System of Particles and Rotational Motion

System of Particles and Rotational MotionNEET Physics · Class 11 · NCERT Chapter 6

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for System of Particles and Rotational Motion. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Centre of mass calculator

Torque calculator (τ = r × F)

Moment of inertia for the seven NEET shapes

Parallel axis theorem

Rotational kinematics solver

Conservation of angular momentum

Rolling without slipping — body comparison

Centre of mass and torque

The CoM is the weighted-average position of a system. Torque is the rotational analog of force.

Centre of mass

Centre of mass calculator

Place 2–4 particles on a number line. The widget computes the centre of mass and shows it as the larger blue marker on the axis.

Particle 1

m: 1 kg

x: 1 m

Particle 2

m: 3 kg

x: 5 m

x_{cm} = \frac{\sum m _{i} x _{i}}{\sum m _{i}} = 4.000 m

Try this

Move the heavier particle and watch the CoM slide toward it. The CoM is always closer to the heavier mass.
Set both particles to equal mass — CoM lands at the midpoint.
Add a third particle and explore: the CoM is the mass-weighted average of all positions.

Torque

Torque calculator (τ = r × F)

Adjust the lever arm, force and angle. Torque is maximum at 90° and vanishes when the force points along the lever.

Lever arm r: 0.50 m

Force F: 20 N

Angle θ between r and F: 60°

Torque

τ = 8.660 N·m

τ = r F sin θ = 0.50 \times 20 \times sin 60° = 8.660 N \cdot m

Effective lever arm r⊥ = r·sin θ = 0.433 m

Try this

Set θ = 90° → maximum torque τ = r·F. The wrench technique.
Set θ = 0° or 180° → torque is zero. Force points along the lever — pure pull, no twist.
Doubling r doubles τ. That is why long wrenches loosen tight bolts.

Moment of inertia and the axis theorems

The seven NEET shapes plus the parallel axis theorem cover almost every moment-of-inertia question on NEET.

Moment of inertia

Moment of inertia for the seven NEET shapes

Pick a shape and tweak its mass and radius/length. The seven cases here are the standard NEET memorisation set.

Uniform disc, axis through centre, perpendicular to plane.

Mass M: 2 kg

R: 0.50 m

Moment of inertia

I = 0.2500 kg·m²

I = \frac{M R ^{2}}{2} = 0.5000 \times 2 \times 0.5 0^{2} = 0.2500

Try this

Compare solid sphere vs hollow sphere at the same M and R: the hollow one has 5/3× the moment of inertia (its mass is concentrated at the rim).
Compare ring vs disc: the ring has twice the moment of the disc — same mass, same R, but mass concentrated at the rim.
Rod about its end is 4× the rod about its centre. The parallel axis theorem explains why.

Axis theorems

Parallel axis theorem

I about any parallel axis = I about the CoM axis + M·d². Slide d to see the moment grow with the offset.

Mass M: 2 kg

L (length): 1.00 m

Parallel-axis offset d: 0.50 m

I_cm

0.1667 kg·m²

+ M·d²

0.5000 kg·m²

I (about parallel axis)

0.6667 kg·m²

I = I_{cm} + M d^{2} = 0.1667 + 2 \times 0.5 0^{2} = 0.6667

Try this

Set d = 0 → I equals I_cm. The parallel axis theorem reduces to identity at the centre.
Doubling d quadruples the added term Md² (it scales as d²).
For a rod with L = 1 and d = 0.5 (axis through end) → I = ML²/12 + M(L/2)² = ML²/3. Classic NEET result.

Rotational dynamics and conservation

The rotational analogs of the kinematic equations, plus the figure-skater showing conservation of angular momentum in action.

Rotational kinematics

Rotational kinematics solver

Pick one of the three equations and leave one variable blank. The widget fills it in and shows the working.

ω = ω_{0} + α t

ω₀ (rad/s)

ω (rad/s)

α (rad/s²)

t (s)

Answer

ω = 10.000 rad/s

ω = 0 + (2) (5) = 10.000

Try this

These are the same equations as linear kinematics (v→ω, x→θ, a→α) — just rotational analogs.
Set ω₀ = 0 → simulate a wheel starting from rest under constant α.
Switch equations to see why the third one (no t) is useful when time is unknown.

Angular momentum

Conservation of angular momentum (figure skater)

No external torque ⇒ I₁·ω₁ = I₂·ω₂. Pulling arms in shrinks I and spins you faster.

Initial moment I₁ (arms out): 4.0 kg·m²

Initial spin ω₁: 2.0 rad/s

Final moment I₂ (arms in): 1.0 kg·m²

New angular speed

ω₂ = 8.00 rad/s

I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = \frac{4 \times 2}{1} = 8.00

Angular momentum L

8.00 kg·m²/s

Initial KE

8.00 J

Final KE

32.00 J

+24.00 J (work done by skater pulling arms in)

Try this

Halve I₂ → ω₂ doubles. Angular momentum is conserved exactly.
KE rises when the skater pulls arms in — the muscles do positive work to bring mass closer to the axis. NOT a violation of energy conservation.
For a planet in orbit: at perihelion (close to Sun), L = mvr is fixed, so v is largest. Same Law.

Rolling motion

Why a solid sphere wins the race down an incline, and how each body splits its energy between translation and rotation.

Rolling motion

Rolling without slipping — body comparison

Four shapes roll down the same incline. The widget shows speed at the bottom and how energy splits between translation and rotation.

Four bodies start from rest at the top of an incline of height h and roll down without slipping. The smaller the rotational inertia, the more energy goes into translation and the higher the linear speed at the bottom.

Drop height h: 2 m (g = 10 m/s²)

Solid sphere

k²/R² = 2/5

5.35 m/s

Solid disc / cylinder

k²/R² = 1/2

5.16 m/s

Hollow sphere

k²/R² = 2/3

4.90 m/s

Hollow ring / cylinder

k²/R² = 1

4.47 m/s

v^{2} = \frac{2 g h}{1 + k ^{2} / R ^{2}} \Rightarrow v = \frac{2 g h}{1 + k ^{2} / R ^{2}}

How energy splits between translation and rotation

Solid sphere

Trans 71%

Rot 29%

Solid disc / cylinder

Trans 67%

Rot 33%

Hollow sphere

Trans 60%

Rot 40%

Hollow ring / cylinder

Trans 50%

Rot 50%

Try this

Order is the same at any height: solid sphere fastest, then disc, then hollow sphere, then ring.
Notice mass and radius drop out completely — only k²/R² matters. A tennis ball and a marble (both solid spheres) reach the bottom with the same speed.
A solid sphere always splits energy 5/7 trans, 2/7 rot. A ring splits 50/50. Memorise these for NEET.

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