Centre of Mass
The centre of mass (COM) of a system is the point where the entire mass of the system can be considered to be concentrated for the purpose of analysing translational motion.
For Two-Particle System
x_(cm) = (m_1 x_1 + m_2 x_2)/(m_1 + m_2)
For n-Particle System
vecR_(cm) = (sum m_i vecr_i)/(sum m_i)
For Continuous Bodies
vecR_(cm) = (int vecr dm)/M
Centre of Mass of Common Bodies
- Uniform rod: Centre
- Uniform sphere: Centre
- Uniform disc: Centre
- Uniform triangular lamina: Centroid
- Hollow sphere: Centre
Motion of Centre of Mass
M veca_(cm) = vecF_ext (COM moves as if all mass and external forces are at COM).
Moment of Inertia (MI)
The rotational analogue of mass. It measures the resistance of a body to rotational motion.
I = sum m_i r_i^2 (for discrete particles)
I = int r^2 dm (for continuous bodies)
Radius of Gyration (k)
I = Mk^2, where k is the radius of gyration — the distance from the axis where all mass can be concentrated to give the same MI.
Theorems of Moment of Inertia
Parallel Axis Theorem: I = I_(cm) + Md^2
MI about any axis = MI about parallel axis through COM + M x (distance between axes)^2.
Perpendicular Axis Theorem: For a planar body, I_z = I_x + I_y
MI about perpendicular axis through a point = sum of MI about two perpendicular axes through that point in the plane.
Moments of Inertia of Common Bodies
| Body | Axis | MI |
|---|---|---|
| Thin ring | Through centre perpendicular to plane | MR^2 |
| Thin ring | Diameter | MR^2/2 |
| Solid disc | Through centre perpendicular to plane | MR^2/2 |
| Solid sphere | Diameter | 2MR^2/5 |
| Hollow sphere | Diameter | 2MR^2/3 |
| Thin rod | Through centre perpendicular to rod | ML^2/12 |
| Thin rod | Through end perpendicular to rod | ML^2/3 |
Torque
The rotational analogue of force.
vectau = vecr x vecF
Magnitude: tau = rF sin theta = r_perp F
SI unit: N m.
Relation between Torque and Angular Acceleration
vectau = I vecalpha (where alpha is angular acceleration)
Angular Momentum
vecL = vecr x vecp = I vecomega
Conservation of Angular Momentum
If net external torque on a system is zero, total angular momentum is conserved.
I_1 omega_1 = I_2 omega_2
Examples:
- A ballet dancer spins faster by pulling arms in (I decreases, omega increases).
- A diver curls into a tuck to rotate faster.
Relation between Torque and Angular Momentum
vectau = dvecL/dt (analogous to F = dp/dt)
Rolling Motion
A combination of translation and rotation without slipping.
Condition for Pure Rolling
v_(cm) = R omega
Kinetic Energy in Rolling
K = (1/2) I_(cm) omega^2 + (1/2) M v_(cm)^2
= (1/2) M v_(cm)^2 (1 + I/(MR^2))
= (1/2) M v_(cm)^2 (1 + k^2/R^2)
Forces in Rolling
Static friction provides torque for rolling, but does no work (no relative motion at point of contact).
Acceleration on an Inclined Plane
a = (g sin theta)/(1 + I/(MR^2))
Order of acceleration: Solid sphere > Solid cylinder > Hollow sphere > Hollow cylinder (smaller I gives larger acceleration).
Worked Examples
Example 1: Find COM of two masses 2 kg at (0,0) and 3 kg at (4,0).
Solution: x_(cm) = (2*0 + 3*4)/(2+3) = 12/5 = 2.4 m. COM is at (2.4, 0).
Example 2: Find MI of a solid disc of mass 2 kg and radius 0.1 m about its axis.
Solution: I = MR^2/2 = 2*(0.1)^2/2 = 0.01 kg m^2.
Example 3: A disc (I = MR^2/2) rolls down a 30-degree incline of height 5 m. Find speed at bottom.
Solution: Using energy: mgh = (1/2)mv^2 + (1/2)Iomega^2. With I = MR^2/2 and v = R omega: gh = v^2/2 + v^2/4 = 3v^2/4 => v = sqrt(4gh/3) = sqrt(4*10*5/3) = sqrt(200/3) = 8.16 m/s.
Common Mistakes
- Force vs torque: Force causes linear acceleration; torque causes angular acceleration.
- MI depends on axis: Never quote MI without specifying the axis.
- Angular momentum conservation: Only if net external torque is zero, not force.
- Rolling without slipping: Friction does no work. Energy conservation applies.
ISC Exam Focus
- Theory (70%): COM, MI definitions, theorems, angular momentum conservation, rolling motion.
- Application (30%): Numerical problems on COM, torque, angular momentum, rolling down incline.
- ISC frequently asks: "Find MI of ... about ... axis using parallel/perpendicular axis theorem."
- Application of conservation of angular momentum is a key exam topic.
Self-Test Questions
Q1: Define centre of mass. Find COM of two particles of masses 4 kg and 6 kg at positions (2, 3) and (4, 5).
Answer: x_(cm) = (4*2 + 6*4)/10 = 32/10 = 3.2. y_(cm) = (4*3 + 6*5)/10 = 42/10 = 4.2. COM = (3.2, 4.2).
Q2: State parallel axis theorem.
Answer: I = I_(cm) + Mh^2, where h is perpendicular distance between axes.
Q3: Find MI of a solid sphere of mass 5 kg and radius 0.2 m about its diameter.
Answer: I = 2MR^2/5 = 2*5*0.04/5 = 0.08 kg m^2.
Q4: A ballet dancer spins at 2 rev/s with arms outstretched (I = 6 kg m^2). She pulls arms in (I = 2 kg m^2). Find new angular speed.
Answer: I_1 omega_1 = I_2 omega_2 => 6*2 = 2*omega_2 => omega_2 = 6 rev/s.
Q5: State the law of conservation of angular momentum. Answer: When net external torque on a system is zero, total angular momentum is conserved.
Q6: Find the rotational KE of a disc (I = 0.5 kg m^2) rotating at 10 rad/s.
Answer: KE_rot = (1/2)I omega^2 = 0.5*0.5*100 = 25 J.
