Every Problem Type
Solved & Dissected
6 problem types. Each decoded with examiner's intent, concept selection, solution, and shortcut.
Direct Formula Problems
These are the easiest. 1–2 step substitution. But wrong formula = zero marks.
Q1. A solid sphere of mass 2 kg and radius 0.5 m rolls without slipping on a flat surface with velocity 3 m/s. Find its total kinetic energy.
M = 2 kg, R = 0.5 m, v_cm = 3 m/s, solid sphere → k²/R² = 2/5
Whether you know to use the rolling KE formula (not just ½mv²). Many students forget the rotational part.
KE_total = ½mv²_cm(1 + k²/R²) for rolling motion. For solid sphere: k²/R² = 2/5
KE = ½ × 2 × 3² × (1 + 2/5) = ½ × 2 × 9 × 7/5 = 9 × 7/5 = 12.6 J
For solid sphere: KE_total = (7/10)mv². Memorise this. KE_rot = (2/7) × KE_total = (2/7) × 12.6 = 3.6 J
Q2. A force F = 10 N acts at the end of a rod of length 2 m. The force makes an angle of 30° with the rod. Find the torque about the pivot at the other end.
F = 10 N, r = 2 m, θ = 30° (angle between F and rod)
Whether students correctly use sin(θ) where θ is angle between r and F vectors.
τ = rF sinθ = 2 × 10 × sin(30°) = 2 × 10 × 0.5 = 10 N·m
Lever arm = r × sinθ = 2 × sin30° = 1 m. Torque = F × lever arm = 10 × 1 = 10 N·m. Both give same answer — use whichever is faster.
For direct formula questions: (1) Identify which quantity is asked. (2) Identify which formula directly gives that quantity. (3) Substitute carefully — watch units. (4) Check order of magnitude. If a torque answer is 10⁶ N·m for a small lab setup, you've made an error.
Conceptual Problems
No calculation needed. But conceptual traps are everywhere. This is where NEET takers lose marks.
Q3. A solid sphere, a hollow sphere, a solid cylinder, and a hollow cylinder — all of the same mass and radius — are released from rest at the top of an incline. Which reaches the bottom first?
Whether you know that acceleration on incline = g sinθ/(1 + k²/R²). The body with smallest k²/R² has highest acceleration.
Solid sphere: k²/R² = 2/5. Solid cylinder: 1/2. Hollow sphere: 2/3. Hollow cylinder: 1. Order of acceleration: SS > SC > HS > HC
Solid sphere reaches first. This result is INDEPENDENT of mass, radius, and angle of incline — it depends only on shape (value of k²/R²).
Remember: S–S–H–H (Solid Sphere, Solid Cylinder, Hollow Sphere, Hollow Cylinder) in order of fastest to slowest. The "solid" always beats the corresponding "hollow."
Many students say "they all reach at the same time" (confusing with free fall or frictionless sliding). This is wrong. Rolling involves rotational KE — bodies with more rotational KE have less translational KE, and thus slower COM speed.
Q4. A planet moves around the sun. Which of the following remains constant? (A) Linear velocity (B) Linear momentum (C) Angular momentum (D) Kinetic energy
Conservation of angular momentum and Kepler's second law. The planet has variable speed — it moves faster when closer to the sun (perihelion) and slower when farther (aphelion).
Gravity acts toward the sun → torque about sun = 0 → L = constant. This directly leads to Kepler's 2nd law (equal areas in equal times).
Multi-Step Problems
If this step is wrong, the entire solution fails. Requires planning before solving.
Q5. A solid sphere of mass 5 kg and radius 0.2 m rolls without slipping down an incline of height 3 m and angle 30°. Find: (a) Speed at bottom, (b) Angular velocity at bottom, (c) Time taken.
PE lost = KE gained: mgh = ½mv²(1 + k²/R²) = ½mv²(1 + 2/5) = (7/10)mv²
v² = 10gh/7 = 10 × 10 × 3/7 = 300/7
v = √(300/7) ≈ 6.55 m/s
ω = v/R = 6.55/0.2 = 32.7 rad/s
a = g sinθ/(1 + k²/R²) = 10 × sin30°/(1 + 2/5) = 5/(7/5) = 25/7 m/s²
Length of incline: L = h/sinθ = 3/0.5 = 6 m
v² = u² + 2aL → v² = 0 + 2 × (25/7) × 6 = 300/7 ✓
v = u + at → t = v/a = 6.55/(25/7) = 6.55 × 7/25 = 1.83 s
In multi-step rolling problems: Start with energy (gives speed). Then angular quantities (ω = v/R). Then kinematics (gives time or force). This sequence never fails for this problem type.
Q6. A bullet of mass 10 g and velocity 300 m/s hits a rod of mass 2 kg and length 1 m (hinged at one end, initially at rest) at its free end and gets embedded. Find the angular velocity of the rod after the collision.
Angular momentum conservation about the hinge. NOT linear momentum (hinge exerts force). This distinction is crucial.
L_bullet = m_bullet × v × L = 0.01 × 300 × 1 = 3 kg·m²/s (L = r × mv, at perp. distance = 1 m from hinge)
I_rod (about hinge) = ML²/3 = 2 × 1²/3 = 2/3 kg·m²
I_bullet = m × L² = 0.01 × 1² = 0.01 kg·m²
I_total = 2/3 + 0.01 ≈ 0.677 kg·m²
L_initial = L_final
3 = I_total × ω
ω = 3/0.677 = 4.43 rad/s
Using linear momentum conservation here is WRONG because the hinge exerts an impulsive force. Angular momentum about the hinge is conserved (hinge force has zero moment arm about the hinge). Always ask: "Is there an impulsive external force? If yes, conserve angular momentum about the point of application of that force."
Graph-Based Problems
JEE twists this concept using graphs. Learn to extract quantities from graphs instantly.
Q7. The ω-t graph of a rotating body is a straight line with y-intercept ω₀ = 4 rad/s and slope = 2 rad/s². Find (a) angular displacement in 3 s, (b) angular acceleration.
In ω-t graph: Slope = α (angular acceleration). Area under graph = angular displacement θ.
(a) α = slope = 2 rad/s²
(b) θ = area under ω-t graph = trapezoid = ½(ω₀ + ω_final) × t
ω_final = 4 + 2×3 = 10 rad/s
θ = ½(4+10)×3 = 21 rad
If θ ∝ t², then α = constant (uniform angular acceleration). Slope at any point gives ω at that instant.
Slope = α. Area = θ. If line passes through origin, ω₀ = 0 (starts from rest).
Area under τ-θ graph = work done by torque = ΔKE (rotational). Very common in JEE graph questions.
If L-t graph is horizontal, τ = 0. If slope is non-zero, slope = τ (net torque). Used in angular momentum problems.
For any rotational graph: (1) Identify what's on each axis. (2) Find slope → gives the rate of change. (3) Find area → gives the integral quantity. (4) Find intercepts → gives initial conditions. Apply this template to every graph problem in 10 seconds.
Assertion & Reason
Used in CBSE. Checks whether you understand the "why" behind the concept.
A: The centre of mass of a body always lies inside the body.
R: Centre of mass is the weighted average position of all particles.
Explanation: COM can lie outside the body (e.g., hollow ring, boomerang, horseshoe). The reason (weighted average formula) is correct but does NOT explain the assertion — in fact, it contradicts it.
A: Angular momentum of a planet revolving around the sun is conserved.
R: Gravitational force on the planet acts along the line joining planet to sun (central force).
Central force acts through the sun → torque about sun = r × F = r × F(r̂) = 0 → L is conserved. This directly explains angular momentum conservation.
The 4 options: A=Both true & R explains A. B=Both true but R doesn't explain A. C=A true, R false. D=A false, R true. E=Both false. Strategy: (1) Verify A independently. (2) Verify R independently. (3) If both true, ask: "Does R logically lead to A?" If yes → Answer A. If no → Answer B.
Case-Based Problems
New CBSE format. 4–5 sub-questions from one long scenario. Read carefully — every detail matters.
Passage: A uniform rod of mass M = 3 kg and length L = 1.5 m is free to rotate about a hinge at one end. A horizontal force F = 12 N is applied at the free end, perpendicular to the rod. The rod starts from rest and rotates in the vertical plane.
(i) What is the moment of inertia of the rod about the hinge?
(ii) Find the torque about the hinge.
(iii) Find the initial angular acceleration.
(iv) Find angular velocity after rotating by π/2 radians.
ω = √(8π) ≈ 5.01 rad/s
Note: This assumes force remains perpendicular throughout — if force is constant in direction, torque changes with angle. This question simplifies by keeping force ⊥ rod.
Read the entire passage ONCE before looking at questions. Identify: masses, lengths, forces, constraints (hinged, pivoted, free). Often sub-questions build on each other (Q1 answer is needed for Q2). If you get Q1 wrong, all subsequent answers are wrong — be extra careful with Q1.