Rotational Motion and Centre of Mass: JEE Main 2026 PYQs

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Key idea: Kinematics, Linear Momentum, Force, Angular Momentum, and Torque in 2D motion

Answer: Option (3)

Key idea: Angular acceleration as second derivative of angular displacement

1. Given: $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$
2. Find angular velocity: $\omega = \frac{d\theta}{dt} = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$
3. Find angular acceleration: $\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$
4. Substitute $t = 10$ s: $\alpha = \frac{3(10)^2}{2} - 2(10) = 150 - 20$
5. Therefore, $\alpha = 130$ rad/s²

Answer: 15

Key idea: Moment of Inertia and Parallel Axis Theorem

1. Calculate moment of inertia of rod about axis AB (through one end): $I_{rod} = \frac{1}{3}ML^2 = \frac{1}{3}(40)(3)^2 = 120$ kg·m²
2. For the sphere, use parallel axis theorem: $I_{sphere} = I_{center} + Md^2 = \frac{2}{5}MR^2 + Md^2$
3. Substitute values: $I_{sphere} = \frac{2}{5}(10)R^2 + (10)(3)^2 = 4R^2 + 90$
4. Equate the moments: $120 = 4R^2 + 90$, solving gives $R^2 = \frac{15}{2}$
5. Given $R = \sqrt{\frac{\alpha}{2}}$, we have $R^2 = \frac{\alpha}{2} = \frac{15}{2}$, therefore $\alpha = 15$

Answer: Option (2)

Key idea: Moment of Inertia and Volume Conservation

1. Use volume conservation with constant density: Since smaller sphere has mass $M/8$ and original has mass $M$, we get $\frac{M/8}{(4/3)\pi r^3} = \frac{M}{(4/3)\pi R^3}$, giving $r = R/2$
2. Calculate $I_1$ for smaller sphere (mass $M/8$, radius $r = R/2$): $I_1 = \frac{2}{5} \times \frac{M}{8} \times \left(\frac{R}{2}\right)^2 = \frac{MR^2}{80}$
3. Calculate $I_2$ for disc (mass $7M/8$, radius $2R$) about diameter: $I_2 = \frac{1}{4} \times \frac{7M}{8} \times (2R)^2 = \frac{7MR^2}{8}$
4. Find the ratio: $\frac{I_2}{I_1} = \frac{7MR^2/8}{MR^2/80} = \frac{7 \times 80}{8} = 70$

Answer: Option (2)

Key idea: Rotational Kinematics - Equations of motion for uniform angular acceleration

1. Given: Initial angular velocity $\omega_0 = 0$, uniform angular acceleration $\alpha$
2. For first 2 s: Using $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$, we get $\theta_1 = \frac{1}{2}\alpha(2)^2 = 2\alpha$
3. For total 4 s: $\theta_{total} = \frac{1}{2}\alpha(4)^2 = 8\alpha$
4. For next 2 s (from t=2s to t=4s): $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$
5. Calculate ratio: $\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$

Answer: Option (4)

Key idea: Energy Conservation in Rolling Motion

1. Write energy conservation: Initial KE (translational + rotational) = Final PE at maximum height
2. Express total initial energy: $\frac{1}{2}mv_o^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_o^2(1+k)$ where $I = kmR^2$
3. Use rolling condition $v_o = R\omega$ and energy conservation: $\frac{1}{2}mv_o^2(1+k) = mgh$
4. Solve for height: $h = \frac{v_o^2(1+k)}{2g}$
5. Equate with given height: $\frac{v_o^2(1+k)}{2g} = \frac{7v_o^2}{10g}$, which gives $k = \frac{2}{5}$
6. Identify object: For solid sphere, $I = \frac{2}{5}mR^2$, so $k = \frac{2}{5}$ ✓

Answer: Option (1)

Key idea: Rotational dynamics: torque, moment of inertia of solid sphere, angular kinematics

1. Convert angular velocity: $\omega_0 = 1200$ rpm $= 40\pi$ rad/s
2. Find angular acceleration: $\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - 40\pi}{10} = -4\pi$ rad/s²
3. Calculate moment of inertia: $I = \frac{2}{5}mR^2 = \frac{2}{5}(5)(0.04)^2 = 0.0032$ kg·m²
4. Find torque magnitude: $\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi$ N·m
5. Calculate angular displacement: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 40\pi(10) - \frac{1}{2}(4\pi)(100) = 200\pi$ rad
6. Number of rotations: $n = \frac{\theta}{2\pi} = \frac{200\pi}{2\pi} = 100$

Answer: Option (1)

Key idea: Center of Mass of a System of Particles

1. Set up coordinate system with point $p$ (midpoint of base) as origin
2. Place masses: 2 kg at $(-a, 0)$, 3 kg at $(a, 0)$, 15 kg at $(0, h)$ where the triangle has appropriate dimensions
3. Apply center of mass formula: $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$ and $y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$
4. Calculate $x_{cm} = \frac{2(-a) + 3(a) + 15(0)}{20} = \frac{a}{20}$
5. Calculate $y_{cm} = \frac{2(0) + 3(0) + 15h}{20} = \frac{3h}{4}$
6. Using the given geometry, determine $x_{cm} = \frac{\sqrt{3}}{4}$ m and $y_{cm} = 1.25$ m

Answer: Option (1)

Key idea: Rolling motion with tangential force; moment of inertia of sphere and shell

1. Write equations for rolling motion: Linear equation $F + f = Ma$ and rotational equation $FR - fR = I\alpha$
2. Use rolling constraint $a = \alpha R$ to get $F - f = \frac{Ia}{R^2}$
3. Solve the two equations to get $a = \frac{2FR^2}{MR^2 + I}$
4. For solid sphere A: $I_A = \frac{2}{5}(5m)R^2 = 2mR^2$, so $a_A = \frac{2FR^2}{7mR^2} = \frac{2F}{7m}$
5. For spherical shell B: $I_B = \frac{2}{3}mR^2$, so $a_B = \frac{2FR^2}{\frac{5mR^2}{3}} = \frac{6F}{5m}$
6. Calculate ratio: $\frac{a_A}{a_B} = \frac{2F/7m}{6F/5m} = \frac{5}{21}$

Answer: Option (3)

Key idea: Center of Mass Motion - When acceleration is parallel to initial velocity, the path is a straight line

1. Find initial velocity of COM: $\vec{v}_{CM} = \frac{m\vec{v}_1 + m\vec{v}_2}{2m} = \frac{4\hat{i} + 4\hat{j}}{2} = 2\hat{i} + 2\hat{j}$ m/s
2. Find acceleration of COM: $\vec{a}_{CM} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{6\hat{i} + 6\hat{j}}{2} = 3\hat{i} + 3\hat{j}$ m/s²
3. Check if $\vec{a}_{CM}$ is parallel to $\vec{v}_{CM}$: $\vec{v}_{CM} = 2(\hat{i} + \hat{j})$ and $\vec{a}_{CM} = 3(\hat{i} + \hat{j})$
4. Since $\vec{a}_{CM} = \frac{3}{2}\vec{v}_{CM}$, acceleration is parallel to initial velocity
5. When acceleration is parallel to velocity, the motion is along a straight line

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.