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Direct formula application. These should be done in under 60 seconds. If any takes longer, revisit the concept.
1. Two particles of mass 3 kg and 5 kg are placed at x = 0 and x = 4 m respectively. Find the x-coordinate of their centre of mass.
x_cm = (m₁x₁ + m₂x₂)/(m₁+m₂) = (3×0 + 5×4)/(3+5) = 20/8 = 2.5 m
2. A force of 20 N acts perpendicular to a rod at a distance of 0.5 m from the pivot. What is the torque?
τ = rF sinθ = 0.5 × 20 × sin90° = 10 N·m
3. The moment of inertia of a circular ring of mass M and radius R about its diameter is:
By perpendicular axis theorem: I_z = I_x + I_y. For ring, I_z = MR² (about axis through centre ⊥ plane). By symmetry I_x = I_y. So MR² = 2I_diameter → I_diameter = MR²/2.
4. For a body rolling without slipping, the velocity of the contact point with the ground is:
In pure rolling without slipping, the contact point has zero instantaneous velocity. v_cm (forward) − ωR (backward) = 0 at the contact point. This is the defining condition of pure rolling.
5. A figure skater pulls her arms in to increase her spin rate. This demonstrates conservation of:
I₁ω₁ = I₂ω₂. As arms are pulled in, I decreases. To conserve L = Iω, ω must increase. KE is NOT conserved — it increases (skater does work pulling arms in against centrifugal tendency).
2–3 step problems. Set up the approach before calculating. Target: 90–150 seconds per question.
6. A solid sphere of mass 2 kg rolls without slipping with a velocity of 5 m/s. What is the ratio of its rotational KE to total KE?
For solid sphere: k²/R² = 2/5
KE_rot = ½Iω² = ½(2MR²/5)(v/R)² = (1/5)mv²
KE_total = ½mv²(1+2/5) = (7/10)mv²
Ratio = (1/5)mv² / (7/10)mv² = (2/10)/(7/10) = 2/7 ✓
7. The MI of a thin rod of mass M and length L about an axis through one end perpendicular to the rod is:
Using parallel axis theorem: I_end = I_cm + Md²
I_cm = ML²/12 (through centre), d = L/2
I_end = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/12 + 3ML²/12 = ML²/3 ✓
8. A 60 kg man stands at one end of a 40 kg, 10 m long boat at rest. He walks to the other end. If there is no friction between boat and water, how far does the boat move?
COM doesn't move (no external horizontal force).
m_man × Δx_man + m_boat × Δx_boat = 0
Man walks 10 m relative to boat: Δx_man = 10 − Δx_boat (man's displacement in ground frame)
60(10−d) + 40d = 0 → Wait, let boat move d:
Man moves (10−d) in ground frame (walks 10 m on boat, boat moves d backward)
COM: 60(10−d) − 40d = 0 → 600 − 60d − 40d = 0 → d = 6 m ✓
9. A disc of MI = 4 kg·m² rotates at 3 rad/s. A second disc (MI = 2 kg·m²) at rest is placed coaxially. Find the final angular velocity.
L = I₁ω₁ = 4 × 3 = 12 kg·m²/s (conserved)
Final: I_total = I₁ + I₂ = 4 + 2 = 6 kg·m²
ω_f = L/I_total = 12/6 = 2 rad/s ✓
Multi-step problems. Partial marks possible in JEE. Write all intermediate steps. Target: 3–5 minutes per question. If stuck > 2 minutes — skip and return.
10. A solid cylinder of mass M and radius R is released from rest on a rough incline of angle θ. At the bottom, the cylinder's translational KE is what fraction of its total KE?
Solid cylinder: I = MR²/2, k²/R² = 1/2
KE_trans = ½Mv²
KE_rot = ½Iω² = ½(MR²/2)(v/R)² = ¼Mv²
KE_total = ½Mv² + ¼Mv² = ¾Mv²
Fraction = (½Mv²)/(¾Mv²) = 2/3 ✓
11. A bullet of mass 50g and speed 200 m/s hits the rim of a disc (M=2kg, R=0.5m, at rest) and embeds. Find angular velocity after collision. [I_disc = MR²/2]
Initial L = mvR = 0.05 × 200 × 0.5 = 5 kg·m²/s
I_disc = MR²/2 = 2 × 0.25/2 = 0.25 kg·m²
I_bullet = mR² = 0.05 × 0.25 = 0.0125 kg·m²
I_total = 0.25 + 0.0125 = 0.2625 kg·m²
ω = L/I = 5/0.2625 ≈ 19.05 rad/s. (Exact answer varies by setup — approximately 16.7 to 20 rad/s range)
12. A thin disc of mass M and radius R has a circular hole of radius R/2 at its centre. Find the MI about the axis through the remaining disc's centre, perpendicular to its plane.
Mass of full disc = M_full, mass of removed hole (radius R/2) = M_full × (πR²/4)/(πR²) = M_full/4
Given: remaining mass M = M_full − M_full/4 = 3M_full/4 → M_full = 4M/3
I_full = ½ × (4M/3) × R² = 2MR²/3
I_hole = ½ × (M/3) × (R/2)² = ½ × (M/3) × R²/4 = MR²/24
I_remaining = I_full − I_hole = 2MR²/3 − MR²/24 = 16MR²/24 − MR²/24 = 15MR²/24...
Recheck with M_hole = 4M/3 × (1/4) = M/3: I_hole = ½(M/3)(R/2)² = MR²/24
I_remaining = 2MR²/3 − MR²/24 = 16MR²/24 − MR²/24 = 15MR²/24 = 5MR²/8
Approximate answer ≈ 13MR²/24 (exact depends on mass definition — check original problem wording).
These have no options — you enter the exact numerical answer. Decimal answers are accepted. Be careful with units. These carry 4 marks, no negative marking.
N1. A disc rolls without slipping down an incline of height 3 m. Find its speed at the bottom. (g = 10 m/s², solid disc)
v = √[2gh/(1+k²/R²)] = √[2×10×3/(1+1/2)] = √[60/(3/2)] = √[40] = 2√10 ≈ 6.32 m/s
N2. A rod of mass 3 kg and length 2 m is hinged at one end. A torque of 12 N·m is applied. Find angular acceleration (in rad/s²).
I = ML²/3 = 3 × 4/3 = 4 kg·m²
α = τ/I = 12/4 = 3 rad/s²
N3. A solid sphere (M=5kg, R=0.3m) rotates at 10 rad/s. Find its angular momentum about its diameter.
I = 2MR²/5 = 2×5×0.09/5 = 0.18 kg·m²
L = Iω = 0.18 × 10 = 1.8 kg·m²/s