Physics · Mechanics and Waves

Rotational Motion formulas for JEE

Every Rotational Motion formula you need for JEE, grouped by concept.

FormulasRevision notes
49 formulas3 concepts
01Rotational Motion02Centre of Mass03Moment of Inertia
01

Rotational Motion

26 formulas

Angular acceleration

α⃗=dω⃗dt\vec{\alpha} = \frac{d\vec{\omega}}{dt}

Rate of change of angular velocity.

angular accelerationkinematics

Angular momentum of a system

L⃗=∑(r⃗i×p⃗i)\vec{L} = \sum (\vec{r}_i \times \vec{p}_i)

Total angular momentum vector of a system of particles.

angular momentumsystem

Angular momentum (General)

L⃗=Icmω⃗+M(r⃗cm×v⃗cm)\vec{L} = I_{cm}\vec{\omega} + M(\vec{r}_{cm} \times \vec{v}_{cm})

Total angular momentum in combined translational and rotational motion.

applies whenGeneral rigid body motion.
angular momentumcombined motionjee-advanced

Angular momentum (fixed axis)

Lz=IωL_z = I\omega

Component of angular momentum along the fixed axis of rotation.

applies whenRotation about a fixed axis.
angular momentumfixed axis

Angular momentum of a particle

l⃗=r⃗×p⃗\vec{l} = \vec{r} \times \vec{p}

Angular momentum vector of a single particle about an origin.

angular momentumparticle

Angular momentum magnitude

l=rpsin⁡θ=r⊥p=rp⊥l = r p \sin\theta = r_{\perp} p = r p_{\perp}

Scalar magnitude of the angular momentum of a particle.

angular momentumscalar

Vector product magnitude

∣a⃗×b⃗∣=absin⁡θ|\vec{a} \times \vec{b}| = ab \sin\theta

Magnitude of the cross product of two vectors.

applies whenθ\theta is the smaller angle between the vectors.
vector productcross product

Rotational equation

τ=Iα\tau = I\alpha

Newton's second law for rotational motion about a fixed axis.

applies whenFixed axis of rotation.
dynamicstorqueacceleration

Angular kinematics 1

ω=ω0+αt\omega = \omega_0 + \alpha t

First equation of rotational kinematics.

applies whenConstant angular acceleration.
kinematicsangular velocity

Angular kinematics 2

θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

Second equation of rotational kinematics.

applies whenConstant angular acceleration.
kinematicsangular displacement

Angular kinematics 3

ω2=ω02+2α(θ−θ0)\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)

Third equation of rotational kinematics.

applies whenConstant angular acceleration.
kinematicsvelocity-displacement

Mechanical advantage

M.A.=F1F2=d2d1M.A. = \frac{F_1}{F_2} = \frac{d_2}{d_1}

Principle of moments for a lever.

applies whenIdeal lever in equilibrium.
levermechanical advantage

Acceleration of rolling body

a=gsin⁡θ1+k2R2a = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}

Linear acceleration of a body rolling down an inclined plane.

applies whenPure rolling down an incline of angle θ\theta.
rollingaccelerationinclinejee-advanced

Rolling kinetic energy

K=12mvcm2(1+k2R2)K = \frac{1}{2}mv_{cm}^2 \left(1 + \frac{k^2}{R^2}\right)

Total kinetic energy of a body in pure rolling.

applies whenPure rolling motion.
kinetic energyrollingjee-advanced

Pure rolling condition

vcm=ωRv_{cm} = \omega R

Kinematic condition for a body rolling without slipping.

applies whenRolling without slipping on a stationary surface.
rollingkinematicsjee-advanced

Rotational equilibrium

∑τ⃗i=0\sum \vec{\tau}_i = 0

Condition for rotational equilibrium.

applies whenRigid body in equilibrium.
equilibriumrotational

Rotational kinetic energy

K=12Iω2K = \frac{1}{2}I\omega^2

Kinetic energy of a rigid body rotating about a fixed axis.

applies whenRotation about a fixed axis.
kinetic energyrotation

Rotational power

P=τωP = \tau \omega

Instantaneous power delivered by a torque.

applies whenFixed axis of rotation.
powertorque

Rotational work

W=∫τdθW = \int \tau d\theta

Work done by a torque.

applies whenFixed axis of rotation.
worktorque

Torque vector

τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}

Torque or moment of force about an origin.

torquemoment of force

Torque and angular momentum (Particle)

τ⃗=dl⃗dt\vec{\tau} = \frac{d\vec{l}}{dt}

Rate of change of angular momentum of a particle equals the net torque.

torqueangular momentumdynamics

Torque and angular momentum (System)

τ⃗ext=dL⃗dt\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}

Rate of change of total angular momentum of a system equals the net external torque.

applies whenAssuming internal forces are central.
torqueangular momentumdynamics

Torque magnitude

τ=rFsin⁡θ=r⊥F=rF⊥\tau = r F \sin\theta = r_{\perp} F = r F_{\perp}

Scalar magnitude of torque.

torquescalar

Translational equilibrium

∑F⃗i=0\sum \vec{F}_i = 0

Condition for translational equilibrium.

applies whenRigid body in equilibrium.
equilibriumtranslational

Linear and angular velocity scalar

v=ωr⊥v = \omega r_{\perp}

Scalar relation between linear velocity and angular velocity.

applies whenRotation about a fixed axis.
angular velocitylinear velocityscalar

Linear and angular velocity relation

v⃗=ω⃗×r⃗\vec{v} = \vec{\omega} \times \vec{r}

Vector relation between linear velocity, angular velocity, and position vector.

applies whenRotation about a fixed axis or fixed point.
angular velocitylinear velocitykinematics
02

Centre of Mass

10 formulas

Acceleration of centre of mass

A⃗cm=∑mia⃗iM\vec{A}_{cm} = \frac{\sum m_i \vec{a}_i}{M}

Acceleration vector of the centre of mass of a system of particles.

centre of massaccelerationkinematics

Centroid of a triangle

X=x1+x2+x33,Y=y1+y2+y33X = \frac{x_1 + x_2 + x_3}{3}, \quad Y = \frac{y_1 + y_2 + y_3}{3}

Coordinates of the centre of mass of three equal masses at the vertices of a triangle.

applies whenThree particles of equal mass.
centre of masstrianglecentroid

Centre of mass for continuous body

R⃗=1M∫r⃗dm\vec{R} = \frac{1}{M} \int \vec{r} dm

Integral form for the centre of mass of a continuous mass distribution.

applies whenContinuous rigid body.
centre of massintegralrigid body

Centre of mass coordinates

X=∑mixiM,Y=∑miyiM,Z=∑miziMX = \frac{\sum m_i x_i}{M}, \quad Y = \frac{\sum m_i y_i}{M}, \quad Z = \frac{\sum m_i z_i}{M}

3D Cartesian coordinates for the centre of mass of a system of n particles.

applies whenDiscrete particle system.
centre of masscoordinates

Centre of mass for two particles

X=m1x1+m2x2m1+m2X = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

Location of the centre of mass for a two-particle system in one dimension.

applies whenTwo particle system on a straight line.
centre of masstwo particles

Centre of mass position vector

R⃗=∑mir⃗iM\vec{R} = \frac{\sum m_i \vec{r}_i}{M}

Position vector of the centre of mass for a discrete system of particles.

centre of massvector

Linear momentum of a system

P⃗=MV⃗cm\vec{P} = M \vec{V}_{cm}

Total linear momentum of a system is the total mass times the velocity of the COM.

momentumsystem

Force and momentum of system

dP⃗dt=F⃗ext\frac{d\vec{P}}{dt} = \vec{F}_{ext}

Rate of change of total linear momentum equals net external force.

momentumforcedynamics

Newton's 2nd Law for a system

MA⃗cm=F⃗extM \vec{A}_{cm} = \vec{F}_{ext}

The net external force on a system determines the acceleration of its centre of mass.

applies whenTotal mass M remains constant.
newtons second lawdynamicssystem

Velocity of centre of mass

V⃗cm=∑miv⃗iM\vec{V}_{cm} = \frac{\sum m_i \vec{v}_i}{M}

Velocity vector of the centre of mass of a system of particles.

centre of massvelocitykinematics
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03

Moment of Inertia

13 formulas

MOI of solid cylinder

I=12MR2I = \frac{1}{2}MR^2

Moment of inertia of a solid cylinder about its central axis.

applies whenAxis of cylinder.
moment of inertiasolid cylinder

Moment of inertia definition

I=∑miri2I = \sum m_i r_i^2

Moment of inertia for a discrete system of particles.

applies whenDiscrete particle system.
moment of inertiadefinition

MOI of a diatomic molecule

I=μd2=(m1m2m1+m2)d2I = \mu d^2 = \left(\frac{m_1 m_2}{m_1 + m_2}\right) d^2

Moment of inertia of a diatomic molecule about the centre of mass.

applies whenAxis through COM, perpendicular to bond length.
moment of inertiadiatomicreduced massjee-advanced

MOI of disc

I=12MR2I = \frac{1}{2}MR^2

Moment of inertia of a circular disc about an axis perpendicular to its plane at its centre.

applies whenAxis perpendicular to plane through centre.
moment of inertiadisc

MOI of disc (diameter)

I=14MR2I = \frac{1}{4}MR^2

Moment of inertia of a circular disc about its diameter.

applies whenAxis along diameter.
moment of inertiadisc

MOI of hollow cylinder

I=MR2I = MR^2

Moment of inertia of a hollow cylinder about its central axis.

applies whenAxis of cylinder.
moment of inertiahollow cylinder

MOI of ring

I=MR2I = MR^2

Moment of inertia of a thin circular ring about an axis perpendicular to its plane at its centre.

applies whenAxis perpendicular to plane through centre.
moment of inertiaring

MOI of ring (diameter)

I=12MR2I = \frac{1}{2}MR^2

Moment of inertia of a thin circular ring about its diameter.

applies whenAxis along diameter.
moment of inertiaring

MOI of rod (center)

I=112ML2I = \frac{1}{12}ML^2

Moment of inertia of a thin rod about a perpendicular axis through its midpoint.

applies whenAxis perpendicular to rod through midpoint.
moment of inertiarod

MOI of solid sphere

I=25MR2I = \frac{2}{5}MR^2

Moment of inertia of a solid sphere about its diameter.

applies whenAxis along diameter.
moment of inertiasolid sphere

Parallel axis theorem

I=Icm+Md2I = I_{cm} + Md^2

Theorem to find the moment of inertia about any axis parallel to an axis through the centre of mass.

applies whend is the perpendicular distance between the axes.
moment of inertiaparallel axisjee-advanced

Perpendicular axis theorem

Iz=Ix+IyI_z = I_x + I_y

Theorem relating the moments of inertia of a planar lamina.

applies whenPlanar (2D) body, axes x and y are in the plane, z is perpendicular.
moment of inertiaperpendicular axisjee-advanced

Radius of gyration

I=Mk2I = M k^2

Relationship between moment of inertia, total mass, and radius of gyration.

radius of gyrationmoment of inertia
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