A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as (rσ0), then theradius of gyration of the disc about its axis passing through the centre is(a) 2a2+b2+ab(b) 2a+b(c) 3a2+b2+ab(d) 3a+b

Exp. (c) Key Idea Radius of gyration K of any structure is given by I=MK2 or K=MI To find K, we need to find both moment of inertia / and mass M of the given structure. Given, variation in mass per unit area (surface mass density), σ=rσ0....................(i) Let us divide whole disc in small area elements, one of them shown at r distance from the centre of the disc with its width as dr. Mass of this element is dm=σ⋅dA ⇒dm=rσ0×2πrdr [from Eq. (i)] Mass of the disc can be calculated by integrating it over the given limits of r, ∫0Mdm=∫abσ0×2π×dr ⇒M=σ02π(b−a) Calculation of Moment of Inertia I=∫0Mr2dm=∫abr2⋅rσ0×2πrdr [from Eq. (i)] =σ02π∫abr2dr=σ02π[3r3]ab ⇒I=31σ02π[b3−a3] Now, radius of gyration, K=MI=2πσ0(b−a)2πσ0(b3−a3) ⇒K=31b−a(b3−a3) As we know, b3−a3=(b−a)(b2+a2+ab) ∴K=31(b2+a2+ab) or K=3a2+b2+ab

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System of Particles and Rotational Motion

Moment of inertia

A circular disc of radius b has a hole of radius a at its cent

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A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $(\frac{σ _{0}}{r})$ , then the
radius of gyration of the disc about its axis passing through the centre is

(a) $\frac{a ^{2} + b ^{2} + ab}{2}$
(b) $\frac{a + b}{2}$
(c) $\frac{a ^{2} + b ^{2} + ab}{3}$
(d) $\frac{a + b}{3}$

Views: 5,831 students

Updated on: Jan 24, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

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Benjamin Discussed

A circular disc of radius

b

has a hole of radius

a

at its centre (see figure). If the mass per unit area of the disc varies as

(\frac{σ _{0}}{r})

, then the
radius of gyration of the disc about its axis passing through the centre is

(a)

\frac{a ^{2} + b ^{2} + ab}{2}

(b)

\frac{a + b}{2}

(c)

\frac{a ^{2} + b ^{2} + ab}{3}

(d)

\frac{a + b}{3}

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Exp. (c) Key Idea Radius of gyration

K

of any structure is given by

I = M K^{2}

K = \frac{I}{M}

To find

K

, we need to find both moment of inertia / and mass

M

of the given structure. Given, variation in mass per unit area (surface mass density),

σ = \frac{σ _{0}}{r}

....................(i) Solution Image

Let us divide whole disc in small area elements, one of them shown at

r

distance from the centre of the disc with its width as

d r

. Mass of this element is

d m = σ \cdot d A

\Rightarrow d m = \frac{σ _{0}}{r} \times 2 π r d r

[from Eq. (i)] Mass of the disc can be calculated by integrating it over the given limits of

r

\int_{0}^{M} d m = \int_{a}^{b} σ_{0} \times 2 π \times d r

\Rightarrow M = σ_{0} 2 π (b - a)

Calculation of Moment of Inertia

I = \int_{0}^{M} r^{2} d m = \int_{a}^{b} r^{2} \cdot \frac{σ _{0}}{r} \times 2 π r d r

[from Eq. (i)]

= σ_{0} 2 π \int_{a}^{b} r^{2} d r = σ_{0} 2 π [\frac{r ^{3}}{3}]_{a}^{b}

\Rightarrow I = \frac{1}{3} σ_{0} 2 π [b^{3} - a^{3}]

Now, radius of gyration,

K = \frac{I}{M} = \frac{2 π σ _{0} ( b ^{3} - a ^{3} )}{2 π σ _{0} ( b - a )}

\Rightarrow K = \frac{1}{3} \frac{( b ^{3} - a ^{3} )}{b - a}

As we know,

b^{3} - a^{3} = (b - a) (b^{2} + a^{2} + ab)

∴ K = \frac{1}{3} (b^{2} + a^{2} + ab)

K = \frac{a ^{2} + b ^{2} + ab}{3}

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Question Text	A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $(\frac{σ _{0}}{r})$ , then the radius of gyration of the disc about its axis passing through the centre is (a) $\frac{a ^{2} + b ^{2} + ab}{2}$ (b) $\frac{a + b}{2}$ (c) $\frac{a ^{2} + b ^{2} + ab}{3}$ (d) $\frac{a + b}{3}$
Updated On	Jan 24, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 5
Upvotes	642
Avg. Video Duration	9 min

Text solutionVerified

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A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as (rσ0​​), then theradius of gyration of the disc about its axis passing through the centre is(a) 2a2+b2+ab​​(b) 2a+b​(c) 3a2+b2+ab​​(d) 3a+b​

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Filo tutor solutions (5)

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