Rotational Motion — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Rotational Motion — moment of inertia, torque, rolling without slipping, angular kinematics and centre of mass — with step-by-step solutions.
Solved JEE Main 2026 questions from the Rotational Motion chapter, covering moment of inertia, torque and angular deceleration, angular kinematics, rolling without slipping, and centre-of-mass motion, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Both parts are made of the same material, so density is conserved.
Radius of the small sphere. Mass scales with volume, so
$$\frac{M/8}{M} = \frac{r^3}{R^3} \implies r^3 = \frac{R^3}{8} \implies r = \frac{R}{2}$$Moment of inertia of the small sphere (about a central axis):
$$I_1 = \frac{2}{5}\left(\frac{M}{8}\right)r^2 = \frac{2}{5}\cdot\frac{M}{8}\cdot\frac{R^2}{4} = \frac{MR^2}{80}$$Moment of inertia of the disc about a diameter. Disc mass $= \dfrac{7M}{8}$, radius $= 2R$:
$$I_2 = \frac{1}{4}\left(\frac{7M}{8}\right)(2R)^2 = \frac{1}{4}\cdot\frac{7M}{8}\cdot 4R^2 = \frac{7MR^2}{8}$$Ratio:
$$\frac{I_2}{I_1} = \frac{7MR^2/8}{MR^2/80} = 7 \times 10 = 70$$Answer: B (70)
Solution
Moment of inertia ($R = 0.04$ m, $m = 5$ kg):
$$I = \frac{2}{5}mR^2 = \frac{2}{5}(5)(0.04)^2 = 3.2\times10^{-3}\ \text{kg·m}^2$$Initial angular velocity:
$$\omega_0 = 1200\ \text{rpm} = \frac{1200\times 2\pi}{60} = 40\pi\ \text{rad/s}$$Angular deceleration (to rest in 10 s):
$$\alpha = \frac{\omega_0}{t} = \frac{40\pi}{10} = 4\pi\ \text{rad/s}^2$$Torque:
$$\tau = I\alpha = 3.2\times10^{-3}\times 4\pi = 0.0128\pi\ \text{Nm}$$Number of rotations (uniform deceleration, average $\omega = \omega_0/2$):
$$\theta = \frac{\omega_0}{2}\,t = \frac{40\pi}{2}\times 10 = 200\pi\ \text{rad} \implies N = \frac{200\pi}{2\pi} = 100$$Answer: D ($0.0128\pi$ Nm, 100)
Solution
Differentiate $\vec{r} = 10t^2\hat{i} + 5t^3\hat{j}$ and evaluate at $t = 1$ s ($m = 0.1$ kg).
Velocity and momentum:
$$\vec{v} = 20t\,\hat{i} + 15t^2\,\hat{j} = 20\hat{i} + 15\hat{j},\qquad \vec{p} = m\vec{v} = 2\hat{i} + 1.5\hat{j}\ \Rightarrow\ \textbf{A correct}$$Acceleration and force:
$$\vec{a} = 20\hat{i} + 30t\,\hat{j} = 20\hat{i} + 30\hat{j},\qquad \vec{F} = m\vec{a} = 2\hat{i} + 3\hat{j}\ \Rightarrow\ \textbf{B correct}$$Position at $t=1$: $\vec{r} = 10\hat{i} + 5\hat{j}$.
Angular momentum $\vec{L} = \vec{r}\times\vec{p}$:
$$L_z = r_x p_y - r_y p_x = (10)(1.5) - (5)(2) = 15 - 10 = 5\ \Rightarrow\ \vec{L} = 5\hat{k}\ (\text{not }15\hat{k})\ \Rightarrow\ \textbf{C wrong}$$Torque $\vec{\tau} = \vec{r}\times\vec{F}$:
$$\tau_z = r_x F_y - r_y F_x = (10)(3) - (5)(2) = 30 - 10 = 20\ \Rightarrow\ \vec{\tau} = 20\hat{k}\ \Rightarrow\ \textbf{D correct}$$Correct statements: A, B and D.
Answer: D (A, B and D only)
Solution
Simplify $\theta = \dfrac{t^4}{8} - \dfrac{t^3}{3}$.
Angular velocity:
$$\omega = \frac{d\theta}{dt} = \frac{t^3}{2} - t^2$$Angular acceleration:
$$\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$$At $t = 10$ s:
$$\alpha = \frac{3(100)}{2} - 2(10) = 150 - 20 = 130\ \text{rad/s}^2$$Answer: C (130)
Solution
The wall pushes the mass inward providing the centripetal force, so the normal reaction is
$$N = m\omega^2 R$$For the mass to just not slide down, limiting friction balances gravity:
$$\mu N = mg \implies \mu\, m\omega^2 R = mg$$Solve for $\mu$ ($\omega = 5$ rad/s, $R = 4$ m, $g = 10$):
$$\mu = \frac{g}{\omega^2 R} = \frac{10}{(5)^2 (4)} = \frac{10}{100} = 0.1$$Answer: A ($0.1$)
Solution
Let the body have mass $M$, radius $R$, and moment of inertia $I = kMR^2$. The force $F$ acts at the top; let $f$ be friction at the contact point.
Translation: $F + f = Ma$
Rotation about centre (both $F$ and $f$ produce torque about the centre):
$$FR - fR = I\alpha = kMR^2\cdot\frac{a}{R} \implies F - f = kMa$$Adding the two equations:
$$2F = Ma(1+k) \implies a = \frac{2F}{M(1+k)}$$Solid sphere $A$ ($M = 5m$, $k = \tfrac{2}{5}$):
$$a_A = \frac{2F}{5m\left(1+\tfrac{2}{5}\right)} = \frac{2F}{5m\cdot\tfrac{7}{5}} = \frac{2F}{7m}$$Spherical shell $B$ ($M = m$, $k = \tfrac{2}{3}$):
$$a_B = \frac{2F}{m\left(1+\tfrac{2}{3}\right)} = \frac{2F}{m\cdot\tfrac{5}{3}} = \frac{6F}{5m}$$Ratio:
$$\frac{a_A}{a_B} = \frac{2/7}{6/5} = \frac{2}{7}\cdot\frac{5}{6} = \frac{10}{42} = \frac{5}{21}$$Answer: A ($5:21$)
Solution
For equal masses, the centre of mass has
Initial velocity:
$$\vec{v}_{cm} = \frac{\vec{v_1} + \vec{v_2}}{2} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$$Acceleration:
$$\vec{a}_{cm} = \frac{\vec{a_1} + \vec{a_2}}{2} = \frac{6\hat{i} + 6\hat{j}}{2} = 3\hat{i} + 3\hat{j}$$Both $\vec{v}_{cm}$ and $\vec{a}_{cm}$ point along the same direction $(\hat{i}+\hat{j})$. When initial velocity and constant acceleration are parallel, the motion is one-dimensional along that line.
Answer: C (straight line)
Solution
Starting from rest with constant angular acceleration $\alpha$, the angle in time $t$ is $\theta = \tfrac{1}{2}\alpha t^2$.
First 2 s:
$$\theta_1 = \frac{1}{2}\alpha(2)^2 = 2\alpha$$First 4 s (total):
$$\theta_{0\to4} = \frac{1}{2}\alpha(4)^2 = 8\alpha$$Second interval (2 s to 4 s):
$$\theta_2 = \theta_{0\to4} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$$Ratio:
$$\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$$Answer: B (3)
Solution
Let $I = kmR^2$. Rolling without slipping gives $\omega = v_0/R$. Conserving energy from base to the highest point:
$$\frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2 = mgh$$$$\frac{1}{2}mv_0^2\left(1 + k\right) = mgh \implies h = \frac{v_0^2(1+k)}{2g}$$Match the given height:
$$\frac{v_0^2(1+k)}{2g} = \frac{7v_0^2}{10g} \implies 1 + k = \frac{7}{5} \implies k = \frac{2}{5}$$$k = \tfrac{2}{5}$ corresponds to a solid sphere $\left(I = \tfrac{2}{5}mR^2\right)$.
Answer: D (solid sphere)
Solution
Initial spin gives a surface speed at the contact point of $\omega_0 R = \dfrac{v_0}{4}$, which is less than $v_0$. So the contact point slides forward, and kinetic friction acts backward.
Translation (friction $f = \mu_K mg$ decelerates):
$$v(t) = v_0 - \mu_K g\,t$$Rotation (friction torque $fR$ about the centre spins it up, $I = \tfrac{1}{2}mR^2$):
$$\alpha = \frac{fR}{I} = \frac{\mu_K mgR}{\tfrac{1}{2}mR^2} = \frac{2\mu_K g}{R} \implies \omega(t) = \frac{v_0}{4R} + \frac{2\mu_K g}{R}\,t$$Rolling condition $v = \omega R$:
$$v_0 - \mu_K g\,t = \frac{v_0}{4} + 2\mu_K g\,t$$$$\frac{3v_0}{4} = 3\mu_K g\,t \implies t = \frac{v_0}{4\mu_K g}$$Substitute $v_0 = 49$, $\mu_K = 0.25$, $g = 9.8$:
$$t = \frac{49}{4(0.25)(9.8)} = \frac{49}{9.8} = 5\ \text{s}$$Answer: B (5)