Kinematics and Dynamics of Rotational Motion — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Kinematics and Dynamics of Rotational Motion MCQs with step-by-step solutions (34 questions). Part of System of Particles and Rotational Motion. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Kinematics and Dynamics of Rotational Motion · medium · numerical
The angular speed of the wheel of a vehicle is increased from 360 rpm to 1200 rpm in 14 s. Its angular acceleration is
A. $2\pi$ rad/s$^2$ ✓ Correct
B. $28\pi$ rad/s$^2$
C. $120\pi$ rad/s$^2$
D. $1$ rad/s$^2$
Solution: Initial angular speed, $\omega_0=2\pi f_0=2\pi\times\dfrac{360}{60}=12\pi$ rad/s. Final angular speed, $\omega=2\pi f=2\pi\times\dfrac{1200}{60}=40\pi$ rad/s, with $t=14$ s. From $\omega=\omega_0+\alpha t$, $\alpha=\dfrac{\omega-\omega_0}{t}=\dfrac{40\pi-12\pi}{14}=\dfrac{28\pi}{14}=2\pi$ rad/s$^2$.
Q2 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A particle starting from rest, moves in a circle of radius $r$. It attains a velocity of $v_0$ m/s in the $n^{th}$ round. Its angular acceleration will be
A. $\dfrac{v_0}{n}$ rad/s$^2$
B. $\dfrac{v_0^2}{2\pi nr^2}$ rad/s$^2$
C. $\dfrac{v_0^2}{4\pi nr^2}$ rad/s$^2$ ✓ Correct
D. $\dfrac{v_0^2}{4\pi nr}$ rad/s$^2$
Solution: Using $\omega^2-\omega_0^2=2\alpha\theta$ with $\omega=\dfrac{v_0}{r}$, $\omega_0=0$ (starts from rest) and $\theta=2\pi n$ (for $n$ rounds): $\left(\dfrac{v_0}{r}\right)^2-0=2\alpha(2\pi n)\Rightarrow \alpha=\dfrac{v_0^2}{4\pi nr^2}$ rad/s$^2$.
Q3 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination $30^\circ$. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the inclined surface will be : (Take $g=10$ m/s$^2$)
A. $2.2$ m
B. $1.6$ m
C. $1.2$ m
D. $2.4$ m ✓ Correct
Solution: For a rolling body the total kinetic energy is $KE=\dfrac{1}{2}mv^2\left(1+\dfrac{k^2}{R^2}\right)$. By conservation of energy $KE=PE$, i.e. $\dfrac{1}{2}mv^2\left(1+\dfrac{k^2}{R^2}\right)=mgh$. For a cylinder $\dfrac{k^2}{R^2}=\dfrac{1}{2}$, so $\dfrac{1}{2}\times2\times16\left(1+\dfrac{1}{2}\right)=2\times10\times h\Rightarrow 8\times\dfrac{3}{2}=10h\Rightarrow h=1.2$ m. From the geometry $\sin\theta=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sin\theta}=\dfrac{1.2}{\sin30^\circ}=1.2\times2=2.4$ m.
Q4 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?
A. $30$ kJ
B. $2$ J
C. $1$ J
D. $3$ J ✓ Correct
Solution: Given $R=2$ m, $m=100$ kg and $v_{CM}=20$ cm/s $=20\times10^{-2}$ m/s. By the work-energy theorem $W=KE_f-KE_i$; the disc finally stops so $KE_f=0$. For a rolling disc $I=\dfrac{1}{2}mR^2$, so $KE_i=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2=\dfrac{1}{4}mR^2\dfrac{v^2}{R^2}+\dfrac{1}{2}mv^2=\dfrac{3}{4}mv^2$. Thus $W=\dfrac{3}{4}mv^2=\dfrac{3}{4}\times100\times(20\times10^{-2})^2=\dfrac{3}{4}\times400\times100\times10^{-4}=3$ J.
Q5 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after $2\pi$ revolutions is
A. $2\times10^{-3}$ N-m
B. $12\times10^{-4}$ N-m
C. $2\times10^{6}$ N-m
D. $2\times10^{-6}$ N-m ✓ Correct
Solution: Given $m=2$ kg, $r=4$ cm $=4\times10^{-2}$ m, $\omega=3$ rpm $=3\times\dfrac{2\pi}{60}=\dfrac{\pi}{10}$ rad/s and $\theta=2\pi$ revolution $=2\pi\times2\pi=4\pi^2$ rad. By the work-energy theorem $-\tau\theta=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$ with $\omega_f=0$, giving $\tau=\dfrac{I\omega_i^2}{2\theta}=\dfrac{1}{4}mr^2\dfrac{\omega_i^2}{\theta}$ (using $I=\dfrac{1}{2}mr^2$). $\tau=\dfrac{1}{4}\times2\times(4\times10^{-2})^2\times\left(\dfrac{\pi}{10}\right)^2\times\dfrac{1}{4\pi^2}=\dfrac{2}{100}\times10^{-4}=2\times10^{-6}$ N-m.
Q6 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Three objects, $A$ (a solid sphere), $B$ (a thin circular disk) and $C$ (a circular ring), each have the same mass $M$ and radius $R$. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest, would satisfy the relation
A. $W_B > W_A > W_C$
B. $W_A > W_B > W_C$
C. $W_C > W_B > W_A$ ✓ Correct
D. $W_A > W_C > W_B$
Solution: Work done to bring a spinning object to rest is $W=\dfrac{1}{2}I\omega^2$, so with the same $\omega$, $W\propto I$. Here $I_A=\dfrac{2}{5}MR^2$ (solid sphere), $I_B=\dfrac{1}{2}MR^2$ (disk), $I_C=MR^2$ (ring). Thus $I_A:I_B:I_C=\dfrac{2}{5}:\dfrac{1}{2}:1=4:5:10$, so $W_A<W_B<W_C$, i.e. $W_C > W_B > W_A$.
Q7 — Kinematics and Dynamics of Rotational Motion · medium · theory
A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
A. Rotational kinetic energy
B. Moment of inertia
C. Angular velocity
D. Angular momentum ✓ Correct
Solution: In free space no external torque acts, so $\tau_{ext}=\dfrac{dL}{dt}=0\Rightarrow L=$ constant. When the radius is increased at constant mass, the moment of inertia $\left(I=\dfrac{2}{5}MR^2\right)$, the angular velocity and the rotational kinetic energy all change; only the angular momentum remains constant.
Q8 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t:(K_t+K_r)$ for the sphere is
A. $10:7$
B. $5:7$ ✓ Correct
C. $7:10$
D. $2:5$
Solution: Translational KE $K_t=\dfrac{1}{2}mv_{CM}^2$. Total KE $=K_t+K_r=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv_{CM}^2$. For a solid sphere $I=\dfrac{2}{5}MR^2$ and $v_{CM}=R\omega$, so $K_t+K_r=\dfrac{1}{5}mv_{CM}^2+\dfrac{1}{2}mv_{CM}^2=\dfrac{7}{10}mv_{CM}^2$. $\therefore \dfrac{K_t}{K_t+K_r}=\dfrac{\frac{1}{2}mv_{CM}^2}{\frac{7}{10}mv_{CM}^2}=\dfrac{5}{7}$, i.e. $5:7$.
Q9 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
A. $\dfrac{1}{2}I(\omega_1+\omega_2)^2$
B. $\dfrac{1}{4}I(\omega_1-\omega_2)^2$ ✓ Correct
C. $I(\omega_1-\omega_2)^2$
D. $\dfrac{I}{8}(\omega_1-\omega_2)^2$
Solution: By conservation of angular momentum, $I\omega_1+I\omega_2=(2I)\omega\Rightarrow\omega=\dfrac{\omega_1+\omega_2}{2}$. Loss of energy $=\dfrac{1}{2}I\omega_1^2+\dfrac{1}{2}I\omega_2^2-\dfrac{1}{2}(2I)\omega^2=\dfrac{1}{4}I(\omega_1-\omega_2)^2$.
Q10 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B\,(I_B>I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then
A. $L_A=\dfrac{L_B}{2}$
B. $L_A=2L_B$
C. $L_B>L_A$ ✓ Correct
D. $L_A>L_B$
Solution: Kinetic energy of rotation $KE=\dfrac{1}{2}I\omega^2=\dfrac{L^2}{2I}$. Equal KE gives $\dfrac{L_A^2}{2I_A}=\dfrac{L_B^2}{2I_B}$, so $L\propto\sqrt{I}$. Since $I_B>I_A$, it follows that $L_B>L_A$.
Q11 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?
A. Sphere ✓ Correct
B. Both reach at the same time
C. Depends on their masses
D. Disc
Solution: Acceleration of a body rolling down an incline is $a=\dfrac{g\sin\theta}{1+I/mr^2}$. For a disc $\dfrac{I}{mr^2}=\dfrac{1}{2}\Rightarrow a_{disc}=\dfrac{2}{3}g\sin\theta$; for a solid sphere $\dfrac{I}{mr^2}=\dfrac{2}{5}\Rightarrow a_{sphere}=\dfrac{5}{7}g\sin\theta$. Since $a_{sphere}>a_{disc}$, the sphere reaches the bottom first.
Q12 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A solid sphere of mass $m$ and radius $R$ is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation $(E_{sphere}/E_{cylinder})$ will be
A. $2:3$
B. $1:5$ ✓ Correct
C. $1:4$
D. $3:1$
Solution: $KE=\dfrac{1}{2}I\omega^2$. Sphere: $K_S=\dfrac{1}{2}\cdot\dfrac{2}{5}mR^2\omega_1^2=\dfrac{1}{5}mR^2\omega_1^2$. Cylinder with $\omega_2=2\omega_1$: $K_C=\dfrac{1}{2}\cdot\dfrac{1}{2}mR^2\omega_2^2=\dfrac{1}{4}mR^2(2\omega_1)^2=mR^2\omega_1^2$. Thus $\dfrac{K_S}{K_C}=\dfrac{1/5}{1}=\dfrac{1}{5}$, i.e. $1:5$.
Q13 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$ and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $P$ on this rod through which the axis should pass, so that the work required to set the rod rotating with angular velocity $\omega_0$ is minimum, is given by
A. $x=\dfrac{m_1 L}{m_1+m_2}$
B. $x=\dfrac{m_1}{m_2}L$
C. $x=\dfrac{m_2}{m_1}L$
D. $x=\dfrac{m_2 L}{m_1+m_2}$ ✓ Correct
Solution: Work $=\dfrac{1}{2}I\omega_0^2$ is minimum when $I=m_1x^2+m_2(L-x)^2$ is minimum. Setting $\dfrac{dI}{dx}=2m_1x-2m_2(L-x)=0$ gives $x(2m_1+2m_2)=2m_2L$, hence $x=\dfrac{m_2 L}{m_1+m_2}$, i.e. $P$ is the centre of mass.
Q14 — Kinematics and Dynamics of Rotational Motion · medium · numerical
The ratio of the accelerations for a solid sphere (mass $m$ and radius $R$) rolling down an incline of angle $\theta$ without slipping and slipping down the incline without rolling is
A. $5:7$ ✓ Correct
B. $2:3$
C. $2:5$
D. $7:5$
Solution: For a solid sphere rolling without slipping, $a_1=\dfrac{g\sin\theta}{1+\dfrac{k^2}{R^2}}=\dfrac{g\sin\theta}{1+\dfrac{2}{5}}=\dfrac{5}{7}g\sin\theta$ (since $k^2=\dfrac{2}{5}R^2$). For a sphere slipping without rolling, $a_2=g\sin\theta$. Hence $\dfrac{a_1}{a_2}=\dfrac{5}{7}$.
Q15 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A small object of uniform density rolls up a curved surface with an initial velocity $v'$. It reaches upto a maximum height of $\dfrac{3v^2}{4g}$ with respect to the initial position. The object is
A. ring
B. solid sphere
C. hollow sphere
D. disc ✓ Correct
Solution: From energy conservation $v=\sqrt{\dfrac{2gh}{1+\dfrac{k^2}{r^2}}}$. Given $h=\dfrac{3v^2}{4g}$, so $v^2=\dfrac{2g\left(\frac{3v^2}{4g}\right)}{1+\frac{k^2}{r^2}}=\dfrac{\frac{3v^2}{2}}{1+\frac{k^2}{r^2}}$. Thus $1+\dfrac{k^2}{r^2}=\dfrac{3}{2}\Rightarrow\dfrac{k^2}{r^2}=\dfrac{1}{2}$, i.e. $k^2=\dfrac{1}{2}r^2$, which is a disc.
Q16 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A circular disc of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disc of moment of inertia $I_b$ is dropped coaxially onto the rotating disc. Initially the second disk has zero angular speed. Eventually both the discs rotate with a constant angular speed $\omega_f$. The energy lost by initially rotating disc due to friction is
A. $\dfrac{1}{2}\dfrac{I_b^2}{(I_t+I_b)}\omega_i^2$
B. $\dfrac{1}{2}\dfrac{I_t^2}{(I_t+I_b)}\omega_i^2$
C. $\dfrac{1}{2}\dfrac{(I_b-I_t)}{(I_t+I_b)}\omega_i^2$
D. $\dfrac{1}{2}\dfrac{I_b I_t}{(I_t+I_b)}\omega_i^2$ ✓ Correct
Solution: By conservation of angular momentum $\omega_f=\dfrac{I_t\omega_i}{I_t+I_b}$. Loss of energy $\Delta E=\dfrac{1}{2}I_t\omega_i^2-\dfrac{1}{2}\dfrac{I_t^2}{(I_t+I_b)}\omega_i^2=\dfrac{1}{2}\dfrac{I_b I_t}{(I_t+I_b)}\omega_i^2$.
Q17 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A wheel has angular acceleration of $3\ \text{rad/s}^2$ and an initial angular speed of $2\ \text{rad/s}$. In a time of $2\ \text{s}$, it has rotated through an angle (in radian) of
A. $6$
B. $10$ ✓ Correct
C. $12$
D. $4$
Solution: Using $\theta=\omega_0 t+\dfrac{1}{2}\alpha t^2$ with $\omega_0=2$ rad/s, $\alpha=3\ \text{rad/s}^2$ and $t=2$ s: $\theta=2\times2+\dfrac{1}{2}\times3\times2^2=4+6=10$ rad.
Q18 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Two bodies have their moments of inertia $I$ and $2I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio
A. $1:2$
B. $\sqrt{2}:1$
C. $2:1$
D. $1:\sqrt{2}$ ✓ Correct
Solution: Rotational $KE=\dfrac{L^2}{2I}$. Equal KE gives $\dfrac{L_1^2}{I_1}=\dfrac{L_2^2}{I_2}\Rightarrow\dfrac{L_1}{L_2}=\sqrt{\dfrac{I_1}{I_2}}=\sqrt{\dfrac{I}{2I}}=\dfrac{1}{\sqrt{2}}$, i.e. $L_1:L_2=1:\sqrt{2}$.
Q19 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $k$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be
A. $\dfrac{k^2}{k^2+R^2}$ ✓ Correct
B. $\dfrac{R^2}{k^2+R^2}$
C. $\dfrac{k^2+R^2}{R^2}$
D. $\dfrac{k^2}{R^2}$
Solution: $K_{rot}=\dfrac{1}{2}Mk^2\dfrac{v^2}{R^2}$ and total energy $E=\dfrac{1}{2}\dfrac{Mv^2}{R^2}(k^2+R^2)$. Hence $\dfrac{K_{rot}}{E}=\dfrac{k^2}{k^2+R^2}$.
Q20 — Kinematics and Dynamics of Rotational Motion · medium · theory
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is $v_{CM}$, then true statement is
A. The velocity of point $A$ is $2v_{CM}$ and velocity of point $B$ is zero ✓ Correct
B. The velocity of point $A$ is zero and velocity of point $B$ is $2v_{CM}$
C. The velocity of point $A$ is $2v_{CM}$ and velocity of point $B$ is $-v_{CM}$
D. The velocities of both $A$ and $B$ are $v_{CM}$
Solution: For rolling without slipping the top point $A$ has $v_A=v_{CM}+R\omega=v_{CM}+v_{CM}=2v_{CM}$, while the bottom contact point $B$ has $v_B=v_{CM}-R\omega=v_{CM}-v_{CM}=0$.
Q21 — Kinematics and Dynamics of Rotational Motion · medium · numerical
If a flywheel makes 120 rev/min, then its angular speed will be
A. $8\pi$ rad/s
B. $6\pi$ rad/s
C. $4\pi$ rad/s ✓ Correct
D. $2\pi$ rad/s
Solution: $\omega=2\pi\nu=\dfrac{2\pi\times120}{60}=4\pi$ rad/s.
Q22 — Kinematics and Dynamics of Rotational Motion · medium · numerical
The angular speed of an engine wheel making 90 rev/min is
A. $1.5\pi$ rad/s
B. $3\pi$ rad/s ✓ Correct
C. $4.5\pi$ rad/s
D. $6\pi$ rad/s
Solution: $\omega=\dfrac{2\pi}{T}=2\pi\nu=\dfrac{2\pi\times90}{60}=3\pi$ rad/s.
Q23 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A spherical ball rolls on a table without slipping. Then, the fraction of its total energy associated with rotation is
A. $\dfrac{2}{5}$
B. $\dfrac{2}{7}$ ✓ Correct
C. $\dfrac{3}{5}$
D. $\dfrac{3}{7}$
Solution: For a sphere $I=\dfrac{2}{5}mr^2$, so total $K=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2=\dfrac{7}{10}mv^2$ and rotational $K_r=\dfrac{1}{5}mv^2$. Thus $\dfrac{K_r}{K}=\dfrac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2}=\dfrac{2}{7}$.
Q24 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A thin uniform circular ring is rolling down an inclined plane of inclination $30^\circ$ without slipping. Its linear acceleration along the inclined plane will be
A. $\dfrac{g}{2}$
B. $\dfrac{g}{3}$
C. $\dfrac{g}{4}$ ✓ Correct
D. $\dfrac{2g}{3}$
Solution: For a ring $I=MR^2$, so $a=\dfrac{g\sin\theta}{1+\dfrac{I}{MR^2}}=\dfrac{g\sin30^\circ}{1+1}=\dfrac{g/2}{2}=\dfrac{g}{4}$.
Q25 — Kinematics and Dynamics of Rotational Motion · medium · theory
A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then
A. solid sphere reaches the bottom first ✓ Correct
B. solid sphere reaches the bottom last
C. disc will reach the bottom first
D. all reach the bottom at the same time
Solution: Acceleration while rolling is $a=\dfrac{g\sin\theta}{1+\dfrac{k^2}{R^2}}$. For a solid sphere $\dfrac{k^2}{R^2}=\dfrac{2}{5}$ giving $a=\dfrac{5}{7}g\sin\theta$, while for the disc and solid cylinder $\dfrac{k^2}{R^2}=\dfrac{1}{2}$ giving $a=\dfrac{2}{3}g\sin\theta$. The solid sphere has the largest acceleration, so it reaches the bottom first.
Q26 — Kinematics and Dynamics of Rotational Motion · medium · numerical
The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$ from rest without sliding is
A. $\sqrt{\dfrac{10}{7}gh}$ ✓ Correct
B. $\sqrt{gh}$
C. $\sqrt{\dfrac{6}{5}gh}$
D. $\sqrt{\dfrac{4}{3}gh}$
Solution: For a rolling sphere the total kinetic energy is $\dfrac{7}{10}mv^2$. On reaching the bottom, PE = total KE: $mgh=\dfrac{7}{10}mv^2\Rightarrow v=\sqrt{\dfrac{10gh}{7}}$.
Q27 — Kinematics and Dynamics of Rotational Motion · medium · numerical
If a sphere is rolling, the ratio of the translational energy to total kinetic energy is given by
A. $7:10$
B. $2:5$
C. $10:7$
D. $5:7$ ✓ Correct
Solution: $K=K_{rot}+K_{trans}=\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2$. With $I=\dfrac{2}{5}mr^2$ and $v=r\omega$, $K=\dfrac{1}{5}mv^2+\dfrac{1}{2}mv^2=\dfrac{7}{10}mv^2$. Translational KE $=\dfrac{1}{2}mv^2$, so the ratio $=\dfrac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2}=\dfrac{5}{7}$, i.e. $5:7$.
Q28 — Kinematics and Dynamics of Rotational Motion · medium · numerical
The moment of inertia of a body about a given axis is $1.2$ kg-m$^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500$ J, an angular acceleration of $25$ rad/s$^2$ must be applied about that axis for a duration of
A. $4$ s
B. $2$ s ✓ Correct
C. $8$ s
D. $10$ s
Solution: $K_{rot}=\dfrac{1}{2}I\omega^2\Rightarrow \omega=\sqrt{\dfrac{2K_r}{I}}=\sqrt{\dfrac{2\times1500}{1.2}}=50$ rad/s. From $\omega=\omega_0+\alpha t$, $t=\dfrac{\omega-\omega_0}{\alpha}=\dfrac{50-0}{25}=2$ s.
Q29 — Kinematics and Dynamics of Rotational Motion · medium · numerical
Moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be
A. $5I$
B. $3I$
C. $6I$ ✓ Correct
D. $4I$
Solution: MOI about a diameter $=I$. By the perpendicular axes theorem, MOI about the axis through the centre perpendicular to the plane $=2I$. By the parallel axes theorem, MOI about a parallel axis through the rim $=2I+mr^2$. Since $2I=\dfrac{1}{2}mr^2\Rightarrow mr^2=4I$, so it $=2I+4I=6I$.
Q30 — Kinematics and Dynamics of Rotational Motion · medium · numerical
A flywheel rotating about a fixed axis has a kinetic energy of $360$ J when its angular speed is $30$ rad/s. The moment of inertia of the wheel about the axis of rotation is
A. $0.6$ kg-m$^2$
B. $0.15$ kg-m$^2$
C. $0.8$ kg-m$^2$ ✓ Correct
D. $0.75$ kg-m$^2$
Solution: Rotational kinetic energy $K_{rot}=\dfrac{1}{2}I\omega^2\Rightarrow I=\dfrac{2K_r}{\omega^2}=\dfrac{2\times360}{(30)^2}=0.8$ kg-m$^2$.