq a \leq 7.0$ mein bracket karte hain.

Ex 4 — Shape ke extremes (Cells C1 & C2)

Worked example Frictionless block aur mass-flung-out wheel

Ek $45°$ slope par: (a) ek frictionless sliding block ( $β = 0$ ); (b) ek hypothetical wheel jiska poora mass kabhi-bade-hote radius par rim par baitha hai, $β \to \infty$ drive karta hai. Accelerations compare karo. $g = 9.8$ , $sin 45° = 0.707$ use karo.

Forecast: block ko koi spinning feed nahi karni, toh woh theoretical maximum $g sin θ$ hit karega. $β \to \infty$ wheel sab kuch spin par waste karta hai — guess karo $a \to 0$ .

Block, $β = 0$ (Cell C1). $a = \frac{g sin 45°}{1 + 0} = g sin 45° = 9.8 \times 0.707 = 6.93 m/s^{2}$ . Yeh step kyun? Koi rotational inertia nahi hone par master formula plain Newton on an incline mein collapse ho jaata hai — yeh sliding-block limit hai. Yeh is slope par sabse tez koi bhi ja sakta hai ki speed hai.
$β \to \infty$ (Cell C2). $a = \frac{g sin 45°}{1 + β} \to \frac{6.93}{\infty} = 0$ . Yeh step kyun? Jab mass axis se infinitely door move hoti hai, toh use spin karna infinitely mushkil ho jaata hai; gravity apna poora budget rotation par spend karti hai, forward motion ke liye kuch nahi bachta.

Verify: Har real rolling object ( $0 < β \leq 1$ ) ko is slope par $0 < a < 6.93$ satisfy karna chahiye. Check ring: $a = 6.93/2 = 3.46$ ✔ — in dono extremes ke beech comfortably. Poori ranking in dono cells se boxed in hai.

Ex 5 — Energy method: final speed (Cell D1)

Worked example Hollow sphere ek 1.2 m drop se neeche

Ek hollow sphere ( $β = 2/3$ ) rest se ek incline ke neeche roll karti hai, vertical height $h = 1.2 m$ drop karti hai. Bottom par uski speed nikalo. $g = 9.8$ use karo.

Forecast: ek sliding block $2 g h = 2 (9.8) (1.2) \approx 4.85$ tak pahunchega. Rolling kuch spin ke liye chura leti hai, toh expect karo kam — guess ~ $3.8$ .

Energy balance. Saari lost height-energy motion ban jaati hai: $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$ . Yeh step kyun? Rolling mein static friction koi kaam nahi karta (Friction in Rolling), toh energy conserve hoti hai — dekho Energy Conservation in Rolling.
Spin ko shape factor mein fold karo. $I = β m R^{2}$ aur $ω = v / R$ use karke, poora right side $\frac{1}{2} m v^{2} (1 + β)$ ban jaata hai. $m$ cancel karo: $g h = \frac{1}{2} v^{2} (1 + β)$ . Yeh step kyun? Wahi $(1 + β)$ "spin tax" jo acceleration formula mein appear karta hai; ise fold karne se ek single unknown $v$ bachta hai.
$v$ ke liye solve karo. Dono sides ko $2/ (1 + β)$ se multiply karo phir square root lo: $v = \frac{2 g h}{1 + β} = \frac{2 ( 9.8 ) ( 1.2 )}{1 + 2/3} = \frac{23.52}{1.667} = 14.11 = 3.76 m/s$ . Yeh step kyun? Hume $v$ akela chahiye, toh hum $\frac{1}{2} (1 + β)$ ko undo karte hain jo $v^{2}$ ko multiply kar raha tha, phir square ko rooting se undo karte hain.

Verify: $3.76 < 4.85$ ✔ (sliding block se slower, jaise forecast tha). Units: $m/s^{2} \cdot m = m^{2} / s^{2} = m/s$ ✔.

Ex 6 — Fixed length descend karne ka time (Cell D2)

Worked example 3 m roll karne mein kitna time lagega?

Ek solid sphere ( $β = 2/5$ ) rest se $θ = 25°$ par $L = 3 m$ slope ke neeche roll karti hai. Bottom tak pahunchne mein kitne seconds lagte hain? $g = 9.8$ , $sin 25° = 0.4226$ use karo.

Forecast: yeh kuch m/s² accelerate karti hai, 3 m mein — guess karo about 1–1.5 seconds.

Pehle acceleration. $a = \frac{g sin 25°}{1 + 2/5} = \frac{9.8 ( 0.4226 )}{7/5} = \frac{4.141}{1.4} = 2.958 m/s^{2}$ . Yeh step kyun? Time is par depend karta hai ki yeh kitni tezi se speed up karta hai, toh pehle master formula se $a$ chahiye.
Rest se kinematics. $L = \frac{1}{2} a t^{2} \Rightarrow t = \frac{2 L}{a}$ . Yeh step kyun? Zero start speed ke saath constant acceleration standard "distance = ½at²" deta hai; hum $t$ nikalne ke liye ise invert karte hain.
Plug in. $t = \frac{2 ( 3 )}{2.958} = 2.029 = 1.424 s$ . Yeh step kyun? Known $L$ aur computed $a$ substitute karo number nikalne ke liye.

Verify: forecast ke andar ✔. Sanity: final speed $v = a t = 2.958 \times 1.424 = 4.21 m/s$ ; aur $v = 2 a L = 2 (2.958) (3) = 17.75 = 4.21$ ✔ — dono routes agree karte hain. Units: $m / (m/s^{2}) = s^{2} = s$ ✔.

Ex 7 — Actually kitna friction chahiye? (Cell E1)

Worked example Rolling disc par friction force

Ek solid cylinder jiska mass $m = 2 kg$ hai, $30°$ slope ke neeche without slipping roll karta hai. Woh friction force $f$ nikalo jo yeh possible banata hai. $g = 9.8$ use karo.

Forecast: yahan friction energy waste nahi kar raha — yeh torque supplier hai. Yeh weight component ka ek modest fraction hona chahiye.

Acceleration. $β = \frac{1}{2}$ , toh $a = \frac{g sin 30°}{1.5} = \frac{9.8 ( 0.5 )}{1.5} = 3.267 m/s^{2}$ . Yeh step kyun? Hume $a$ chahiye taaki pata chale rotation mein kitna force "use up" hua.
Rotation equation se friction. Parent derivation mein $f = β ma$ . Yeh step kyun? Torque $f R = I α$ ko $α = a / R$ ke saath rearrange karne par exactly woh force milta hai jo body ko spin karta hai (Torque and Angular Acceleration).
Plug in. $f = \frac{1}{2} (2) (3.267) = 3.267 N$ . Yeh step kyun? $β$ , $m$ , aur computed $a$ ko $f = β ma$ mein substitute karo number nikalne ke liye.

Verify: translation law se cross-check karo $m g sin θ - f = ma$ : $2 (9.8) (0.5) - 3.267 = 9.8 - 3.267 = 6.533$ , aur $ma = 2 (3.267) = 6.533$ ✔. Units: $kg \cdot m/s^{2} = N$ ✔.

Ex 8 — Rolling kab break hoti hai? Slip threshold (Cell E2, exam twist)

Worked example Hoop ko rolling rakhne ke liye minimum coefficient of friction

Ek ring/hoop ( $β = 1$ ) ek angle $θ$ ke slope ke neeche without slipping roll karta hai. Available static friction $f_{m a x} = μ m g cos θ$ hai. Woh sabse chhota $μ$ nikalo jo slipping rokta hai, $θ = 40°$ ke liye.

Forecast: steeper slopes ko zyada friction chahiye; ek hoop ko bahut torque chahiye, toh expect karo $μ$ kuch tenths ka.

Required friction. $f_{req} = β ma$ jahan $a = \frac{g sin θ}{1 + β}$ hai, toh $f_{req} = \frac{β m g sin θ}{1 + β}$ . Yeh step kyun? Yeh woh friction hai jo rolling motion demand karti hai (Ex 7 ki logic se).
Demand ≤ supply set karo. Rolling tab hold karti hai jab $f_{req} \leq μ m g cos θ$ . Threshold equality hai: $μ_{m i n} = \frac{β t a n θ}{1 + β} .$ Yeh step kyun? Dono sides ko $m g cos θ$ se divide karo; $m g$ cancel ho jaata hai aur $sin / cos = tan$ . Hum $tan$ use karte hain kyunki yeh exactly "pull-along ÷ press-into-slope" ka ratio hai, yaani gravity kitni zyada ise skid karne par majaboot karti hai versus kitna zyada yeh pin down hai.
Plug in $β = 1$ , $θ = 40°$ , $tan 40° = 0.8391$ : $μ_{m i n} = \frac{1 \cdot 0.8391}{2} = 0.4196$ . Yeh step kyun? Hoop ka $β$ aur angle substitute karo numeric threshold nikalne ke liye.

Verify: dimensionless (ek coefficient) ✔. Shape par sanity: ek solid sphere ( $β = 2/5$ ) ko sirf $μ_{m i n} = \frac{( 2/5 ) ( 0.8391 )}{7/5} = 0.2397$ chahiye — hoop se kam ✔, kyunki use kam spin-torque chahiye. In $μ$ values se neeche object slip karega, aur poora $a = g sin θ / (1 + β)$ analysis valid rehna band ho jaayega.

Ex 9 — Real-world word problem (Cell F1)

Worked example Ek ramp par soup ki can

"Ek sealed can of soup (treat as a solid cylinder, $β = 1/2$ ) ko rest se ek $15°$ loading ramp ke neeche roll kiya jaata hai jo $2.5 m$ lamba hai. Jab yeh floor par pahunchti hai toh kitni tezi se ja rahi hai, aur kitna time laga?" $g = 9.8$ , $sin 15° = 0.2588$ use karo.

Forecast: gentle slope, short ramp — slow-ish, maybe 2 m/s aur ek do seconds.

Words ko symbols mein translate karo. "solid cylinder" $\Rightarrow β = \frac{1}{2}$ ; "from rest" $\Rightarrow v_{0} = 0$ ; " $2.5$ m long" $\Rightarrow L = 2.5$ (slope ke along distance). Yeh step kyun? Word problems 90% translation hote hain — har phrase ko ek symbol se pin karo.
Acceleration. $a = \frac{g sin 15°}{1.5} = \frac{9.8 ( 0.2588 )}{1.5} = 1.691 m/s^{2}$ . Yeh step kyun? Speed aur time dono ko $a$ chahiye, toh pehle master formula se compute karo.
Final speed. $v = 2 a L = 2 (1.691) (2.5) = 8.454 = 2.907 m/s$ . Yeh step kyun? Rest se start karke, $v^{2} = 2 a L$ speed ko directly distance se link karta hai bina time ke; $L$ ramp ke along hai toh yeh $a$ ke saath pair karta hai.
Time. $t = \frac{v}{a} = \frac{2.907}{1.691} = 1.719 s$ . Yeh step kyun? $v = a t$ se (rest se start kiya), abhi-mila $v$ se $t$ nikalne ke liye invert karo.

Verify: forecast match hua (≈2.9 m/s, ≈1.7 s) ✔. Cross-check $L = \frac{1}{2} a t^{2} = \frac{1}{2} (1.691) (1.71 9^{2}) = \frac{1}{2} (1.691) (2.955) = 2.50$ ✔.

Ex 10 — Mystery shape back-solve karo (Cell F2)

Worked example Acceleration se unknown object identify karo

Ek unknown uniform round object rest se ek $30°$ incline ke neeche roll karta hai aur measure kiya jaata hai ki yeh $a = 3.5 m/s^{2}$ par accelerate karta hai. Uska $β$ kya hai, aur yeh kaunsa standard shape hai? $g = 9.8$ use karo.

Forecast: $g sin 30° = 4.9$ , aur $a = 3.5$ kaafi close hai iske → chhota $β$ → probably solid sphere.

Master formula invert karo. $a = \frac{g sin θ}{1 + β}$ se, dono sides ko $(1 + β)$ se multiply karo aur $a$ se divide karo: $1 + β = \frac{g sin θ}{a}$ , toh $β = \frac{g sin θ}{a} - 1$ . Yeh step kyun? Hume $a$ diya gaya hai aur $β$ chahiye, toh hum $β$ ko algebraically isolate karte hain usual direction mein plug karne ki bajaye.
Plug in. $g sin 30° = 9.8 (0.5) = 4.9$ . $β = \frac{4.9}{3.5} - 1 = 1.4 - 1 = 0.4$ . Yeh step kyun? Known angle aur measured $a$ substitute karo numeric shape factor nikalne ke liye.
Table se match karo. $β = 0.4 = 2/5$ → solid sphere. Yeh step kyun? Computed $β$ ko pinned shape table se compare karo object ka naam lene ke liye.

Verify: forward-check — solid sphere deta hai $a = \frac{4.9}{1.4} = 3.5 m/s^{2}$ ✔, exactly measured value. Forecast confirmed.

Coverage check

Scenario matrix ka har cell ab ek worked home ke saath hai:

Recall Kaun sa formula kaun se question ke liye?

$a$ nikalna hai ::: $a = \frac{g sin θ}{1 + β}$ Final speed nikalni hai ::: $v = 2 g h / (1 + β)$ ya $v = 2 a L$ Rest se time nikalna hai ::: $t = 2 L / a$ Required friction nikalni hai ::: $f = β ma$ Slip threshold nikalna hai ::: $μ_{m i n} = \frac{β tan θ}{1 + β}$ Shape back-solve karna hai ::: $β = \frac{g sin θ}{a} - 1$

Connections

Parent topic — woh master formula jise yeh examples drill karte hain.
Moment of Inertia — jahan se har $β$ value (aur $I$ khud) aata hai.
Rolling Without Slipping — $v = ω R$ , $a = α R$ glue (Ex 8 mein break ho jaata hai).
Energy Conservation in Rolling — Ex 5 ke peeche.
Torque and Angular Acceleration — $f = β ma$ ke peeche (Ex 7).
Friction in Rolling — kyun friction koi kaam nahi karta (Ex 5) phir bhi essential hai (Ex 7–8).
Newton's Second Law on Inclines — $β = 0$ sliding limit (Ex 4).