Applications — mass, centre of mass, moments of inertia: Question bank

4.4.22 · D5 · Maths › Multivariable Calculus › Applications — mass, centre of mass, moments of inertia

True ya false — justify karo

Har claim ek aisi sentence hai jo koi maanta hai. True/false decide karo aur kyun bolo — justification hi asli cheez hai.

Kisi lamina ka centre of mass hamesha region ke andar hota hai.

False. Non-convex ya ring-shaped region ke liye (jaise L-shape ya annulus) balance point khaali jagah mein pad sakta hai — yeh mass se weighted average position hai, koi physical material point nahi.

Agar

ρ

constant hai, toh centre of mass centroid ke barabar hota hai.

True. Constant

ρ

dono

\iint x ρ d A

aur

\iint ρ d A

se bahar aa jaata hai aur cancel ho jaata hai, sirf ek purely geometric average bachta hai — yahi centroid ki definition hai.

Moment of inertia negative ho sakta hai agar region third quadrant mein ho.

False.

I_{x} = \iint y^{2} ρ d A

mass ko

y^{2} \geq 0

aur

ρ \geq 0

se multiply karta hai, isliye integrand kabhi negative nahi hota —

I

spinning ke resistance ko measure karta hai aur quadrant chahe jo bhi ho, hamesha

\geq 0

rehta hai.

Moment

M_{x}

negative ho sakta hai.

True.

M_{x} = \iint y ρ d A

mein

y

pehli power par hai; agar plate

x

-axis ke neeche ho (

y < 0

) toh har tile ek negative lever arm contribute karta hai, isliye

M_{x} < 0

— isi tarah balance point axis ke neeche aata hai.

I_{0} = I_{x} + I_{y}

kisi bhi body ke liye valid hai.

False. Yeh perpendicular-axis theorem hai, jo sirf flat lamina ke liye valid hai

x y

-plane mein, kyunki isme

r^{2} = x^{2} + y^{2}

chahiye bina kisi

z

contribution ke. Ek 3-D solid mein

z^{2}

terms hote hain jo ise tod dete hain.

Har density value ko double karne se centre of mass coordinates double ho jaate hain.

False.

\overset{x}{ˉ} = \frac{\iint x ρ d A}{\iint ρ d A}

—

2

ka global factor numerator aur denominator dono ko multiply karta hai aur cancel ho jaata hai. Balance point unchanged rehta hai; sirf

ρ

ki distribution ise move karti hai.

Har density value ko double karne se mass double aur

I_{0}

double hota hai.

True. Dono

m = \iint ρ d A

aur

I_{0} = \iint r^{2} ρ d A

ρ

mein linear hain, isliye ek constant factor unhe directly scale karta hai — CoM ke unlike, yeh factor ko divide nahi karte.

Ek uniform disc ke liye,

I_{0}

a^{2}

(radius squared) ke proportional hai.

False.

I_{0} = \iint r^{2} ρ d A

aur

d A \sim (length)^{2}

, jisse

I_{0} \propto a^{4}

milta hai. Sirf per unit mass,

I_{0} / m \propto a^{2}

hota hai, kyunki

m \propto a^{2}

do powers absorb kar leta hai.

Radius of gyration

R_{g} = I / m

centre of mass ki axis se distance hai.

False.

R_{g}

woh single distance hai jis par poora mass concentrated hone par same

I

deta; yeh generally CoM distance se bada hota hai kyunki

I

door wale mass ko

r^{2}

se weight karta hai.

Error dhundo

Kisine neeche di gayi line likhi. Galti dhundo aur sahi idea batao.

"Polar mein, mass

= ρ \int_{0}^{2 π} \int_{0}^{a} d r d θ

Area element galat hai: ek polar tile ek curved wedge hoti hai jiska area

r d r d θ

hai,

d r d θ

nahi. Jacobian factor

r

mandatory hai — sahi integrand hai

ρ r d r d θ

\overset{x}{ˉ} = \frac{M _{x}}{m}

Indices swap ho gaye hain.

\overset{x}{ˉ}

y

-axis ke baare mein balance karta hai jiska lever arm

x

hai, isliye

\overset{x}{ˉ} = M_{y} / m

M_{x}

use karna (lever arm

y

)

\overset{y}{ˉ}

deta.

x

-axis ke baare mein:

I_{x} = \iint x^{2} ρ d A

Galat lever arm. $x$ -axis tak ki distance vertical gap

y

hai, isliye

I_{x} = \iint y^{2} ρ d A

x^{2}

wala version

I_{y}

hai (

y

-axis tak ki distance).

"Kyunki yeh

x

-axis ke baare mein hai,

M_{x} = \iint x ρ d A

x

-axis ke baare mein" kehne se lever arm

y

fix hota hai,

x

nahi:

M_{x} = \iint y ρ d A

. Naam us axis ko refer karta hai jis par tum balance kar rahe ho, aur arm us tak ki perpendicular distance hai.

"Plate

y

-axis ke baare mein symmetric hai, isliye uska mass zero hai kyunki negative

x

, positive

x

ko cancel karta hai."

Cancellation moment

M_{y} = \iint x ρ d A

ke liye hoti hai (odd integrand), jisse

\overset{x}{ˉ} = 0

milta hai. Mass integrand

ρ \geq 0

use karta hai bina kisi

x

factor ke, isliye kuch cancel nahi hota — mass strictly positive hota hai.

x y

-plane mein ek lamina ke liye,

I_{z} = \iint z^{2} ρ d A

, jo

0

hai kyunki

z = 0

z

-axis ke baare mein moment of inertia us axis se distance use karta hai,

x^{2} + y^{2}

, coordinate

z

nahi. Isliye

I_{z} = I_{0} = \iint (x^{2} + y^{2}) ρ d A \neq = 0

Why questions

"Kyun" ka jawab genuinely soch kar ek ya do sentences mein do.

Moment of inertia

r^{2}

kyun use karta hai lekin moment (CoM ke liye)

r^{1}

use karta hai?

Balance/leverage distance mein linear hai, isliye moment ko

r^{1}

chahiye; rotational energy

\frac{1}{2} (d m) v^{2}

hai jisme

v = ω r

hai, isliye energy

r^{2}

carry karta hai — alag physics, alag power.

Centre of mass paane ke liye moment ko mass se kyun divide karte hain?

Hum yeh demand karte hain ki

(\overset{x}{ˉ}, \overset{y}{ˉ})

par rakha single point mass

m

same moments reproduce kare:

m \overset{x}{ˉ} = M_{y}

. Solve karne par

\overset{x}{ˉ} = M_{y} / m

milta hai — division total leverage ko ek location mein convert karta hai.

Varying-density plate ke CoM integral mein

ρ

andar kyun rakhna padta hai?

Position ko kitna mass wahan hai us se weight karna padta hai; heavy regions balance point ko apni taraf khichte hain.

ρ

hata do toh sab tiles equal ho jaayenge aur geometric centroid milega.

Extra

r

sirf polar coordinates mein kyun aata hai, Cartesian mein nahi?

Ek Cartesian tile

d x d y

actually

d x d y

area ka rectangle hai. Ek polar tile ki sides

d r

aur arc

r d θ

hoti hain, isliye uska area radius ke saath badhta hai — change of variables is stretch ko factor

r

ke roop mein record karta hai.

Spinning ke liye moment of inertia relevant kyun hai lekin moment (CoM) nahi?

Spinning kinetic energy

\frac{1}{2} I ω^{2}

aur angular momentum

I ω

store karta hai, dono

I = \iint r^{2} ρ d A

se governed hain; dekho Rotational Kinetic Energy and Angular Momentum. CoM sirf balance aur translational motion locate karta hai.

Same mass wali do plates ke

I_{0}

itne alag kyun ho sakte hain?

I_{0}

depend karta hai ki mass kahan hai, door ke mass ko

r^{2}

se weight karta hai. Same mass ko bahar spread karne se

I_{0}

tezi se badhta hai jabki

m

unchanged rehta hai.

Perpendicular-axis theorem sirf flat plates tak restricted kyun hai?

Yeh

r^{2} = x^{2} + y^{2}

par rely karta hai taaki

I_{0} = \iint x^{2} + \iint y^{2} = I_{y} + I_{x}

ho sake. Ek solid mein

z^{2}

terms add ho jaate hain, isliye yeh clean split nahi rehta.

Edge cases

Boundary aur degenerate inputs — kabhi inhe surprise mat hone do.

Ek plate ka centre of mass kya hoga jiska density sirf ek point par non-zero ho?

Undefined (

0/0

form): dono

\iint x ρ d A

aur

\iint ρ d A

zero area ke set par vanish karte hain. Ek point koi area nahi carry karta, isliye koi mass nahi, isliye koi balance point exist nahi karta.

Origin ke baare mein symmetric region jisme even density

ρ (- x, - y) = ρ (x, y)

ho,

M_{x}

aur

M_{y}

kya honge?

Dono

0

honge: integrand

x ρ

(aur

y ρ

) symmetry ke under odd hai, isliye opposite tiles se contributions cancel ho jaate hain. Isliye

\overset{x}{ˉ} = \overset{y}{ˉ} = 0

— CoM origin par hota hai.

Agar ek plate poori tarah

x

-axis par ho (ek thin rod along

y = 0

), toh

I_{x}

kya hai?

I_{x} = \iint y^{2} ρ d A = 0

, kyunki har tile ka

y = 0

hai. Ek line mein koi thickness nahi hoti jo rotation us axis ke baare mein lad sake jis par woh khud lie karti hai.

Ek homogeneous plate ko dono dimensions mein

k

factor se scale kiya jaata hai (density fixed).

m

aur

I_{0}

kaise change honge?

Area

k^{2}

se scale hota hai isliye

m \to k^{2} m

; distances bhi

k

se scale hoti hain, jisse

I_{0} \to k^{4} I_{0}

milta hai (do powers area se, do

r^{2}

se). Isliye

I_{0} / m \to k^{2} (I_{0} / m)

Jab region ek single point

(x_{0}, y_{0})

par shrink ho jaata hai toh

\overset{x}{ˉ}

ka kya hoga?

Yeh

x_{0}

approach karta hai: dono integrals us point par concentrate ho jaate hain, aur ek single location ka weighted average woh location khud hai — limit well-behaved hai chahe mass

\to 0

ho.

Agar

ρ

kahin negative ho (ek modelling error), toh pehle kya toota?

Mass

0

ya negative ho sakta hai, isliye

\overset{x}{ˉ} = M_{y} / m

blow up ya sign flip kar sakta hai, aur

I_{0}

negative ho sakta hai — sab physically meaningless. Poore framework ko sahi rakhne ke liye density

ρ \geq 0

satisfy karni chahiye.

3-D solid ke liye, kya centre-of-mass recipe abhi bhi total mass se divide karti hai?

Haan — pattern Triple Integrals ke saath identical hai:

\overset{x}{ˉ} = \frac{∭ x ρ d V}{∭ ρ d V}

. Sirf integral ki dimension aur density (mass per volume) change hoti hai.

Recall One-line self-test

Right side cover karo aur har sentence apne words mein complete karo. " $M$ uses distance to the ___ power; $I$ uses the ___ power." ::: First; second. "Polar mein, har area integral ek extra ___ pehnta hai." ::: $r$ (Jacobian se). "CoM mass se divide karta hai; $I$ mass se ___ divide karta." ::: nahi karta.