Rotational Motion
25 Board Physics previous year questions on Rotational Motion — options free on every question; 2 include the answer & explanation free, the rest unlock with PYQ Pass.
Torque on a uniform disk of mass \(2\text{ }\text{kg}\), radius \(1\text{ }\text{m}\), is given as \(\tau (t)=5{t}^{2}−8t\). If the disk was initially at rest, find the power by torque at \(t=1\text{ }\text{s}\).
7 W
(3)
\(\begin{aligned}& \rho=\vec{\tau} \cdot \vec{\omega} \\& \tau=l \alpha \\& I=\frac{M R^2}{2}=1 \\& \alpha=5 t^2-8 t \\& \omega=\frac{5}{3} t^3-4 t^2 \\& \text { At } t=1, \omega=\frac{5}{3}-4=-\frac{7}{3} \\& t(3)=5-8=-3 \\& \rho=\vec{\tau} \cdot \vec{\omega}=7 \mathrm{~W}\end{aligned}\)
Two identical masses, each \(1\text{ }\text{kg}\), having velocity vectors:\(\vec{{\mathrm{V}}_{\mathrm{A}}}={\alpha }_{1}{\mathrm{t}}^{2}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}},\) \(\vec{{\mathrm{V}}_{\mathrm{B}}}={\alpha }_{1}\hat{\mathrm{i}}+{\alpha }_{2}\hat{\mathrm{j}}+{\alpha }_{3}{\mathrm{t}}^{2}\hat{\mathrm{k}},\)where \({\alpha }_{1}=2,\text{ }{\alpha }_{2}=3n,\text{ }{\alpha }_{3}=4p\), and \(n,p\)are constants.
At \(t=1\text{ }\text{s}\), the velocities of \(A\) and \(B\) are orthogonal to each other and the magnitudes of their velocities are equal: \(\left|\vec{{\mathrm{V}}_{\mathrm{A}}}\right|=\left|\vec{{\mathrm{V}}_{\mathrm{B}}}\right|\).
Find the relative displacement between \(A\) and \(B\) at \(t=1\text{ }\text{s}\), and calculate the angular momentum of \(A\) with respect to \(B\).
0
\(\vec{{\mathrm{V}}_{\mathrm{A}}}={\alpha }_{1}{\mathrm{t}}^{2}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}},\) \(\vec{{\mathrm{V}}_{\mathrm{B}}}={\alpha }_{1}\hat{\mathrm{i}}+{\alpha }_{2}\hat{\mathrm{j}}+{\alpha }_{3}{\mathrm{t}}^{2}\hat{\mathrm{k}},\) \({\alpha }_{1}=2,\text{ }{\alpha }_{2}=3n,\text{ }{\alpha }_{3}=4p\)
\(\mathrm{For}\mathrm{the}\mathrm{velocities}\mathrm{to}\mathrm{be}\mathrm{orthogonal}:\\ \vec{{\mathrm{V}}_{\mathrm{A}}}.\vec{{\mathrm{V}}_{\mathrm{B}}}=0\\ ({\alpha }_{1}{\mathrm{t}}^{2}){\alpha }_{1}+({\alpha }_{2}\mathrm{t}){\alpha }_{2}+({\alpha }_{3})({\alpha }_{3}{\mathrm{t}}^{2})=0\\ \mathrm{At},\mathrm{t}=1\\ {\alpha }_{1}^{2}+{\alpha }_{2}^{2}+{\alpha }_{3}^{2}=0\\ \mathrm{Substitute}{\alpha }_{1}=2,{\alpha }_{2}=3\mathrm{n},{\alpha }_{3}=4\mathrm{p}\\ {2}^{2}+(3\mathrm{n}{)}^{2}+(4\mathrm{p}{)}^{2}=0\\ 4+9{\mathrm{n}}^{2}+16{\mathrm{p}}^{2}=0....(1)\)
\(\vec{{\mathrm{r}}_{\mathrm{A}}}=\int \vec{{\mathrm{V}}_{\mathrm{A}}}\mathrm{dt}=\int ({\alpha }_{1}{\mathrm{t}}^{2}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}})\mathrm{dt}={\alpha }_{1}\frac{{\mathrm{t}}^{3}}{3}\hat{\mathrm{i}}+{\alpha }_{2}\frac{{\mathrm{t}}^{2}}{2}\hat{\mathrm{j}}+{\alpha }_{3}\mathrm{t}\hat{\mathrm{k}}\\ \vec{{\mathrm{r}}_{\mathrm{B}}}=\int \vec{{\mathrm{V}}_{\mathrm{B}}}\mathrm{dt}=\int ({\alpha }_{1}\hat{\mathrm{i}}+{\alpha }_{2}\hat{\mathrm{j}}+{\alpha }_{3}{\mathrm{t}}^{2}\hat{\mathrm{k}})\mathrm{dt}={\alpha }_{1}\mathrm{t}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\frac{{\mathrm{t}}^{3}}{3}\hat{\mathrm{k}}\\ \mathrm{At}\mathrm{t}=1\mathrm{s}\\ \vec{{\mathrm{r}}_{\mathrm{A}}}=\frac{{\alpha }_{1}}{3}\hat{\mathrm{i}}+\frac{{\alpha }_{2}}{2}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}}\\ \vec{{\mathrm{r}}_{\mathrm{B}}}={\alpha }_{1}\hat{\mathrm{i}}+{\alpha }_{2}\hat{\mathrm{j}}+\frac{{\alpha }_{3}}{3}\hat{\mathrm{k}}\)
\(\mathrm{Relative}\mathrm{displacement}:\\ r{⃗}_{AB}=r{⃗}_{A}-r{⃗}_{B}=\left(\frac{{\alpha }_{1}}{3}-{\alpha }_{1}\right)\hat{\mathrm{i}}+\left(\frac{{\alpha }_{2}}{2}-{\alpha }_{2}\right)\hat{\mathrm{j}}+\left({\alpha }_{3}-\frac{{\alpha }_{3}}{3}\right)\hat{\mathrm{k}}\\ r{⃗}_{AB}=\left(-\frac{2{\alpha }_{1}}{3}\right)\hat{\mathrm{i}}+\left(-\frac{{\alpha }_{2}}{2}\right)\hat{\mathrm{j}}+\left(\frac{2{\alpha }_{3}}{3}\right)\hat{\mathrm{k}}\\ \mathrm{Substitute}{\alpha }_{1}=2,{\alpha }_{2}=3\mathrm{n},{\alpha }_{3}=4\mathrm{p}\\ r{⃗}_{AB}=\left(-\frac{4}{3}\right)\hat{\mathrm{i}}+\left(-\frac{3\mathrm{n}}{2}\right)\hat{\mathrm{j}}+\left(\frac{8\mathrm{p}}{3}\right)\hat{\mathrm{k}}\)
Relative velocity is given by
\(\vec{{\mathrm{V}}_{\mathrm{AB}}}=\vec{{\mathrm{V}}_{\mathrm{A}}}-\vec{{\mathrm{V}}_{\mathrm{B}}}=\\ \vec{{\mathrm{V}}_{\mathrm{AB}}}={\alpha }_{1}{\mathrm{t}}^{2}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}}-{\alpha }_{1}\hat{\mathrm{i}}-{\alpha }_{2}\hat{\mathrm{j}}-{\alpha }_{3}{\mathrm{t}}^{2}\hat{\mathrm{k}}\\ \mathrm{At}\mathrm{t}=1\mathrm{s}\\ \vec{{\mathrm{V}}_{\mathrm{AB}}}=0\hat{\mathrm{i}}+0\hat{\mathrm{j}}+0\hat{\mathrm{k}}\)
The angular momentum of A with respect to B s givne by
\(\vec{\mathrm{L}}=\mathrm{m}\left(\vec{{\mathrm{r}}_{\mathrm{AB}}}\times \vec{{\mathrm{V}}_{\mathrm{AB}}}\right)=\vec{0}\mathrm{As}\vec{{\mathrm{V}}_{\mathrm{AB}}}=0\)
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
[JEE Main 2025, 24 Jan (Shift 2)]
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A solid sphere rolls without slipping on a horizontal plane. What is ratio of translational kinetic energy to the rotational kinetic energy of the sphere.
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A body of mass \(m\) is projected with an initial velocity \({v}_{0}\) at an angle of \(4{5}^{∘}\) to the horizontal in the \(X−Y\) plane. Find the angular momentum of the body at the highest point with respect to the point of projection.
(Shift I Memory Based)
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A disc of mass \(M\) and radius \(R\) is rotating about its axis. If the angle rotated about it as a function of time \(t\)t is \(\theta =a{t}^{2}+bt+c\), where \(a,b,\) and \(c\) are constants, find the power derived to the disc as a function of time.
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A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
[JEE Main 2025, 24 Jan (Shift 2)]
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The torque due to the force \((2\hat{i}+\hat{j}+2\hat{k})\) about the origin, acting on a particle whose position vector is \((\hat{i}+\hat{j}+\hat{k})\), would be
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A solid sphere and hollow sphere rolls down purely equal distances on same inclined plane (starting from rest) in time \(t_1\) and \(t_2\) then
(Shift - II Memory based)
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A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
[JEE Main 2025, 24 Jan (Shift 2)]
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A solid sphere of mass ' \(m\) ' and radius ' \(r\) ' is allowed to roll without slipping from the highest point of an inclined plane of length ' \(L\) ' and makes an angle \(30^{\circ}\) with the horizontal. The speed of the particle at the bottom of the plane is \(v_1\). If the angle of inclination is increased to \(45^\circ\) while keeping \(L\) constant. Then the new speed of the sphere at the bottom of the plane is \(v_2\). The ratio\({v}_{1}^{2}:{v}_{2}^{2}\) is
[JEE Main 2025, 23 Jan (Shift 1)]
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A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be \({t}_{1}\) and \({t}_{2}\), respectively, then
[JEE Main 2025, 24 Jan (Shift 2)]
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Torque on a uniform disk of mass \(2\text{ }\text{kg}\), radius \(1\text{ }\text{m}\), is given as \(\tau (t)=5{t}^{2}−8t\). If the disk was initially at rest, find the power by torque at \(t=1\text{ }\text{s}\).
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A solid cylinder of mass \(m\) and radius \(r\) is released from rest at the top of a rough inclined plane making an angle of \(45^{\circ}\) with the horizontal. Assuming the cylinder rolls without slipping, find the acceleration of the axis of the cylinder.
(Shift I - Memory Based)
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Two uniform discs of radius \(R\) and \(2R\) have moments of inertia \({I}_{1}\) and \({I}_{2}\) respectively about their central axes. If both discs have the same surface mass density, determine the ratio \({I}_{1}\mathrm{/}{I}_{2}\).
(Shift - I Memory Based)
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A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that \(\theta (t)=5{t}^{2}-8t\), where \(\theta (t)\) is the angular position of the rotating disc as a function of time \(t\).
How much power is delivered by the applied torque, at \(t=2\mathrm{s}\)?
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A uniform solid cylinder of mass '𝑚' and radius '𝑟' rolls along an inclined rough plane of inclination \(45^\circ .\) If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be
[JEE Main 2025, 24 Jan (Shift 1)]
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A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
[JEE Main 2025, 24 Jan (Shift 2)]
Options are free to see. Unlock the correct answer and full explanation with Pass.
A solid cylinder of mass \(m\) and radius \(r\) is released from rest at the top of a rough inclined plane making an angle of \(45^{\circ}\) with the horizontal. Assuming the cylinder rolls without slipping, find the acceleration of the axis of the cylinder.
(Shift I - Memory Based)
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Two uniform discs of radius \(R\) and \(2R\) have moments of inertia \({I}_{1}\) and \({I}_{2}\) respectively about their central axes. If both discs have the same surface mass density, determine the ratio \({I}_{1}\mathrm{/}{I}_{2}\).
(Shift - I Memory Based)
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A force \(\vec{\mathrm{F}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+2\hat{\mathrm{k}}\text{ N}\) is acting at a point \((1,1,1)\). Find the torque of this force about the origin \((0,0,0)\).
(Shift II Memory Based)
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A solid sphere rolls without slipping on a horizontal plane. What is ratio of translational kinetic energy to the rotational kinetic energy of the sphere.
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The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is \(2400\mathrm{g}{\mathrm{cm}}^{2}\). The length of the 400 g rod is nearly :
[NEET 2024]
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Two identical masses, each \(1\text{ }\text{kg}\), having velocity vectors:\(\vec{{\mathrm{V}}_{\mathrm{A}}}={\alpha }_{1}{\mathrm{t}}^{2}\hat{\mathrm{i}}+{\alpha }_{2}\mathrm{t}\hat{\mathrm{j}}+{\alpha }_{3}\hat{\mathrm{k}},\) \(\vec{{\mathrm{V}}_{\mathrm{B}}}={\alpha }_{1}\hat{\mathrm{i}}+{\alpha }_{2}\hat{\mathrm{j}}+{\alpha }_{3}{\mathrm{t}}^{2}\hat{\mathrm{k}},\)where \({\alpha }_{1}=2,\text{ }{\alpha }_{2}=3n,\text{ }{\alpha }_{3}=4p\), and \(n,p\)are constants.
At \(t=1\text{ }\text{s}\), the velocities of \(A\) and \(B\) are orthogonal to each other and the magnitudes of their velocities are equal: \(\left|\vec{{\mathrm{V}}_{\mathrm{A}}}\right|=\left|\vec{{\mathrm{V}}_{\mathrm{B}}}\right|\).
Find the relative displacement between \(A\) and \(B\) at \(t=1\text{ }\text{s}\), and calculate the angular momentum of \(A\) with respect to \(B\).
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A force \( \vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) N \) acts at a point \( (4 \hat{i}+3 \hat{j}-\hat{k}) m \). Then the magnitude of torque about the point \( (\hat{i}+2 \hat{j}+\hat{k}) \) \( m \) will be \( \sqrt{x} N-m \). The value of \( x \) is
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