Rotational Motion Formula Sheet for JEE Main

Core Rotational Kinematics

• Angular displacement $\theta$, angular velocity $\omega = d\theta/dt$, angular acceleration $\alpha = d\omega/dt$.
• Constant $\alpha$: $\omega = \omega_0 + \alpha t$, $\theta = \omega_0 t + \tfrac12 \alpha t^2$, $\omega^2 = \omega_0^2 + 2\alpha\theta$.
• Linear-angular link: $v = r\omega$, $a_t = r\alpha$, $a_c = r\omega^2 = v^2/r$.

Moment of Inertia (I)

• Parallel axis: $I = I_{cm} + Md^2$.
• Perpendicular axis (lamina): $I_z = I_x + I_y$.

Torque, Angular Momentum, Energy

• Torque $\vec\tau = \vec r \times \vec F = I\vec\alpha$.
• Angular momentum $\vec L = \vec r \times m\vec v = I\vec\omega$.
• $\vec\tau_{ext} = d\vec L/dt$; if $\vec\tau_{ext}=0$, $L$ conserved.
• Rotational KE $= \tfrac12 I\omega^2$; total KE of rolling body $= \tfrac12 Mv^2 + \tfrac12 I\omega^2 = \tfrac12 Mv^2(1 + I/MR^2)$.

Pure Rolling

• Condition: $v_{cm} = R\omega$; no sliding at contact.
• Acceleration down incline: $a = g\sin\theta/(1 + I/MR^2)$.
• Minimum friction coefficient: $\mu_{min} = \tan\theta/(1 + MR^2/I)$.

Common Rolling Accelerations

Quick Tricks

• Ranking by finish-line speed on same incline: solid sphere > disc > hollow sphere > ring (smaller $I/MR^2$ wins).
• For a body pivoted at end and released from horizontal: $\omega = \sqrt{3g/L}$ when vertical.
• Precession of top: $\Omega = Mgr/I\omega$.

Practice MCQs

Q1. A solid sphere and a hollow sphere of same mass and radius roll without slipping down the same incline. Which reaches the bottom first? A) Hollow sphere B) Solid sphere C) Both together D) Depends on incline angle

Q2. Moment of inertia of a uniform rod of mass M and length L about an axis perpendicular to it and passing through one end is A) ML²/12 B) ML²/3 C) ML²/6 D) ML²/2

Q3. A disc rotates at 300 rpm and comes to rest in 20 s under constant deceleration. Its angular acceleration is A) −π/2 rad/s² B) −π rad/s² C) −π/4 rad/s² D) −2π rad/s²

Q4. For a body rolling without slipping down an incline of angle θ, the friction force is A) Zero B) Kinetic, μₖMg cosθ C) Static, less than μₛMg cosθ D) Equal to Mg sinθ

Q5. Two children sit at the ends of a rotating merry-go-round. When they move to the centre, the angular velocity A) Decreases B) Remains the same C) Increases D) Becomes zero

Q6. A ring of radius R and mass M rolls without slipping with velocity v. Its total kinetic energy is A) ½Mv² B) Mv² C) 3/2 Mv² D) 2Mv²

Q7. Angular momentum has the same dimensions as A) Torque B) Power C) Planck's constant D) Impulse

Q8. A uniform disc of radius R is pivoted at its rim and released from horizontal. Angular velocity when vertical is A) √(2g/R) B) √(4g/3R) C) √(3g/R) D) √(g/R)

Q9. The radius of gyration of a solid sphere about a tangent is A) R√(2/5) B) R√(7/5) C) R√(3/5) D) R

Q10. If external torque about an axis is zero, which of the following is conserved about that axis? A) Kinetic energy B) Angular momentum C) Linear momentum D) Moment of inertia

Answers: 1-B (Smaller I/MR² → larger a; solid sphere has 2/5 vs 2/3.), 2-B (Parallel axis: ML²/12 + M(L/2)² = ML²/3.), 3-A (ω₀ = 10π rad/s; α = −10π/20 = −π/2.), 4-C (Static friction adjusts to prevent slipping; not necessarily maximum.), 5-C (L = Iω is conserved; I decreases → ω rises.), 6-B (½Mv²(1 + 1) = Mv² since I/MR² = 1.), 7-C (Both L and h have units J·s.), 8-B (Energy conservation with I about rim = 3MR²/2 gives ω²=4g/3R.), 9-B (I_tan = 2MR²/5 + MR² = 7MR²/5 = Mk².), 10-B (Direct statement of angular impulse–momentum theorem.)

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