System of Particles and Rotational Motion — JEE Mains Physics Formulas

Centre of Mass and Its Motion

Locating the centre of mass

Centre of mass

x_{cm} = \frac{\sum m_i x_i}{\sum m_i} \qquad x_{cm} = \frac{\int x\,dm}{\int dm} \qquad x_{cm} = \frac{Mx_1 - mx_2}{M - m}

Motion of the centre of mass, explosions and recoil

Velocity of the centre of mass, and Newton's second law for a system

\vec v_{cm} = \frac{\sum m_i\vec v_i}{\sum m_i} \qquad M\vec a_{cm} = \vec F_{ext}

Common traps

Averaging positions without the masses

The centre of mass of 1 kg at x = 0 and 3 kg at x = 4 m is at 3 m, not at the midpoint 2 m. Each position must be multiplied by its mass before dividing by the total mass.

A hole's mass goes with its area

A circular hole of half the disc's radius removes a quarter of the mass, not half. For a uniform solid, mass goes with volume, so half the radius removes an eighth.

Semicircular ring and semicircular disc differ

A ring has all its mass on the rim, so its centre of mass is further out: 2R/π ≈ 0.64R. A disc has mass spread inwards: 4R/3π ≈ 0.42R.

Equal momenta, not equal speeds

After a body at rest explodes into two pieces, the pieces have equal and opposite MOMENTA. The lighter piece is faster and carries more kinetic energy, in the inverse ratio of the masses.

Signs in an Atwood machine

One block goes up while the other comes down, so their accelerations enter a_cm with opposite signs. Adding them as if both moved down gives a_cm = a, which is too large.

An accelerating centre of mass need not curve

A constant acceleration gives a straight path when the velocity of the centre of mass is along it, or zero. Only an acceleration at an angle to the velocity gives a parabola.

Rotational Kinematics, Torque and Equilibrium

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Rotational kinematics with constant and varying angular acceleration

Equations of rotation with constant angular acceleration

\omega = \omega_0 + \alpha t \qquad \theta = \omega_0t + \tfrac{1}{2}\alpha t^{2} \qquad \omega^{2} = \omega_0^{2} + 2\alpha\theta

Torque as the cross product of position and force

Torque

\vec\tau = \vec r \times \vec F = \begin{vmatrix} \hat i & \hat j & \hat k \\ x & y & z \\ F_x & F_y & F_z \end{vmatrix}

Equilibrium of a rigid body

Conditions for equilibrium

\sum\vec F = 0 \qquad \sum\vec\tau_{\text{about any point}} = 0

Common traps

rpm left unconverted

The equations need ω in rad/s. 600 rpm is 20π rad/s, not 600 and not 10 (that is revolutions per second, still to be multiplied by 2π).

The angle in an interval is a difference

The angle turned 'in the next 2 s' is θ(4) − θ(2), not θ(2) again with a new start. From rest, consecutive equal intervals give angles in the ratio 1 : 3 : 5.

F × r has the wrong sign

The torque is r × F. Writing F × r reverses every component. Keep r in the middle row of the determinant and F in the bottom row.

Torque about a point that is not the origin

Use the vector from that point to where the force acts, r − r_P. Using the force point's own position vector gives the torque about the origin instead.

The middle term of the determinant

The ĵ component is −(xF_z − zF_x). Dropping that minus sign is the most common slip in a three-dimensional torque.

Forgetting the rod's own weight

A metre scale pivoted anywhere but its centre has its own weight acting off the pivot. Leaving it out balances only the hanging masses and gives a wrong answer that is usually among the options.

Distances from the end, not from the pivot

A mark on a scale is a position. The lever arm is the distance from that mark to the pivot: a mass at the 10 cm mark with the pivot at 40 cm has a 30 cm arm.

Moment of Inertia and Radius of Gyration

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Moment of inertia when a body is resized or reshaped

Disc in terms of density, and mass from volume

I_{\text{disc}} = \tfrac{1}{2}MR^{2} = \tfrac{1}{2}\rho\pi R^{4}t \qquad M = \rho V

Moments of inertia of standard bodies

Body	Axis	Moment of inertia	Radius of gyration
Thin ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$
Thin ring, radius R	A diameter	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	Through the centre, normal to the plane	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	A diameter	$\tfrac{1}{4}MR^{2}$	$R/2$ Half the normal-axis value, by the perpendicular-axis theorem.
Solid cylinder, radius R	Its own axis	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Thin-walled hollow cylinder, radius R	Its own axis	$MR^{2}$	$R$
Solid cylinder, radius R, length L	Through the centre, normal to its axis	$M\left(\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}\right)$	$\sqrt{\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}}$
Solid sphere, radius R	A diameter	$\tfrac{2}{5}MR^{2}$	$\sqrt{2/5}\,R$
Thin hollow sphere, radius R	A diameter	$\tfrac{2}{3}MR^{2}$	$\sqrt{2/3}\,R$
Thin rod, length L	Through the centre, normal to the rod	$\tfrac{1}{12}ML^{2}$	$L/(2\sqrt{3})$
Thin rod, length L	Through one end, normal to the rod	$\tfrac{1}{3}ML^{2}$	$L/\sqrt{3}$
Rectangular plate, sides a and b	Through the centre, normal to the plate	$\tfrac{1}{12}M(a^{2} + b^{2})$	$\sqrt{(a^{2} + b^{2})/12}$
Semicircular ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$

M is the body's mass. Every value is about an axis through the centre of mass, except the rod about its end.

Common traps

Disc about a diameter is not MR²/2

MR²/2 is a disc about its normal axis. About a diameter it is MR²/4. A ring about a diameter is MR²/2, which is easy to confuse with the disc.

The radius of gyration is squared in I

I = Mk², so k = √(I/M). A ratio of radii of gyration is the square root of a ratio of moments of inertia (for equal masses), not the ratio itself.

Read the radius off the stem

If the bodies have radius 2R, every I picks up a factor of 4. A stem that gives a diameter needs halving before it goes into MR².

The new piece keeps the old mass

A piece cut or remoulded from a body has its own mass, found from its volume. Putting the original M into the new piece's formula is the usual wrong answer.

I ∝ R⁴ only for the same material and thickness

For equal masses, I goes as R². The fourth power appears only when the mass itself grows with R², as for discs cut from the same sheet.

Parallel and Perpendicular Axes and Composite Bodies

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Parallel-axis and perpendicular-axis theorems

Axis theorems

I = I_{cm} + Md^{2} \qquad I_z = I_x + I_y \qquad k^{2} = k_{cm}^{2} + d^{2}

Composite bodies and bodies with a piece removed

Adding parts and removing a hole

I_{\text{system}} = \sum I_{\text{part}} \qquad I_{\text{rest}} = I_{\text{full}} - \left(I_{\text{hole}} + m_{\text{hole}}d^{2}\right)

Common traps

The parallel-axis theorem starts at the centre of mass

I = I_cm + Md² holds only when one of the two axes passes through the centre of mass. To move between two other axes, go back to the centre-of-mass axis first, then out again.

The perpendicular-axis theorem is for flat bodies

I_z = I_x + I_y works for a disc, a ring or a plate. It fails for a sphere or a cylinder: a sphere's three diameters all give 2MR²/5, not 4MR²/5 for one of them.

A diameter given in place of a radius

A ring 'of diameter r' has radius r/2. Its tangent in the plane gives (3/2)M(r/2)² = 3Mr²/8, not 3Mr²/2.

Parts about different axes

Each part's I must be about the system's axis before they are added. Adding each sphere's 2mR²/5 about its own centre leaves out the md² that usually dominates.

The removed piece has its own moment of inertia

Subtracting only m_hole·d² treats the hole as a point. Its own (1/2)m_hole·r² about its centre must go too.

Distance from the axis, not from the origin

For a point mass, r is the perpendicular distance to the axis. A mass on the axis contributes nothing, however far it is from the origin.

Torque, Angular Acceleration and Rotational Energy

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Torque and angular acceleration about a fixed axis

Rotational second law, heavy pulley, and power

\tau = I\alpha \qquad a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^{2}} \qquad P = \tau\omega

Rotational kinetic energy and energy conservation

Rotational kinetic energy, and energy for a block on a wheel

K_{\text{rot}} = \tfrac{1}{2}I\omega^{2} \qquad mgh = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2},\ v = \omega R

Common traps

T = mg for an accelerating block

If the block accelerates downward, the string pulls with less than its weight: T = m(g − a). Using T = mg makes the pulley's acceleration too large.

Equal tensions over a heavy pulley

Equal tensions on both sides give zero net torque, so a pulley with mass could never start turning. With a massive pulley the tensions differ, and I/R² joins the masses in the denominator.

Angular speed in rpm

τ = Iω₀/t needs ω₀ in rad/s. 1200 rpm is 40π rad/s; leaving it as 1200, or as 20 revolutions a second, gives a torque off by a factor of 2π/60 or of 2π.

Leaving out the block's kinetic energy

The falling block is moving too. mgh = ½Iω² alone gives the wheel all the energy; the block's ½mv² must be on the same side.

A rod's centre of mass falls half as far

A rod falling from upright to flat about its foot lowers its centre of mass by L/2, not L. Using L gives √(6g/L) in place of √(3g/L).

Moment of inertia about the pivot

A rod turning about its end has I = mL²/3 about that end. Using mL²/12, the value about its centre, ignores the motion of the centre of mass.

Angular Momentum and Its Conservation

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Angular momentum of a particle

Angular momentum of a particle, and of a projectile at its highest point

\vec L = \vec r \times m\vec v,\ |\vec L| = mvd \qquad L_{\text{top}} = \frac{mu^{3}\sin^{2}\theta\cos\theta}{2g}

Conservation of angular momentum

Conservation of angular momentum, and energy lost on locking

I_1\omega_1 = I_2\omega_2 \qquad \Delta K = \frac{I_1I_2(\omega_1 - \omega_2)^{2}}{2(I_1 + I_2)}

Common traps

Distance to the point, not to the line

L = mvd uses the perpendicular distance from the point to the LINE of motion. The particle's straight-line distance from the point changes as it moves; d does not.

Angular momentum about a moving body

For one car about another, use the relative velocity. Using the first car's own speed gives its angular momentum about a fixed point beside the track.

A projectile's L is not conserved about the launch point

Gravity has a torque mgx about the launch point, so L changes along the path. Only about a point on the line of the force could L stay constant.

Kinetic energy is not conserved when bodies stick

When a disc lands on a spinning disc, or beads stick to a ring, angular momentum is conserved but kinetic energy is lost. Equating the energies gives ω' = ω√(I₁/(I₁ + I₂)), which is wrong.

Day length goes with the radius squared

With fixed mass, I ∝ R², so T ∝ R². A volume change must first become a radius change: a sixty-fourth of the volume is a quarter of the radius, and T falls to a sixteenth.

About which point is L conserved?

For a body striking a rod on a fixed pivot, the pivot pushes during the impact, so linear momentum is not conserved. Angular momentum about the pivot is, because the pivot's force has no lever arm there.

Rolling: Velocities and Kinetic Energy

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The rolling condition and the speeds of points on a rolling body

Rolling without slipping

v_{cm} = \omega R \qquad v_P = \omega\, r_P \quad (r_P = \text{distance of P from the contact point})

Kinetic energy of rolling bodies

Body	k²/R²	Total kinetic energy	Rotational share	Translational : rotational
Ring or thin hollow cylinder	1	$mv^{2}$	$\tfrac{1}{2}$	1 : 1
Disc or solid cylinder	$\tfrac{1}{2}$	$\tfrac{3}{4}mv^{2}$	$\tfrac{1}{3}$	2 : 1
Solid sphere	$\tfrac{2}{5}$	$\tfrac{7}{10}mv^{2}$	$\tfrac{2}{7}$	5 : 2
Thin hollow sphere (shell)	$\tfrac{2}{3}$	$\tfrac{5}{6}mv^{2}$	$\tfrac{2}{5}$	3 : 2 The shell's rotational share, 2/5, is the same number as the solid sphere's k²/R². Keep the two apart.

Every row follows from ½mv²(1 + k²/R²); only k²/R² changes from shape to shape.

Common traps

The top of a rolling wheel moves at 2v

The centre moves at v and the top also turns forward at ωR = v about the centre, so the top moves at 2v. Only the centre moves at v.

Friction acts only while the body slips

Kinetic friction μmg acts until v = ωR. After that the body rolls on a level floor with no friction and no further change in speed.

½mv² alone for a rolling body

A body rolling at speed v has more than ½mv²: the spin adds ½Iω². Finding v from K = ½mv² overestimates the speed of every rolling body.

Share of the total, or ratio of the parts

For a solid sphere, rotational over total is 2/7, but rotational over translational is 2/5. Read whether the question divides by the total or by the other part.

Rolling on Inclines: Acceleration, Speed and Friction

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Speed and height of a body rolling on a slope

Speed at the foot of a slope, and height climbed

v = \sqrt{\frac{2gh}{1 + k^{2}/R^{2}}} \qquad h_{\max} = \frac{v^{2}}{2g}\left(1 + \frac{k^{2}}{R^{2}}\right)

Acceleration and friction of a rolling body

Acceleration and friction when rolling down a slope

a = \frac{g\sin\theta}{1 + k^{2}/R^{2}} \qquad f = \frac{mg\sin\theta\,(k^{2}/R^{2})}{1 + k^{2}/R^{2}}

Common traps

The slope's length is not the height

The energy equation needs the vertical drop h. A slope of length l at angle θ drops h = l sin θ; using l gives too large a speed.

A smooth slope does not stop the spin

Without friction nothing can slow the rotation, so only the translational energy ½mv² becomes height. Using the rolling formula on a smooth slope overstates the height.

Mass and radius do not decide the race

Two solid spheres of different sizes reach the bottom together. Only the shape, through k²/R², changes the speed.

A rolling body does not accelerate at g sin θ

g sin θ is the frictionless sliding value. Rolling divides it by 1 + k²/R²: a solid sphere gets 5/7 of it, a ring only half.

Friction pushes forward when the force is at the top

A force at the top tends to spin the body faster than v = ωR allows, so friction acts forward, adding to F. Taking it backward gives an acceleration below F/m for a sphere, which is wrong.

Needed friction is not μN

Static friction takes whatever value rolling needs, up to μN. Put f = μmg cos θ only when asking for the least μ, or when the body slips.

System of Particles and Rotational Motion formulas

Centre of Mass and Its Motion

Locating the centre of mass

Motion of the centre of mass, explosions and recoil

Common traps

Rotational Kinematics, Torque and Equilibrium

Rotational kinematics with constant and varying angular acceleration

Torque as the cross product of position and force

Equilibrium of a rigid body

Common traps

Moment of Inertia and Radius of Gyration

Moment of inertia when a body is resized or reshaped

Moments of inertia of standard bodies

Common traps

Parallel and Perpendicular Axes and Composite Bodies

Parallel-axis and perpendicular-axis theorems

Composite bodies and bodies with a piece removed

Common traps

Torque, Angular Acceleration and Rotational Energy

Torque and angular acceleration about a fixed axis

Rotational kinetic energy and energy conservation

Common traps

Angular Momentum and Its Conservation

Angular momentum of a particle

Conservation of angular momentum

Common traps

Rolling: Velocities and Kinetic Energy

The rolling condition and the speeds of points on a rolling body

Kinetic energy of rolling bodies

Common traps

Rolling on Inclines: Acceleration, Speed and Friction

Speed and height of a body rolling on a slope

Acceleration and friction of a rolling body

Common traps

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