Rotational Motion: JEE Main Physics formulas and notes

Q: How do I solve “Centre of mass of a system” questions in JEE Main?

Several masses or simple shapes are joined together. Find the position of the centre of mass. 1. Choose the axes and the origin. 2. Find each coordinate: Σmx ÷ Σm. 3. For a shape, put its mass at its own centre. Example: 1, 2 and 3 kg at x = 0, 1, 2 m: x(cm) = 4/3 m.

Q: How do I solve “Man on a boat and explosions” questions in JEE Main?

People move on a boat or a body breaks up. Find the displacement or the speed. 1. Check that no external force acts along the line. 2. The centre of mass does not move, so Σm Δx = 0. 3. Use the relative displacement to find each shift. Example: A 50 kg man walks 3 m on a 100 kg boat: the boat moves 1 m the other way.

Q: How do I solve “Body with a hole” questions in JEE Main?

A disc or a sphere has a piece removed. Find the new centre of mass or the moment of inertia. 1. Treat the hole as a negative mass. 2. Write x(cm) = (M x − m x₀) ÷ (M − m). 3. For I, subtract the I of the hole about the same axis. Example: A disc with a hole of radius R/2 at R/2: x(cm) = −R/6.

Q: How do I solve “Torque and equilibrium” questions in JEE Main?

A rod, a beam or a body is held at rest by several forces. Find an unknown force or the position. 1. Write ΣF = 0 in two directions. 2. Take torques about the point where unknown forces meet. 3. Solve and check the signs. Example: A 4 m beam of 100 N with 200 N at 1 m: R(right) = 100 N, R(left) = 200 N.

Q: How do I solve “Ladder and friction” questions in JEE Main?

A ladder rests against a wall. Find the reactions or the least angle without slipping. 1. Balance the forces: N₁ = W, f = N₂. 2. Take torques about the foot of the ladder. 3. Use f ≤ μN₁ for no slipping. Example: μ = 0.5: tanθ ≥ 1, so θ ≥ 45°.

Q: How do I solve “Moment of inertia and the axis theorems” questions in JEE Main?

Find the moment of inertia of a body about an axis not through its centre, or of a composite body. 1. Write I(cm) from the standard formula. 2. Add Md² for a parallel axis. 3. Use the perpendicular axis theorem for flat bodies. Example: Rod 3 kg, 2 m, axis ⟂ at 0.5 m from the centre: I = 1.75 kg m².

Q: How do I solve “Torque and angular acceleration” questions in JEE Main?

A torque acts on a wheel, a rod or a pulley. Find α, the time, ω or the angle. 1. Write the net torque about the axis. 2. Use τ = Iα for α. 3. Then apply the equations of rotation. Example: I = 0.5 kg m², τ = 2 N m: α = 4 rad/s²; after 3 s ω = 12 rad/s.

Q: How do I solve “Massive pulley” questions in JEE Main?

Masses hang over a pulley that has mass. Find the acceleration and the tensions. 1. Write ΣF = ma for each mass. 2. Write τ = Iα for the pulley with a = αr. 3. Solve: the two tensions differ. Example: 2 kg on a 4 kg disc pulley: a = 5 m/s², T = 10 N.

Q: How do I solve “Conservation of angular momentum” questions in JEE Main?

A skater, a person on a turntable or a planet changes its radius. Find the new ω or the energy change. 1. Check that the external torque is zero. 2. Write I₁ω₁ = I₂ω₂. 3. Find the change in kinetic energy ½Iω². Example: I from 4 to 2 kg m²: ω doubles from 3 to 6 rad/s, K doubles from 18 J to 36 J.

Q: How do I solve “Collision with a rotating body” questions in JEE Main?

A bullet or a lump of clay hits a rod, a disc or a door. Find the angular speed after the collision. 1. Take angular momentum about the pivot. 2. Write m v d = (I + m d²)ω. 3. Find the energy lost if asked. Example: 0.1 kg at 5 m/s on a rim of 0.4 m, I = 0.5: ω = 0.39 rad/s.

Rotational Motion is one of the heaviest chapters in mechanics, and JEE Mains usually asks 2 to 3 questions from it. The favourites are moment of inertia with the axis theorems, rolling on an incline, conservation of angular momentum and the centre of mass. Aim for 90 seconds per question. About 12 hours in total is enough.

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Quick revision

x(cm) = (Σmx)/(Σm), M a(cm) = F(ext)
Man on a boat: Σm Δx = 0
Rod (ML²)/(12) (centre), (ML²)/(3) (end); disc ½MR²; ring MR²
Solid sphere (2)/(5)MR², hollow (2)/(3)MR²
I = I(cm) + Md², I(z) = I(x) + I(y) (flat bodies)
τ = r⃗ × F⃗ = Iα = (dL)/(dt)
Massive pulley: a = ((m₁ − m₂)g)/(m₁ + m₂ + I/r²)
K = ½Iω² = (L²)/(2I), P = τω
L = Iω = mvr sinθ, conserved if τ = 0
Rolling: v = ωR, K = ½Mv²(1 + (k²)/(R²))
Incline: a = (g sinθ)/(1 + I/MR²); order: sphere, disc, hollow sphere, ring
Rod hinged at the end: α = (3g)/(2L), ω = √(3g)/(L)
Ladder: tanθ ≥ (1)/(2μ)
Slip to roll: sphere (5)/(7)v₀, disc (2)/(3)v₀, ring ½v₀

Formulas with solved examples

Centre of mass

Centre of mass of particles and bodies

x(cm) = (Σmx)/(Σm)
a continuous body: x(cm) = (∫x dm)/(M)
uniform rod: L/2
semicircular ring: (2R)/(π) from the centre
semicircular disc: (4R)/(3π)
solid hemisphere: (3R)/(8)
hollow hemisphere: (R)/(2)
solid cone: h/4 from the base (hollow cone: h/3)

How to apply. Choose an origin and axes, then find each coordinate separately. For a symmetric body the centre of mass lies on the axis of symmetry. Break a body into simple parts and treat each as a point mass at its own centre.

Example. Masses 1, 2 and 3 kg at x = 0, 1 and 2 m: x(cm) = (0 + 2 + 6) ÷ 6 = 4/3 m.

Motion of the centre of mass

M a(cm) = F(ext)
v(cm) = (Σmv)/(M)
no external force: v(cm) is constant
internal forces do not move the centre of mass
explosion in flight: the pieces’ centre of mass follows the original path

How to apply. The centre of mass behaves as if all the mass sat there and all external forces acted there. If it starts at rest and no external force acts along a line, it stays where it is, so the bodies move so that Σm Δx = 0.

Example. A 50 kg man walks 3 m along a 100 kg boat that is at rest on still water: the boat moves 1 m the other way and the man moves 2 m relative to the water.

Body with a hole

treat the hole as a negative mass at its own centre
disc of radius R with a hole of radius R/2 whose centre is at R/2: x(cm) = −(R)/(6)
mass of the removed part = mass per area × area

How to apply. Take the full body, then subtract the removed piece: x(cm) = (M x − m x₀) ÷ (M − m). The centre of mass of the remaining body moves away from the hole.

Example. A disc of mass M with a hole of mass M/4 centred at R/2: x(cm) = −(M/4)(R/2) ÷ (3M/4) = −R/6 from the disc’s centre.

Torque and equilibrium

Torque

τ⃗ = r⃗ × F⃗
τ = rF sinθ = F × (perpendicular distance from the axis)
τ = Iα
τ = (dL)/(dt)
a force through the axis has no torque

How to apply. Take the perpendicular distance from the axis to the line of the force. Counter-clockwise torque is positive. A couple is two equal and opposite forces: its torque is F × (separation) about any point.

Example. A 20 N force at 30° to a 0.5 m spanner: τ = 0.5 × 20 × sin30° = 5 N m.

Equilibrium of a rigid body

ΣF = 0 and Στ = 0 about any point
take torques about the point where most unknown forces meet
a beam on two supports: find the reactions from the torque balance

How to apply. Write the force balance in two directions and the torque balance about a smart point. Three non-parallel forces in equilibrium must meet at one point.

Example. A 4 m beam of weight 100 N rests on its ends and carries 200 N at 1 m from the left: torque about the left end: R × 4 = 100 × 2 + 200 × 1, so R(right) = 100 N and R(left) = 200 N.

Ladder against a wall

smooth wall, rough floor: N₁ = W and f = N₂
torque about the foot: N₂ L sinθ = W (L)/(2) cosθ, so N₂ = (W cotθ)/(2)
no slipping: f ≤ μN₁ gives tanθ ≥ (1)/(2μ)
rough wall as well: add the friction at the wall

How to apply. The wall is smooth, so its force is horizontal. Balance the forces, then take torques about the foot of the ladder to remove the floor forces.

Example. μ = 0.5 between ladder and floor: tanθ ≥ 1 ÷ (2 × 0.5) = 1, so the ladder stays up for θ ≥ 45°.

Moment of inertia

Moment of inertia of common bodies

ring: MR²
disc or solid cylinder: ½MR²
solid sphere: (2)/(5)MR²
hollow sphere: (2)/(3)MR²
rod about the centre ⟂: (ML²)/(12)
rod about one end: (ML²)/(3)
rectangular plate, axis ⟂ through the centre: (M(a² + b²))/(12)

How to apply. These are for the axis through the centre unless stated. The more the mass is far from the axis, the larger I is: ring > disc, hollow sphere > solid sphere.

Example. A disc of 2 kg and radius 0.5 m: I = ½ × 2 × 0.25 = 0.25 kg m².

Parallel and perpendicular axes theorems

I = I(cm) + Md² (parallel axes)
I(z) = I(x) + I(y) (perpendicular axes, flat bodies only)
disc about a diameter: (MR²)/(4)
ring about a diameter: (MR²)/(2)
disc about a tangent in its plane: (5MR²)/(4)
disc about a tangent ⟂: (3MR²)/(2)

How to apply. The parallel axis theorem needs the axis through the centre of mass as the starting point. The perpendicular axis theorem works only for a plane body with the three axes meeting at one point.

Example. A rod of 3 kg and length 2 m about a ⟂ axis 0.5 m from its centre: I = 3 × 4 ÷ 12 + 3 × 0.25 = 1.75 kg m².

Radius of gyration and composite bodies

I = Mk²
k = √(I)/(M)
rod about one end: k = (L)/(√3)
disc: k = (R)/(√2)
solid sphere: k = R√(2)/(5)
add the I of parts about the same axis
remove a part: subtract its I

How to apply. Reduce a composite body to simple pieces. Convert each piece to the common axis with the parallel axis theorem, then add. A body with a hole has I(full) − I(hole).

Example. Solid sphere of radius R: k = R√(2/5) ≈ 0.63R.

Frames and plates made of rods

square frame of 4 rods, side a, axis ⟂ through the centre: (4ma²)/(3) (m = mass of one rod)
square plate, axis ⟂ through the centre: (Ma²)/(6), about a diagonal or a side through the centre: (Ma²)/(12)
thin rod bent into a shape: add the I of each part about the same axis

How to apply. Treat each rod with the parallel axis theorem: I(rod about the centre) + m d², with d the distance of the rod’s centre from the axis.

Example. Square frame of four rods of mass m and side a, axis ⟂ through the centre: I = 4[ma²/12 + m(a/2)²] = 4ma²/3.

Rotational dynamics

Rotational kinematics

ω = ω₀ + αt
θ = ω₀t + ½αt²
ω² = ω₀² + 2αθ
v = ωr
a(t) = αr
a(c) = ω²r
revolutions = (θ)/(2π)

How to apply. The equations are the same as for straight-line motion with x → θ, v → ω, a → α. Convert revolutions per minute to rad/s by multiplying by 2π/60.

Example. A wheel starts from rest with α = 2 rad/s²: after 5 s ω = 10 rad/s and θ = 25 rad, about 4 revolutions.

Torque, angular acceleration and rotational energy

τ = Iα
rotational kinetic energy K = ½Iω²
work done by a torque W = τθ
power P = τω
K = (L²)/(2I)

How to apply. Use τ(net) = Iα about a fixed axis, with the torque of every force on that axis. Energy methods work well when the angle turned is given.

Example. A flywheel with I = 0.5 kg m² and a torque of 2 N m: α = 4 rad/s², after 3 s ω = 12 rad/s and K = ½ × 0.5 × 144 = 36 J.

Massive pulley with hanging masses

a = ((m₁ − m₂)g)/(m₁ + m₂ + I/r²)
T₁ − T₂ = (Ia)/(r²)
for a disc pulley I/r² = M/2
single mass m on a disc pulley of mass M: a = (mg)/(m + M/2)

How to apply. The two tensions are different when the pulley has mass. Write ΣF = ma for each mass and τ = Iα for the pulley, and use a = αr for a string that does not slip.

Example. A 2 kg mass hangs from a string on a disc pulley of mass 4 kg (g = 10): a = 20 ÷ (2 + 2) = 5 m/s² and T = 2 × (10 − 5) = 10 N.

Rod hinged at one end

released from horizontal: α = (3g)/(2L)
acceleration of the centre of mass = (3g)/(4)
hinge force at release = (Mg)/(4)
at the vertical: ω = √(3g)/(L) and the tip speed = √(3gL)

How to apply. Use I = ML²/3 about the hinge. Energy: Mg(L/2) = ½(ML²/3)ω². At the instant of release, the torque is Mg × L/2 about the hinge.

Example. A rod of 1 m released from horizontal (g = 10): ω = √30 ≈ 5.5 rad/s at the vertical, and the tip moves at 5.5 m/s. For M = 4 kg the hinge force at release is 10 N.

Angular momentum

L⃗ = r⃗ × p⃗
a particle: L = mvr sinθ
a rigid body about a fixed axis: L = Iω
τ = (dL)/(dt)
for a particle in a straight line, L about a point is constant

How to apply. Angular momentum depends on the point chosen. For a particle, use the perpendicular distance of its line of motion from that point.

Example. A 2 kg particle moving at 3 m/s along a line 4 m from the origin: L = 2 × 3 × 4 = 24 kg m²/s.

Conservation of angular momentum

no external torque: L = Iω = constant
I₁ω₁ = I₂ω₂
a skater who pulls in the arms spins faster and gains kinetic energy
a person who steps on a turntable: I₁ω₁ = (I₁ + I₂)ω

How to apply. Check that no external torque acts about the axis. Kinetic energy is not conserved when I changes, because the person does work.

Example. A skater turns at 3 rad/s with I = 4 kg m² and pulls the arms in to I = 2 kg m²: ω = 6 rad/s and K goes from 18 J to 36 J.

Collision with a rotating body

a mass sticking on the rim: m v r = (I + m r²) ω
a bullet hitting a hanging rod at distance d from the hinge: m v d = Iω (bullet passes through or I includes it); if it embeds: m v d = (I + m d²)ω
linear momentum is not conserved when the axle gives an impulse
kinetic energy is lost

How to apply. Take angular momentum about the fixed axis (or the hinge) so that the unknown reaction from the support has no torque.

Example. A 0.1 kg clay ball at 5 m/s sticks to the rim (r = 0.4 m) of a disc with I = 0.5 kg m²: ω = 0.2 ÷ (0.5 + 0.016) = 0.39 rad/s.

Angular impulse

angular impulse = ∫τ dt = ΔL
average torque = (ΔL)/(Δt)
for a constant torque: Δω = (τt)/(I)

How to apply. The change in angular momentum equals the angular impulse. It is the rotational counterpart of the impulse–momentum theorem.

Example. A torque of 4 N m acts for 2 s on a body with I = 2 kg m² starting from rest: ΔL = 8 kg m²/s and ω = 4 rad/s.

Areal velocity and central forces

areal velocity dA/dt = (L)/(2m)
for a central force (gravity): L is constant, so dA/dt is constant (Kepler’s second law)
r v sinθ = constant, e.g. r₁v₁ = r₂v₂ at perihelion and aphelion

How to apply. A central force passes through the centre, so its torque about that centre is zero and L is conserved.

Example. A planet moves at 30 km/s at r = 1 unit and the aphelion is 3 units: v = 30 × 1 ÷ 3 = 10 km/s (at perihelion and aphelion r ⟂ v).

Rolling motion

Rolling without slipping

v(cm) = ωR
a(cm) = αR
K = ½Mv² + ½Iω² = ½Mv²(1 + (k²)/(R²))
share of K that is rotational: ring ½, disc ⅓, solid sphere (2)/(7), hollow sphere (2)/(5)
the contact point is at rest, the top point moves at 2v

How to apply. Rolling is a combination of a translation of the centre and a rotation about it. Friction is static, so it does no work and no energy is lost.

Example. A solid sphere rolling at 7 m/s: K = ½mv²(1 + 2/5) = 0.7mv², and the rotational part 0.2mv² is 2/7 of it.

Rolling down an incline

a = (g sinθ)/(1 + I/MR²)
ring: ½g sinθ
disc: ⅔g sinθ
solid sphere: (5)/(7)g sinθ
hollow sphere: (3)/(5)g sinθ
the smaller I/MR², the faster it arrives
v at the bottom = √(2gh)/(1 + I/MR²)
least friction for rolling: μ ≥ (tanθ)/(1 + MR²/I)

How to apply. Use energy for the speed at the bottom and Newton’s law with torque for the acceleration. Without friction a body would only slide; with enough friction it rolls.

Example. Solid sphere on a 30° incline (g = 10): a = (5/7) × 10 × 0.5 = 3.57 m/s² and μ ≥ 0.577 ÷ (1 + 5/2) = 0.165.

Force on a rolling body

a force F at the centre of a disc that rolls: a = (2F)/(3M) and friction = (F)/(3) (backward)
a force at the top: a = (4F)/(3M)
friction acts forward or backward as needed for pure rolling
static friction does no work

How to apply. Write F − f = Ma and fR = Iα with a = αR. The friction comes out of the equations, so check its direction from the sign.

Example. A disc of 3 kg pulled by 12 N at the centre rolls: a = 12 ÷ (3 + 1.5) = 8/3 m/s² and the friction is 4 N, backward.

From slipping to rolling

a body thrown with v₀ and no spin on a rough floor: rolling begins when v = ωR
angular momentum about the contact point is conserved: v = (v₀)/(1 + k²/R²)
solid sphere: (5)/(7)v₀
disc: (2)/(3)v₀
ring: ½v₀
friction does negative work while it slips

How to apply. While the surface slips, kinetic friction acts. Use conservation of angular momentum about any point on the floor, because the friction passes through it.

Example. A solid sphere given 7 m/s with no spin on a rough floor ends up rolling at 5 m/s.

Velocity of points and the instantaneous axis

contact point = instantaneous axis
speed of a point at distance r from the contact point: v = ωr
top point: 2v
point at angle φ from the vertical through the contact point: v = 2v(cm) cos(φ/2)
K = ½I(P)ω² about the contact point, I(P) = I(cm) + MR²
L about the contact point = I(P)ω

How to apply. In pure rolling every point moves, at that instant, as if rotating about the contact point with the same ω. The point at the level of the centre has speed √2 v. Using the contact point gives K and L in one step.

Example. A ring of mass M rolling at v: K = ½(MR² + MR²)(v/R)² = Mv².

Rolling time and work-energy on an incline

time to roll a length s down an incline from rest: t = √(2s)/(a) with a = (g sinθ)/(1 + k²/R²)
friction f = (Mg sinθ)/(1 + R²/k²) (up the slope)
a body that slips on a rough incline (μ < the least value): a = g(sinθ − μ cosθ), friction μMg cosθ

How to apply. Compare rolling and sliding: rolling arrives later because part of the energy goes into rotation. Friction does no work in pure rolling, but does negative work on the translation and positive work on the rotation.

Example. Disc on a 30° incline (g = 10), length 3 m: a = (2/3) × 5 = 10/3 m/s² and t = √(6 ÷ 10/3) = 1.34 s.

Most asked question types

How do I solve “Centre of mass of a system” questions in JEE Main?

Several masses or simple shapes are joined together. Find the position of the centre of mass. 1. Choose the axes and the origin. 2. Find each coordinate: Σmx ÷ Σm. 3. For a shape, put its mass at its own centre. Example: 1, 2 and 3 kg at x = 0, 1, 2 m: x(cm) = 4/3 m.

How do I solve “Man on a boat and explosions” questions in JEE Main?

People move on a boat or a body breaks up. Find the displacement or the speed. 1. Check that no external force acts along the line. 2. The centre of mass does not move, so Σm Δx = 0. 3. Use the relative displacement to find each shift. Example: A 50 kg man walks 3 m on a 100 kg boat: the boat moves 1 m the other way.

How do I solve “Body with a hole” questions in JEE Main?

A disc or a sphere has a piece removed. Find the new centre of mass or the moment of inertia. 1. Treat the hole as a negative mass. 2. Write x(cm) = (M x − m x₀) ÷ (M − m). 3. For I, subtract the I of the hole about the same axis. Example: A disc with a hole of radius R/2 at R/2: x(cm) = −R/6.

How do I solve “Torque and equilibrium” questions in JEE Main?

A rod, a beam or a body is held at rest by several forces. Find an unknown force or the position. 1. Write ΣF = 0 in two directions. 2. Take torques about the point where unknown forces meet. 3. Solve and check the signs. Example: A 4 m beam of 100 N with 200 N at 1 m: R(right) = 100 N, R(left) = 200 N.

How do I solve “Ladder and friction” questions in JEE Main?

A ladder rests against a wall. Find the reactions or the least angle without slipping. 1. Balance the forces: N₁ = W, f = N₂. 2. Take torques about the foot of the ladder. 3. Use f ≤ μN₁ for no slipping. Example: μ = 0.5: tanθ ≥ 1, so θ ≥ 45°.

How do I solve “Moment of inertia and the axis theorems” questions in JEE Main?

Find the moment of inertia of a body about an axis not through its centre, or of a composite body. 1. Write I(cm) from the standard formula. 2. Add Md² for a parallel axis. 3. Use the perpendicular axis theorem for flat bodies. Example: Rod 3 kg, 2 m, axis ⟂ at 0.5 m from the centre: I = 1.75 kg m².

How do I solve “Torque and angular acceleration” questions in JEE Main?

A torque acts on a wheel, a rod or a pulley. Find α, the time, ω or the angle. 1. Write the net torque about the axis. 2. Use τ = Iα for α. 3. Then apply the equations of rotation. Example: I = 0.5 kg m², τ = 2 N m: α = 4 rad/s²; after 3 s ω = 12 rad/s.

How do I solve “Massive pulley” questions in JEE Main?

Masses hang over a pulley that has mass. Find the acceleration and the tensions. 1. Write ΣF = ma for each mass. 2. Write τ = Iα for the pulley with a = αr. 3. Solve: the two tensions differ. Example: 2 kg on a 4 kg disc pulley: a = 5 m/s², T = 10 N.

How do I solve “Conservation of angular momentum” questions in JEE Main?

A skater, a person on a turntable or a planet changes its radius. Find the new ω or the energy change. 1. Check that the external torque is zero. 2. Write I₁ω₁ = I₂ω₂. 3. Find the change in kinetic energy ½Iω². Example: I from 4 to 2 kg m²: ω doubles from 3 to 6 rad/s, K doubles from 18 J to 36 J.

How do I solve “Collision with a rotating body” questions in JEE Main?

A bullet or a lump of clay hits a rod, a disc or a door. Find the angular speed after the collision. 1. Take angular momentum about the pivot. 2. Write m v d = (I + m d²)ω. 3. Find the energy lost if asked. Example: 0.1 kg at 5 m/s on a rim of 0.4 m, I = 0.5: ω = 0.39 rad/s.

How do I solve “Rolling energy and speed” questions in JEE Main?

A body rolls without slipping. Find the kinetic energy, the share that is rotational or the speed at the bottom. 1. Write K = ½Mv²(1 + k²/R²). 2. For a ramp, equate it to Mgh. 3. Use the share: ring ½, disc ⅓, sphere 2/7. Example: A solid sphere rolling at v: K = 0.7Mv², and 2/7 of it is rotational.

How do I solve “Rolling down an incline” questions in JEE Main?

Bodies roll down a slope. Find the acceleration, the order of arrival or the least friction. 1. Use a = g sinθ ÷ (1 + I/MR²). 2. Compare I/MR²: smaller wins. 3. μ(min) = tanθ ÷ (1 + MR²/I). Example: Solid sphere on 30°: a = 3.57 m/s², μ ≥ 0.165.

How do I solve “Slipping and rolling” questions in JEE Main?

A body is thrown or spun on a rough floor. Find the speed when it begins to roll. 1. Conserve angular momentum about the contact line. 2. Rolling starts when v = ωR. 3. For a sphere, v = 5v₀/7. Example: 7 m/s thrown with no spin: it rolls at 5 m/s.

Common myths and the truth

Myth: Torque and force are the same kind of quantity. Truth: Torque is force times perpendicular distance. A large force through the axis gives no torque.
Myth: The centre of mass is always inside the body. Truth: For a ring or a bent rod it lies outside the material.
Myth: Moment of inertia depends only on the mass. Truth: It depends on how the mass is spread about the axis, so one body has different values for different axes.
Myth: The perpendicular axis theorem works for every body. Truth: Only for flat (plane) bodies.
Myth: A rolling body has no friction. Truth: Static friction acts, but it does no work in pure rolling.
Myth: A heavier body rolls down a slope faster. Truth: The acceleration depends on I/MR², not on the mass or the radius.
Myth: Angular momentum conservation means the kinetic energy is conserved. Truth: When I changes, the kinetic energy changes. The skater does work when pulling in the arms.
Myth: The contact point of a rolling wheel moves forward fastest. Truth: It is momentarily at rest. The top point moves at 2v.
Myth: A hinged rod released from horizontal falls with acceleration g. Truth: Its centre falls with 3g/4 at that instant, and the hinge supports Mg/4.
Myth: A massless-pulley formula works for a pulley with mass. Truth: The tensions on the two sides differ, and the acceleration is smaller because I/r² is added to the inertia in the denominator.

Mistakes to avoid

Take torque and angular momentum about the same point.
The parallel axis theorem needs the axis through the centre of mass first.
The perpendicular axis theorem works only for flat bodies.
Kinetic energy of a rolling body has two parts: translation and rotation.
Static friction in pure rolling does no work and is not μN.
Angular momentum is conserved only if the external torque is zero, but the energy may change.
A pulley with mass has different tensions on the two sides.
Use I of the whole system about the axis when something sticks to it.

Simulations for Rotational Motion

Conservation of angular momentum — Pull two rotating masses inward and watch the angular speed increase.
Centre of mass of three masses — Change three masses on a light triangular plate. The plate balances only on its centre of mass.
Moment of inertia of shapes — Compare how mass distribution changes rotational inertia.
Parallel-axis theorem — Move a rotation axis away from the centre of a uniform disc and feel its growing inertia.
A rod as a physical pendulum — Swing a uniform rod about one end and connect rotational inertia with its oscillation period.
Rolling without slipping — Compare the motion of the centre and a point on a wheel.
Rolling race: ring, disc, spheres — Release four shapes of equal radius down the same incline. The moment of inertia decides the winner.
Torque and rotation — Apply a tangential force at different distances from a pivot and spin the disc.
Topple or slide? — Push a tall block with a growing force. It either slides first or tips over its front edge first.

More: Class 11 Physics simulations · all JEE Main chapters.