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Chapter 5 Rigid-Body Dynamics Previous Year Questions with Solutions

Pravegaa PYQ Series · Volume 2: Classical Mechanics & Special Relativity

CM05: Rigid-Body Dynamics

57 previous-year questions with complete solutions, from IIT JAM, GATE, CSIR NET-JRF, JEST and TIFR, arranged from IIT JAM and GATE through to TIFR and from the most recent paper backwards. Click Show solution under any question.

  • IIT JAM: 17
  • GATE: 5
  • CSIR NET: 6
  • JEST: 11
  • TIFR GS: 18
CM05.1IIT JAM 2025, Q41NAT | 1 mark

Two solid cylinders of the same density are found to have the same moment of inertia about their respective principal axes. The length of the second cylinder is 16 times that of the first cylinder. If the radius of the first cylinder is 4 cm, the radius of the second cylinder is cm. (in integer)

Show solution

For a solid cylinder about its axis, \(I=\frac12MR^2\) with \(M=\rho\pi R^2l\), so \(I=\frac12\rho\pi R^4l\). Equal moments of inertia give \[R_1^4l_1=R_2^4l_2\;\Rightarrow\;\frac{R_1}{R_2}=\left(\frac{l_2}{l_1}\right)^{1/4}=(16)^{1/4}=2\;\Rightarrow\;R_2=\frac{4}{2}=\boxed{2\ \text{cm}}\]

Answer: 2

CM05.2IIT JAM 2024, Q44NAT | 1 mark

A particle of mass \(1\ \mathrm{kg}\), initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\pi/6\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2 seconds is \(\mathrm{N\text{-}m}\). (Rounded off to nearest integer)Figure(Take \(g=10\ \mathrm{m/s^2}\))

Show solution

Along the frictionless incline, the acceleration is \(a=g\sin30^\circ=5\ \mathrm{m/s^2}\). In \(t=2\) s the particle slides \[ s=\tfrac12at^2=\tfrac12\times5\times4=10\ \mathrm m \] down the incline, so its horizontal distance from the vertical line through O is \[ x=s\cos30^\circ=10\times\frac{\sqrt3}{2}=5\sqrt3\ \mathrm m . \] The torque about O is due to the weight \(mg\), which acts vertically downward at horizontal distance \(x\) from O. Its magnitude is independent of \(h\): \[ \tau=|\vec r\times m\vec g|=mg\,x=1\times10\times5\sqrt3=50\sqrt3\approx\boxed{86.6\ \mathrm{N\,m}} \]

Answer: 87 (range 85 to 88)

CM05.3IIT JAM 2024, Q45NAT | 1 mark

The moment of inertia of a solid hemisphere (mass \(M\) and radius \(R\)) about the axis passing through the hemisphere and parallel to its flat surface is \(\frac25MR^2\). The distance of the axis from the center of mass of the hemisphere (in units of \(R\)) is .

(Rounded off to two decimal places)

Show solution

About a diameter of the flat face (an axis through the centre O of the flat face, parallel to it), the hemisphere has the same moment of inertia as a full sphere per unit mass, \(I_O=\frac25MR^2\). So the required axis is this diameter (or its mirror image on the other side of the centre of mass).

The centre of mass lies on the symmetry axis at a distance \[ z_{\mathrm{cm}}=\frac{3R}{8} \] from the flat face. Check with the parallel-axis theorem: \(I_{\mathrm{cm}}=\frac25MR^2-M\left(\frac{3R}{8}\right)^2=\frac{83}{320}MR^2\), and the axis with \(I=\frac25MR^2\) satisfies \(Md^2=\frac25MR^2-\frac{83}{320}MR^2=\frac{9}{64}MR^2\). Hence \[ d=\frac38R=\boxed{0.375\,R}\approx0.38\,R \]

Answer: 0.38 (range 0.36 to 0.40)

CM05.4IIT JAM 2023, Q14MCQ | 2 marks

A rotating disc is held in front of a plane mirror in two different orientations which are (i) angular momentum parallel to the mirror and (ii) angular momentum perpendicular to the mirror. Which of the following schematic figures correctly describes the angular momentum (solid arrow) and its mirror image (shown by dashed arrows) in the two orientations?

  1. Figure

  2. Figure

  3. Figure

  4. Figure

Show solution

Angular momentum \(\vec L=\vec r\times\vec p\) is an axial vector, set by the sense of rotation through the right-hand rule. A mirror reverses the velocity component normal to the mirror and keeps the parallel components.

(i) \(\vec L\) parallel to the mirror: the disc lies in a plane perpendicular to the mirror. In the image the sense of rotation is reversed, so the image of \(\vec L\) points opposite to \(\vec L\) (solid arrow up, dashed arrow down).

(ii) \(\vec L\) perpendicular to the mirror: the disc lies parallel to the mirror, and its velocities are all parallel to the mirror and are unchanged in the image. The image disc rotates in the same sense, so the image \(\vec L\) points in the same direction as \(\vec L\) (both arrows point the same way).

This is figure (B). \[\boxed{\text{(B)}}\]

Answer: (B)

CM05.5IIT JAM 2023, Q34MSQ | 2 marks

A rod of mass \(M\), length \(L\) and non-uniform mass per unit length \(\lambda(x)=\dfrac{3Mx^2}{L^3}\), is held horizontally by a pivot, as shown in the figure, and is free to move in the plane of the figure. For this rod, which of the following statements are true?Figure

  1. Moment of inertia of the rod about an axis passing through the pivot is \(\tfrac35ML^2\)

  2. Moment of inertia of the rod about an axis passing through the pivot is \(\tfrac13ML^2\)

  3. Torque on the rod about the pivot is \(\tfrac34MgL\)

  4. If the rod is released, the point at a distance \(\tfrac{2L}{3}\) from the pivot will fall with acceleration \(g\)

Show solution

With \(x\) measured from the pivot, \(dm=\lambda\,dx\) (and \(\int_0^L\lambda\,dx=M\)).

Moment of inertia: \[ I_0=\int_0^L x^2\,\frac{3Mx^2}{L^3}\,dx=\frac{3M}{L^3}\cdot\frac{L^5}{5}=\frac35ML^2 , \] so (A) is true and (B) is false.

Torque: the centre of mass is at \[ x_{\rm cm}=\frac1M\int_0^L x\,\frac{3Mx^2}{L^3}\,dx=\frac{3}{L^3}\cdot\frac{L^4}{4}=\frac{3L}{4}, \] so the torque of gravity about the pivot is \(\tau=Mg\,x_{\rm cm}=\tfrac34MgL\), and (C) is true.

Initial acceleration: \(\alpha=\dfrac{\tau}{I_0}=\dfrac{\tfrac34MgL}{\tfrac35ML^2}=\dfrac{5g}{4L}\). The point at \(x=\tfrac{2L}{3}\) has acceleration \(\alpha x=\tfrac{5g}{4L}\cdot\tfrac{2L}{3}=\tfrac56g\neq g\), so (D) is false. (The point falling with acceleration \(g\) is at \(x=4L/5\).) \[\boxed{\text{(A), (C)}}\]

Answer: (A), (C)

CM05.6IIT JAM 2022, Q26MCQ | 2 marks

A square laminar sheet with side \(a\) and mass \(M\), has mass per unit area given by \(\sigma(x)=\sigma_0\left[1-\dfrac xa\right]\), (see figure). Moment of inertia of the sheet about \(y\)-axis isFigure

  1. \(\dfrac{Ma^2}{2}\)

  2. \(\dfrac{Ma^2}{4}\)

  3. \(\dfrac{Ma^2}{6}\)

  4. \(\dfrac{Ma^2}{12}\)

Show solution

The sheet occupies \(0\le x\le a\), \(-a/2\le y\le a/2\), and \(dm=\sigma(x)\,dx\,dy\). \[ M=\int_0^a\sigma_0\left(1-\frac xa\right)dx\int_{-a/2}^{a/2}dy=\sigma_0\,a\cdot\frac a2=\frac{\sigma_0a^2}{2}\;\Rightarrow\;\sigma_0=\frac{2M}{a^2}. \] \[ I_{y}=\int x^2\,dm=a\sigma_0\int_0^a\left(x^2-\frac{x^3}{a}\right)dx=a\sigma_0\left(\frac{a^3}{3}-\frac{a^3}{4}\right)=\frac{\sigma_0a^4}{12}=\boxed{\frac{Ma^2}{6}} \]

Answer: (C) \(Ma^2/6\)

CM05.7IIT JAM 2021, Q10MCQ | 1 mark

The moment of inertia of a solid sphere (radius \(R\) and mass \(M\)) about the axis which is at a distance of \(\dfrac{R}{2}\) from the center is

  1. \(\dfrac{3}{20}MR^2\)

  2. \(\dfrac{1}{2}MR^2\)

  3. \(\dfrac{13}{20}MR^2\)

  4. \(\dfrac{9}{10}MR^2\)

Show solution

By the parallel-axis theorem with \(I_{\rm cm}=\frac25MR^2\) and \(d=R/2\), \[ I=\frac25MR^2+M\left(\frac R2\right)^2=\left(\frac{8}{20}+\frac{5}{20}\right)MR^2 \] \[ \boxed{I=\frac{13}{20}MR^2} \]

Answer: (C) \(\dfrac{13}{20}MR^2\)

CM05.8IIT JAM 2021, Q29MCQ | 2 marks

A thin circular disc lying in the \(xy\)-plane has a surface mass density \(\sigma\), given by \[ \sigma(r)=\begin{cases}\sigma_0\left(1-\dfrac{r^2}{R^2}\right) & \text{if } r\le R\\[4pt] 0 & \text{if } r>R\end{cases} \] where \(r\) is the distance from its center. Its moment of inertia about the \(z\)-axis, passing through its center is

  1. \(\dfrac{\sigma_0R^4}{4}\)

  2. \(\dfrac{\pi\sigma_0R^4}{6}\)

  3. \(\sigma_0R^4\)

  4. \(2\pi\sigma_0R^4\)

Show solution

With \(dm=\sigma(r)\,r\,dr\,d\theta\), \[ I_z=\int_0^{2\pi}\!\!\int_0^R r^2\sigma_0\left(1-\frac{r^2}{R^2}\right)r\,dr\,d\theta =2\pi\sigma_0\left(\frac{R^4}{4}-\frac{R^4}{6}\right) \] \[ \boxed{I_z=\frac{\pi\sigma_0R^4}{6}} \]

Answer: (B) \(\dfrac{\pi\sigma_0R^4}{6}\)

CM05.9IIT JAM 2020, Q14MCQ | 2 marks

A wheel is rotating at a frequency \(f_0\) Hz about a fixed vertical axis. The wheel stops in \(t_0\) seconds, with constant angular deceleration. The number of turns covered by the wheel before it comes to rest is given by

  1. \(f_0t_0\)

  2. \(2f_0t_0\)

  3. \(\dfrac{f_0t_0}{2}\)

  4. \(\dfrac{f_0t_0}{\sqrt2}\)

Show solution

\(\omega_0=2\pi f_0\) and \(\omega=\omega_0-\alpha t\); the wheel stops at \(t_0\), so \(\alpha=\omega_0/t_0\). The angle turned is \[ \theta=\omega_0t_0-\frac12\alpha t_0^2=\omega_0t_0-\frac12\omega_0t_0=\frac{\omega_0t_0}{2}=\pi f_0t_0 \] \[ \text{Number of turns}=\frac{\theta}{2\pi}=\boxed{\frac{f_0t_0}{2}} \]

Answer: (C) \(\dfrac{f_0t_0}{2}\)

CM05.10IIT JAM 2020, Q53NAT | 2 marks

The angular momentum of a particle relative to origin varies with time \((t)\) as \(\vec L=\left(4\hat x+at^2\hat y\right)kg\,m^2/s\), where \(a=1\,kg\,m^2/s^3\). The angle between \(\vec L\) and the torque acting on the particle becomes \(45^\circ\) after a time of \(s\).

Show solution

\[ \vec\tau=\frac{d\vec L}{dt}=2at\,\hat y \] \[ \cos45^\circ=\frac{\vec L\cdot\vec\tau}{|\vec L||\vec\tau|}\ \Rightarrow\ \frac{1}{\sqrt2}=\frac{2a^2t^3}{\sqrt{16+a^2t^4}\;2at}=\frac{at^2}{\sqrt{16+a^2t^4}} \] \[ 16+a^2t^4=2a^2t^4\ \Rightarrow\ a^2t^4=16\ \Rightarrow\ t^4=16\ (a=1)\ \Rightarrow\ \boxed{t=2\,\text{s}} \]

Answer: 2 (range 1.9 to 2.1)

CM05.11IIT JAM 2019, Q18MCQ | 2 marks

The mass per unit length of a rod (length \(2~\text{m}\)) varies as \(\rho=3x~\text{kg/m}\). The moment of inertia (in \(\text{kg m}^2\)) of the rod about a perpendicular-axis passing through the tip of the rod (at \(x=0\))

  1. 10

  2. 12

  3. 14

  4. 16

Show solution

\[ I=\int_0^2 x^2\rho\,dx=\int_0^2 3x^3\,dx=\left.\frac{3x^4}{4}\right|_0^2=\frac{3\times16}{4} \;\Rightarrow\;\boxed{I=12~\text{kg m}^2} \]

Answer: (B) 12

CM05.12IIT JAM 2019, Q57NAT | 2 marks

If the diameter of the Earth is increased by \(4\%\) without changing the mass, then the length of the day is hours.

(Take the length of the day before the increment as \(24\) hours. Assume the Earth to be a sphere with uniform density) (Round off to 2 decimal places)

Show solution

No external torque acts, so the angular momentum \(L=I\omega=\tfrac{2}{5}MR^2\cdot\dfrac{2\pi}{T}\) is conserved. With \(M\) fixed, \(T\propto R^2\): \[ \frac{T_2}{T_1}=\left(\frac{R_2}{R_1}\right)^2=(1.04)^2=1.0816 \;\Rightarrow\;T_2=24\times1.0816=25.958~\text{h} \] \[ \boxed{T_2=25.96~\text{hours}} \]

Answer: 25.96 (range 25.95 to 25.97)

CM05.13IIT JAM 2018, Q22MCQ | 2 marks

A disc of radius \(R_1\) having uniform surface density has a concentric hole of radius \(R_2<R_1\). If its mass is \(M\), the principal moments of inertia are

  1. \(\dfrac{M\left(R_1^2-R_2^2\right)}{2},\dfrac{M\left(R_1^2-R_2^2\right)}{4},\dfrac{M\left(R_1^2-R_2^2\right)}{4}\)

  2. \(\dfrac{M\left(R_1^2+R_2^2\right)}{2},\dfrac{M\left(R_1^2+R_2^2\right)}{4},\dfrac{M\left(R_1^2+R_2^2\right)}{4}\)

  3. \(\dfrac{M\left(R_1^2+R_2^2\right)}{2},\dfrac{M\left(R_1^2+R_2^2\right)}{4},\dfrac{M\left(R_1^2+R_2^2\right)}{8}\)

  4. \(\dfrac{M\left(R_1^2-R_2^2\right)}{2},\dfrac{M\left(R_1^2-R_2^2\right)}{4},\dfrac{M\left(R_1^2-R_2^2\right)}{8}\)

Show solution

The surface density is \(\sigma=\dfrac{M}{\pi\left(R_1^2-R_2^2\right)}\). About the symmetry axis, \[ I_{zz}=\int r^2\,dm=\frac{M}{\pi\left(R_1^2-R_2^2\right)}\int_{R_2}^{R_1}2\pi r\cdot r^2\,dr=\frac{M}{\pi\left(R_1^2-R_2^2\right)}\cdot\frac{\pi\left(R_1^4-R_2^4\right)}{2}=\frac{M\left(R_1^2+R_2^2\right)}{2}. \] For a plane lamina \(I_{xx}+I_{yy}=I_{zz}\), and by symmetry \(I_{xx}=I_{yy}\), so \[ \boxed{I_{zz}=\frac{M\left(R_1^2+R_2^2\right)}{2},\quad I_{xx}=I_{yy}=\frac{M\left(R_1^2+R_2^2\right)}{4}} \]

Answer: (B)

CM05.14IIT JAM 2017, Q12MCQ | 2 marks

Consider a uniform thin circular disk of radius \(R\) and mass \(M\). A concentric square of side \(R/2\) is cut out from the disk (see figure). What is the moment of inertia of the resultant disk about an axis passing through the centre of the disk and perpendicular to it?Figure

  1. \(I=\dfrac{MR^2}{4}\left[1-\dfrac{1}{48\pi}\right]\)

  2. \(I=\dfrac{MR^2}{2}\left[1-\dfrac{1}{48\pi}\right]\)

  3. \(I=\dfrac{MR^2}{4}\left[1-\dfrac{1}{24\pi}\right]\)

  4. \(I=\dfrac{MR^2}{2}\left[1-\dfrac{1}{24\pi}\right]\)

Show solution

Surface density \(\sigma=M/(\pi R^2)\). The square of side \(a=R/2\) has mass \[ m=\sigma a^2=\frac{M}{\pi R^2}\cdot\frac{R^2}{4}=\frac{M}{4\pi}. \] The moment of inertia of a square plate about the perpendicular axis through its centre is \(m(a^2+a^2)/12=ma^2/6\): \[ I_{sq}=\frac{M}{4\pi}\cdot\frac{R^2}{4}\cdot\frac16=\frac{MR^2}{96\pi}. \] Subtracting from the full disk, \[ I=\frac{MR^2}{2}-\frac{MR^2}{96\pi}=\boxed{\frac{MR^2}{2}\left[1-\frac{1}{48\pi}\right]} \]

Answer: (B) \(\dfrac{MR^2}{2}\left[1-\dfrac{1}{48\pi}\right]\)

CM05.15IIT JAM 2017, Q16MCQ | 2 marks

A uniform rigid meter-scale is held horizontally with one of its end at the edge of a table and the other supported by hand. Some coins of negligible mass are kept on the meter scale as shown in the figure.FigureAs the hand supporting the scale is removed, the scale starts rotating about its edge on the table and the coins start moving. If a photograph of the rotating scale is taken soon after, it will look closest to:Figure

Show solution

For a uniform rod of length \(L\) pivoted at one end, just after release \[ \tau=I\alpha\;\Rightarrow\;Mg\frac{L}{2}=\frac{ML^2}{3}\alpha\;\Rightarrow\;\alpha=\frac{3g}{2L}. \] A point of the scale at distance \(x\) from the pivot has downward acceleration \[ a(x)=\alpha x=\frac{3g}{2L}\,x . \] A free coin can fall with acceleration at most \(g\). So

  • for \(x<\dfrac{2L}{3}\): \(a(x)<g\), the coins stay in contact and move down with the scale;

  • for \(x>\dfrac{2L}{3}\): \(a(x)>g\), the scale falls away from the coins, which then fall freely with acceleration \(g\).

All the freely falling coins drop the same distance \(\tfrac12gt^2\), so they stay on a horizontal line above the scale. \[ \boxed{\text{coins lie on the scale up to }2L/3,\ \text{then on a horizontal line above it}} \] This is picture (B).

Answer: (B)

CM05.16IIT JAM 2017, Q37MSQ | 2 marks

The linear mass density of a rod of length \(L\) varies from one end to the other as \(\lambda_0\left(1+\dfrac{x^2}{L^2}\right)\), where \(x\) is the distance from one end with tensions \(T_1\) and \(T_2\) in them (see figure), and \(\lambda_0\) is a constant. The rod is suspended from a ceiling by two massless strings. Then, which of the following statement(s) is (are) correct?Figure

  1. The mass of the rod is \(\dfrac{2\lambda_0L}{3}\)

  2. The centre of gravity of the rod is located at \(x=\dfrac{9L}{16}\)

  3. The tension \(T_1\) in the left string is \(\dfrac{7\lambda_0Lg}{12}\)

  4. The tension \(T_2\) in the right string is \(\dfrac{3\lambda_0Lg}{2}\)

Show solution

\[ M=\int_0^L\lambda_0\left(1+\frac{x^2}{L^2}\right)dx=\lambda_0\left(L+\frac L3\right)=\frac{4\lambda_0L}{3}. \] (A) gives \(\tfrac{2}{3}\lambda_0L\). Incorrect. \[ \int_0^Lx\,\lambda\,dx=\lambda_0\left(\frac{L^2}{2}+\frac{L^2}{4}\right)=\frac{3\lambda_0L^2}{4} \;\Rightarrow\;x_{cg}=\frac{3\lambda_0L^2/4}{4\lambda_0L/3}=\frac{9L}{16}. \] (B) Correct.

Torques about \(x=0\) and vertical equilibrium: \[ T_2L=Mg\,x_{cg}\;\Rightarrow\;T_2=\frac{9}{16}\cdot\frac{4\lambda_0Lg}{3}=\frac{3\lambda_0Lg}{4},\qquad T_1=Mg-T_2=\frac{4\lambda_0Lg}{3}-\frac{3\lambda_0Lg}{4}=\frac{7\lambda_0Lg}{12}. \] (C) Correct. (D) gives \(\tfrac32\lambda_0Lg\) instead of \(\tfrac34\lambda_0Lg\). Incorrect. \[ \boxed{(B),\ (C)} \]

Answer: (B), (C)

CM05.17IIT JAM 2016, Q37

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