JEE Advanced PYQ System of Particles and Rotational Motion – Moment of Inertia and Rotational K.E.

Q 1 :

A solid sphere of mass $M$ and radius $R$ having moment of inertia $I$ about its diameter is recast into a solid disc of radius $r$ and thickness $t$ . The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains $I$ . Then $R$ and $r$ are related as [2006]

$r = \sqrt{\frac{2}{15}} R$
$r = \frac{2}{\sqrt{15}} R$
$r = \frac{2}{15} R$
$r = \frac{\sqrt{2}}{15} R$

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(2)

For solid sphere, moment of inertia about diameter

$I_{A B} = \frac{2}{5} M R^{2} = I$

For solid disc

$I_{A' B'} = I_{y y'} + M r^{2} = \frac{1}{2} M r^{2} + M r^{2} = \frac{3}{2} M r^{2}$

$I_{A B} = I_{A' B'} (given)$

$∴$ $\frac{2}{5} M R^{2} = \frac{3}{2} M r^{2} \Rightarrow r = \frac{2}{\sqrt{15}} R$

Q 2 :

From a circular disc of radius $R$ and mass 9M, a small disc of radius $\frac{R}{3}$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [2005]

$4 M R^{2}$
$\frac{40}{9} M R^{2}$
$10 M R^{2}$
$\frac{37}{9} M R^{2}$

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4

(1)

Q 3 :

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is [2001]

$\frac{1}{2} M R^{2}$
$\frac{1}{4} M R^{2}$
$\frac{1}{8} M R^{2}$
$\sqrt{2} M R^{2}$

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4

(1)

$Mass of disc = 4 M$

$I_{disc} = \frac{1}{2} M R^{2}$

$∴$ $Moment of inertia of one quarter of disc$

$= \frac{1}{4} [\frac{1}{2} (4 M) R^{2}] = \frac{1}{2} M R^{2}$

Q 4 :

A thin wire of length $L$ and uniform linear mass density $ρ$ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is [2000]

$\frac{ρ L^{3}}{8 π^{2}}$
$\frac{ρ L^{3}}{16 π^{2}}$
$\frac{5 ρ L^{3}}{16 π^{2}}$
$\frac{3 ρ L^{3}}{8 π^{2}}$

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(4)

About the diameter of the circular loop (ring)

$I = \frac{1}{2} M R^{2}$

Using parallel axis theorem

Moment of inertia of the loop about XX' axis

$I_{X X'} = \frac{M R^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

Here mass $M = L ρ$ and radius $R = \frac{L}{2 π}$

$∴$ $I_{X X'} = \frac{3}{2} (L ρ) {(\frac{L}{2 π})}^{2} = \frac{3 L^{3} ρ}{8 π^{2}}$

Q 5 :

A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse P is imparted to the rod at a distance $x = \frac{L}{n}$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of $n$ is _______. [2024]

(18)

$Linear impulse, \int F d t = Δ momentum = m (V_{c m} - 0)$

$or, P = m (ω r_{c m}) = m ω (L + \frac{L}{2})$

$or, P = m ω (\frac{3 L}{2}) \dots (i)$

$And angular impulse \int τ d t = Δ angular momentum$

$\int r \times F d t = Δ L \Rightarrow r \times \int F d t = I (ω - 0) Here I = moment of inertia about axis of rotation.$

$(L + \frac{L}{2} + x) P = (I_{c m} + m d^{2}) ω$

$= (\frac{m L^{2}}{12} + m {(L + \frac{L}{2})}^{2}) ω$

$(\frac{3 L}{2} + x) P = m L^{2} (\frac{1}{12} + {(\frac{3}{2})}^{2}) ω$

$(\frac{3 L}{2} + x) P = m L^{2} (\frac{7}{3}) ω \dots (i i)$

$Dividing equation (ii) by (i)$

$(\frac{3 L}{2} + x) = \frac{L (\frac{7}{3})}{(\frac{3}{2})} \Rightarrow \frac{3 L}{2} + x = L (\frac{14}{9})$

$or, x = \frac{L}{18}$ $∴$ $n = 18$

Q 6 :

The densities of two solid spheres A and B of the same radius R vary with radial distance $r$ as $ρ_{A} (r) = k (\frac{r}{R})$ and $ρ_{B} (r) = k {(\frac{r}{R})}^{5},$ respectively, where $k$ is a constant. The moments of inertia of the individual spheres about axes passing through their centres are $I_{A}$ and $I_{B}$ , respectively. If $\frac{I_{B}}{I_{A}} = \frac{n}{10},$ the value of $n$ is _______. [2015]

(6)

For solid sphere A, density

$ρ_{A} (r) = k (\frac{r}{R})$

$And ρ_{B} (r) = k {(\frac{r}{R})}^{s}$

Consider a spherical shell of radius $x$ and thickness $d x$ .

$Mass of the shell, d m = density \times volume$ $= (k \frac{x}{R}) (4 π x^{2} d x)$

So, moment of inertia of shell about its diameter,

$d I = \frac{2}{3} (d m) x^{2} = \frac{2}{3} (k \frac{x}{R}) (4 π x^{2} d x) x^{2} = (\frac{8 π k}{3 R}) x^{5} d x$

$∴$ $Moment of inertia of the sphere A, I_{A} = \int_{0}^{R} d I$

$= \frac{8 π k}{3 R} \int_{0}^{R} x^{5} d x = \frac{8 π k}{3 R} {[\frac{x^{6}}{6}]}_{0}^{R}$

$∴$ $I_{A} = (\frac{8 π k}{18}) R^{5} (i)$

$Similarly, for sphere B$

$I_{B} = \frac{8 π k}{3 R^{5}} \int_{0}^{R} x^{9} d x = (\frac{8 π k}{3 R^{5}}) {[\frac{x^{10}}{10}]}_{0}^{R}$

$∴$ $I_{B} = \frac{8 π k}{30} R^{5}$

$From eqns. (i) and (ii)$

$\frac{I_{B}}{I_{A}} = \frac{18}{30} = \frac{6}{10} = \frac{n}{10}$ $∴$ $n = 6$

Q 7 :

A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_{O}$ and $I_{P}$ respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $I_{P} / I_{O}$ to the nearest integer is [2012]

(3)

Let $σ$ be the surface mass density.

Moment of inertia of the lamina about axes passing through 'O'

$I_{O} = \frac{1}{2} σ [π {(2 R)}^{2}] \times {(2 R)}^{2} - [\frac{1}{2} {(σ π R^{2})}^{2} + σ (π R^{2}) \times R^{2}]$

$= \frac{13}{2} π σ R^{4}$

Moment of inertia of the lamina about axes passing through P

$I_{P} = 8 π σ R^{4} + σ π {(2 R)}^{2} \times {(2 R)}^{2} - [\frac{1}{2} σ (π R^{2}) R^{2} + σ (π R^{2}) {(\sqrt{{(2 R)}^{2} + R^{2}})}^{2}]$

$= 24 π σ R^{4} - 5.5 π σ R^{4} = 18.5 π σ R^{4}$

$∴$ $\frac{I_{P}}{I_{O}} = \frac{18.5 π σ R^{4}}{\frac{13}{2} π σ R^{4}} = \frac{37}{13} \approx 3$

Q 8 :

Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]

(9)

Let the four spheres be A, B, C & D

$m_{A} = m_{B} = m_{C} = m_{D} = 0.5 kg$

Moment of inertia of the system about the diagonal of the square

$I_{X Y} = I_{A} + I_{B} + I_{C} + I_{D} = 2 I_{A} + 2 I_{B}$

$= 2 [\frac{2}{5} M R^{2} + M a^{2}] + 2 [\frac{2}{5} M R^{2}]$

$= 4 \times \frac{2}{5} M R^{2} + 2 M a^{2} = M [\frac{8}{5} R^{2} + 2 a^{2}]$

$= 0.5 [\frac{8}{5} \times {(\frac{\sqrt{5}}{2})}^{2} + 2 \times 8] \times 10^{- 4}$

$= 0.5 [2 + 16] \times 10^{- 4}$

$= 9 \times 10^{- 4} = N \times 10^{- 4} {kg m}^{2}$ $∴$ $N = 9$

Topic Question Set

Q 2 :

From a circular disc of radius $R$ and mass 9M, a small disc of radius $\frac{R}{3}$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [2005]

(1)

Q 3 :

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is [2001]

Q 4 :

A thin wire of length $L$ and uniform linear mass density $ρ$ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is [2000]

Q 8 :

Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]

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Topic Question Set

Q 1 : A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains I. Then R and r are related as [2006]

Q 2 : From a circular disc of radius R and mass 9M, a small disc of radius R3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [2005]

(1)

Q 3 : One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is [2001]

Q 4 : A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is [2000]

Q 8 : Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N×10-4 kg m2, then N is [2011]

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Q 2 :

From a circular disc of radius $R$ and mass 9M, a small disc of radius $\frac{R}{3}$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is [2005]

Q 3 :

One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is [2001]

Q 4 :

A thin wire of length $L$ and uniform linear mass density $ρ$ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is [2000]

Q 8 :

Four solid spheres each of diameter $\sqrt{5} cm$ and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} {kg m}^{2}$ , then $N$ is [2011]