Physical pendulum — compound pendulum: Question bank

1.6.7 · D5 · Physics › Oscillations & Waves › Physical pendulum — compound pendulum

Neeche use hone wale symbols ki reminders (sab parent note mein build hue hain):

$m$ = rigid body ki total mass.
$g$ = acceleration due to gravity (neeche ki taraf pull ki strength, $\approx 9.8 m/s^{2}$ Earth par).
$I$ = moment of inertia pivot ke baare mein (dekho Moment of inertia).
$d$ = pivot se center of mass (CM) ki doori.
$k$ = CM ke baare mein Radius of gyration, toh $I_{c m} = m k^{2}$ .
$L_{eq} = I / (m d) = k^{2} / d + d$ = equivalent Simple pendulum length.
$T = 2 π I / (m g d)$ , sirf small $θ$ ke liye valid aur sirf tab jab CM pivot ke neeche ho.

Spot the error

"I'll plug

I_{c m} = \frac{1}{2} m R^{2}

straight into

T = 2 π I / m g d

for a disk hung from its rim."

Wrong

I

— body physically pivot ke baare mein rotate karti hai, toh tumhe

I = I_{c m} + m d^{2}

chahiye (Parallel axis theorem); correct value

\frac{3}{2} m R^{2}

hai,

\frac{1}{2} m R^{2}

nahi.

"The effective length is just

d

, the pivot-to-CM distance."

Effective length

L_{eq} = k^{2} / d + d

hai, jo strictly

d

se badi hai;

L = d

use karna assume karta hai ki sari mass CM par hai aur

k^{2} / d

jo spread measure karta hai use ignore karta hai.

"As I slide the pivot toward the CM, the period keeps shrinking because there's less inertia to swing."

Yeh sirf

d = k

tak shrink hota hai; minimum ke baad restoring torque

m g d sin θ

I

se zyada tezi se khatam hota hai, toh

T

turn around kar ke CM par infinity tak blow up ho jaata hai.

"The lever arm of gravity is

d

, so the torque is

m g d

Perpendicular lever arm

d sin θ

hai, jo restoring torque

τ = - m g d sin θ

deta hai;

m g d

sirf coefficient hai, aur

sin θ

(aur sign) drop karne se restoring behaviour poora delete ho jaata hai.

"Since

sin θ \approx θ

is just an approximation, the formula is really exact if I'm careful."

Nahi — SHM form sirf isliye appear hoti hai kyunki

sin θ

ko linearise kiya gaya; large amplitude par equation genuinely nonlinear hai aur period sach mein depend karta hai ki tum kitna door swing karte ho.

"A door is not a pendulum because it swings about a vertical axis."

Sahi pakda — gravity vertical axis ke baare mein koi restoring torque provide nahi karti (CM ki height change nahi hoti), toh door gravity pendulum ki tarah bilkul oscillate nahi karta.

Why questions

Why must

I

be taken about the pivot and not the CM?

Kyunki body physically fixed pivot axis ke baare mein rotate karti hai, aur rotational Newton's law

τ = I α

moment of inertia actual axis of rotation ke baare mein use karta hai.

Why does the period go to infinity as the pivot approaches the CM?

Restoring torque

m g d sin θ

hai, jo

d \to 0

hone par

\to 0

ho jaata hai, jabki

I = m k^{2}

finite rehta hai; vanishing restoring "spring" ka matlab infinitely slow return hai.

Why is torque, not force, the right starting law here?

Motion fixed axis ke baare mein rotation hai, toh relevant balance twisting effect (torque) aur rotational sluggishness (

I

) ke beech hai, jo

τ = I α

se capture hota hai,

F = ma

se nahi.

Why does a minimum period exist at all?

Do competing effects — CM ke paas restoring torque khatam hota hai (period badhta hai), aur CM se door

I

d^{2}

ki tarah badhta hai (period badhta hai) — toh beech mein ek sweet spot

T

ko minimize karta hai, jo

d = k

par hota hai.

Why does mass cancel out of the period?

Har restoring term mein

m

ka ek factor hota hai (gravity

m g

) aur har inertial term mein

m

ka ek factor hota hai (mass distribution

I \propto m

); ratio

I / (m g d)

mein

m

kahin nahi bachta, jo sabhi gravity pendulums ki ek hallmark hai.

Why is the equivalent length useful rather than just quoting

T

L_{eq}

sari messy mass distribution ko ek single number mein collapse kar deta hai, jo tumhe kisi bhi rigid swinger ko Simple pendulum ki tarah treat karne aur bodies ko ek nazar mein compare karne deta hai.

Why does the parallel-axis theorem show up everywhere in this topic?

Tables

I_{c m}

dete hain, lekin physics ko

I_{pivot}

chahiye; Parallel axis theorem

I = I_{c m} + m d^{2}

dono ke beech ka bridge hai, aur yeh directly

L_{eq}

k^{2} / d + d

form produce karta hai.

Edge cases

What happens to the period if the pivot is placed exactly at the CM (

d = 0

Gravity ka zero lever arm hai, toh koi restoring torque nahi hai; body ki koi preferred orientation nahi hoti aur yeh simply oscillate nahi karti —

T \to \infty

What happens if the pivot lies below the center of mass (an inverted, top-heavy setup)?

Equilibrium unstable ho jaata hai: thodi si tilt par gravity ab

θ

ke same sign wala torque produce karti hai (yaani

τ = + m g d sin θ

), jo body ko vertical se aur door push karta hai na ki wapas. Koi oscillation nahi hoti, formula

T = 2 π I / m g d

real motion describe nahi karta (formally

ω^{2} < 0

), aur body topple ho jaati hai — bilkul inverted pendulum ki tarah.

What is the smallest possible period for a body of given radius of gyration

k

T_{m i n} = 2 π 2 k / g

, tab achieve hota hai jab pivot CM se

d = k

ki doori par ho, jisse

L_{eq} = 2 k

milta hai.

If a body's entire mass really were concentrated at one point (so

k = 0

), what does the compound formula reduce to?

L_{eq} = 0/ d + d = d

, jisse

T = 2 π d / g

milta hai — bilkul Simple pendulum, jo confirm karta hai ki compound result mein point-mass case included hai.

What limits the accuracy of

T = 2 π I / m g d

as amplitude grows?

Approximation

sin θ \approx θ

break down ho jaati hai; true period amplitude ke saath badhta hai, toh formula sirf small-oscillation limit hai.

If a body has two possible pivot distances giving the same period, how are they related?

Yeh conjugate points

d_{1}

aur

d_{2} = k^{2} / d_{1}

hain jo CM ke opposite sides par hain aur

d_{1} + d_{2} = L_{eq}

ke saath; yeh pairing (pivot ↔ center of oscillation) Kater's pendulum ko underlie karta hai.

Does the compound-pendulum analysis apply to a Torsional pendulum?

Nahi — torsional pendulum ka restoring torque twisted wire se aata hai (

τ = - κ θ

), gravity ke lever arm se nahi, toh uska period

2 π I / κ

mein koi

d

aur koi

g

nahi hai.

Recall Har trap ki one-line summary

Lagbhag har mistake teen mein se ek hai: (1) $I_{pivot}$ ki jagah $I_{c m}$ use karna, (2) $d$ ko $L_{eq}$ samajh lena, ya (3) yeh bhool jaana ki $sin θ \approx θ$ assume kiya gaya tha (aur ki CM pivot ke neeche hona chahiye). Inhe seedha rakho aur topic disarm ho jaata hai.

Question bank — Physical pendulum — compound pendulum

True or false — justify

Spot the error

Why questions

Edge cases