Centre of Mass & Rotational Motion
1. Centre of Mass (CoM)
Point where the entire mass of the system can be assumed to be concentrated.
Discrete System:
RCM = (m1r1 + m2r2 + ...) / (m1 + m2 + ...)
Two Particles: CoM divides line joining them inversely as ratio of masses (r1/r2 = m2/m1).
RCM = (m1r1 + m2r2 + ...) / (m1 + m2 + ...)
Two Particles: CoM divides line joining them inversely as ratio of masses (r1/r2 = m2/m1).
Motion of CoM
M ACM = Fext
If Fext = 0, then ACM = 0 and VCM = Constant.
(Conservation of Linear Momentum applies to CoM).
If Fext = 0, then ACM = 0 and VCM = Constant.
(Conservation of Linear Momentum applies to CoM).
(Asked in NEET 2019, 2021)
2. Rotational Variables
Analogue to Linear Motion variables.
Torque (τ): Moment of Force.
τ = r × F = rF sinθ Ŏ
Angular Momentum (L): Moment of Momentum.
L = r × p = m(r × v)
τ = r × F = rF sinθ Ŏ
Angular Momentum (L): Moment of Momentum.
L = r × p = m(r × v)
Relation & Conservation
τ = dL / dt
Conservation of Angular Momentum:
If τext = 0, then L = Constant.
I1ω1 = I2ω2 (Example: Ballet Dancer).
Conservation of Angular Momentum:
If τext = 0, then L = Constant.
I1ω1 = I2ω2 (Example: Ballet Dancer).
(Asked in NEET 2016, 2018, 2022)
3. Moment of Inertia (I)
Rotational Inertia. Resists change in rotational state.
I = Σ mi ri2
Radius of Gyration (k): I = Mk2.
Radius of Gyration (k): I = Mk2.
Theorems
Parallel Axes: I = ICM + Md2
Perpendicular Axes (Laminar objects): Iz = Ix + Iy
Perpendicular Axes (Laminar objects): Iz = Ix + Iy
4. Rolling Motion
Combination of Translation + Rotation.
Condition for Pure Rolling (No slipping): vCM = Rω
Total KE = KEtrans + KErot
K = ½Mv2 (1 + k2/R2)
Total KE = KEtrans + KErot
K = ½Mv2 (1 + k2/R2)
Rolling on Incline
Acceleration a = (g sinθ) / (1 + k2/R2)
Friction acts up the incline to provide torque.
Friction acts up the incline to provide torque.
(Asked in NEET 2017, 2020, 2023)
Numericals: Rotational Motion
Q1. Three particles of masses 1 kg, 2 kg, and 3 kg are placed at (0,0), (1,2), and (2,1) respectively. Find
coordinate of Centre of Mass.
Solution:
Total Mass M = 1 + 2 + 3 = 6 kg.
XCM = (1×0 + 2×1 + 3×2) / 6 = (0 + 2 + 6) / 6 = 8/6 = 1.33
YCM = (1×0 + 2×2 + 3×1) / 6 = (0 + 4 + 3) / 6 = 7/6 = 1.16
CoM is at (1.33, 1.16).
Total Mass M = 1 + 2 + 3 = 6 kg.
XCM = (1×0 + 2×1 + 3×2) / 6 = (0 + 2 + 6) / 6 = 8/6 = 1.33
YCM = (1×0 + 2×2 + 3×1) / 6 = (0 + 4 + 3) / 6 = 7/6 = 1.16
CoM is at (1.33, 1.16).
Q2. A force F = (2i + 3j - k) N acts at a point r = (i - j
+ 2k) m. Calculate Torque.
Solution:
τ = r × F.
Using determinant:
| i j k |
| 1 -1 2 |
| 2 3 -1 |
= i(1 - 6) - j(-1 - 4) + k(3 - (-2))
= i(-5) - j(-5) + k(5) = -5i + 5j + 5k Nm.
τ = r × F.
Using determinant:
| i j k |
| 1 -1 2 |
| 2 3 -1 |
= i(1 - 6) - j(-1 - 4) + k(3 - (-2))
= i(-5) - j(-5) + k(5) = -5i + 5j + 5k Nm.
Q3. A disc rotates at 100 rpm. A blob of wax falls on it and mass moment of inertia becomes 1.5 times. Find
new speed.
Solution:
I1 = I, ω1 = 100.
I2 = 1.5 I.
L1 = L2 → I1ω1 = I2ω2
I × 100 = 1.5 I × ω2
ω2 = 100 / 1.5 = 200/3 = 66.67 rpm.
I1 = I, ω1 = 100.
I2 = 1.5 I.
L1 = L2 → I1ω1 = I2ω2
I × 100 = 1.5 I × ω2
ω2 = 100 / 1.5 = 200/3 = 66.67 rpm.
Q4. Calculate Moment of Inertia of a solid sphere of mass 10 kg and radius 0.5 m about a tangent.
Solution:
ICM (Diameter) = (2/5)MR2.
Using Parallel Axis Theorem: Itangent = ICM + MR2.
I = (2/5) MR2 + MR2 = (7/5) MR2.
I = (7/5) × 10 × (0.5)2
I = 1.4 × 10 × 0.25 = 14 × 0.25 = 3.5 kg m2.
ICM (Diameter) = (2/5)MR2.
Using Parallel Axis Theorem: Itangent = ICM + MR2.
I = (2/5) MR2 + MR2 = (7/5) MR2.
I = (7/5) × 10 × (0.5)2
I = 1.4 × 10 × 0.25 = 14 × 0.25 = 3.5 kg m2.
Q5. A flywheel has MI of 4 kg m2 and rotates at 120 rpm. Find its Kinetic energy.
Solution:
ω = 2πf = 2π(120/60) = 4π rad/s.
K = ½ I ω2
K = 0.5 × 4 × (4π)2
K = 2 × 16 π2 = 32 × 10 = 320 J (approx).
ω = 2πf = 2π(120/60) = 4π rad/s.
K = ½ I ω2
K = 0.5 × 4 × (4π)2
K = 2 × 16 π2 = 32 × 10 = 320 J (approx).
Q6. A solid cylinder and a hollow cylinder of same mass and radius roll down an incline. Ratio of their
acceleration?
Solution:
a = g sinθ / (1 + k2/R2).
For Solid Cyl (Disc): k2/R2 = 1/2. Term = 1+0.5 = 1.5.
For Hollow Cyl (Ring): k2/R2 = 1. Term = 1+1 = 2.
Ratio aS / aH = (1/1.5) / (1/2) = 2 / 1.5 = 4:3.
a = g sinθ / (1 + k2/R2).
For Solid Cyl (Disc): k2/R2 = 1/2. Term = 1+0.5 = 1.5.
For Hollow Cyl (Ring): k2/R2 = 1. Term = 1+1 = 2.
Ratio aS / aH = (1/1.5) / (1/2) = 2 / 1.5 = 4:3.
Q7. If Earth suddenly shrinks to half its radius without change in mass, what is new duration of day?
Solution:
L = Const. I1ω1 = I2ω2.
(2/5)MR2 (2π/T1) = (2/5)M(R/2)2 (2π/T2)
R2 / 24 = (R2/4) / T2
1/24 = 1 / 4T2 → 4T2 = 24 → T2 = 6 Hours.
L = Const. I1ω1 = I2ω2.
(2/5)MR2 (2π/T1) = (2/5)M(R/2)2 (2π/T2)
R2 / 24 = (R2/4) / T2
1/24 = 1 / 4T2 → 4T2 = 24 → T2 = 6 Hours.
Q8. A wheel of MI 2 kg m2 starts from rest and rotates 10 rad in 2 s. Find constant torque
applied.
Solution:
θ = ω0t + ½αt2
10 = 0 + 0.5 α (2)2
10 = 2 α → α = 5 rad/s2.
τ = I α = 2 × 5 = 10 Nm.
θ = ω0t + ½αt2
10 = 0 + 0.5 α (2)2
10 = 2 α → α = 5 rad/s2.
τ = I α = 2 × 5 = 10 Nm.
Q9. 3m ladder weighing 20kg leans on smooth wall. Foot is 1m from wall. Find reaction forces.
Solution:
This is a torque Balance problem.
Key Concept: Στ = 0 about foot.
Weight torque = Wall Reaction torque.
(Not fully solved, just logic: mg × 0.5 = Nwall × height).
Complex calculation for quick sheet. Answer conceptually: Torque balance.
This is a torque Balance problem.
Key Concept: Στ = 0 about foot.
Weight torque = Wall Reaction torque.
(Not fully solved, just logic: mg × 0.5 = Nwall × height).
Complex calculation for quick sheet. Answer conceptually: Torque balance.
Q10. A disc rolls down height h. Find its velocity at bottom.
Solution:
PE = Total KE.
mgh = ½ mv2 (1 + k2/R2)
For disc (k2/R2 = 0.5):
gh = 0.5 v2 (1.5) = 0.75 v2 = (3/4) v2.
v2 = 4gh/3 → v = √(4gh/3).
PE = Total KE.
mgh = ½ mv2 (1 + k2/R2)
For disc (k2/R2 = 0.5):
gh = 0.5 v2 (1.5) = 0.75 v2 = (3/4) v2.
v2 = 4gh/3 → v = √(4gh/3).
Important Formulae
20 NEET Golden Facts
- 1. CoM Location: Can be outside the body (e.g., Ring, L-shaped rod). Depends on mass distribution.
- 2. Internal Forces: Cannot change velocity of CoM. Only external forces affect CoM motion.
- 3. Projectile Explosion: If a projectile explodes in mid-air, its CoM continues to follow the same parabolic path.
- 4. Torque Direction: Perpendicular to plane containing r and F. Given by Right Hand Rule.
- 5. Couple: Two equal and opposite forces not falling on same line. Produces pure rotation (Net Force = 0).
- 6. Angular Momentum: Conserved if Net External Torque is Zero. Examples: Planetary motion, Diver folding hands in air.
- 7. Moment of Inertia: Depends on Mass, Axis of rotation, and Distribution of mass (Shape/Size). Not a Vector, Not a Scalar (Tensor).
- 8. Flywheel: High MOI flywheel is used in engines to minimize fluctuations in speed (stores rotational energy).
- 9. Radius of Gyration: Distance from axis where entire mass can be assumed concentrated to give same I. Depends on axis.
- 10. Parallel Axis: Applicable to any shape. I = ICM + Md2. I is minimum about CoM axis.
- 11. Perpendicular Axis: Only for 2D (Planar/Laminar) bodies. Iz = Ix + Iy (Z is perpendicular to plane).
- 12. Rolling Friction: In pure rolling, point of contact is at rest. Static friction acts (work done by static friction is zero).
- 13. Fastest Descent: Solid Sphere reaches bottom of incline first (least k, max acceleration). Ring reaches last.
- 14. Angular Speed: v = rω. All particles on a rotating rigid body have same ω but different v (v ∝ r).
- 15. Vector Product: A × B = - (B × A). Order matters.
- 16. L and v: For a particle moving in a straight line, angular momentum about a point on the line is zero. offset point is constant (mvr).
- 17. Hollow vs Solid: Hollow Cylinder has greater I than Solid Cylinder of same mass and radius (mass is far from axis).
- 18. Work-Energy: Work done by torque = ∫ τ dθ = Change in Rotational KE.
- 19. Toppling: Toppling happens when vertical line through CoM falls outside the base.
- 20. Pure Translation: All points have same velocity. Pure Rotation: All points move in circles about axis.
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