NEET Physics — Chapter 7

System of Particles & Rotational Motion

Centre of mass, angular kinematics, moment of inertia, torque, angular momentum, rolling motion — complete NEET notes with all standard formulas.

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1. Centre of Mass — Definition and Motion

The Centre of Mass (CM) is the point where the entire mass of a system can be assumed to be concentrated for the purpose of translational analysis. For a system of particles:

x_{cm} = rac{m_1 x_1 + m_2 x_2 + cdots}{m_1 + m_2 + cdots} = rac{sum m_i x_i}{sum m_i}

Similarly for $y_{cm}$ and $z_{cm}$ . For continuous bodies, replace sums with integrals.

CM of common uniform bodies:

Body	CM position
Uniform rod	Midpoint
Uniform triangle	Centroid (intersection of medians)
Semicircular ring (radius R)	$2R/pi$ from centre
Semicircular disc (radius R)	$4R/3pi$ from centre
Hemispherical shell (radius R)	$R/2$ from flat face

Motion of CM: The CM of a system moves as if all external forces act on the total mass at that point:

ec{F}_{external} = M_{total} ec{a}_{cm}

Internal forces between particles do not affect CM motion. If net external force = 0, CM moves with constant velocity (or stays at rest).

NEET tip: When an exploding shell or a gun fires a bullet, the CM of the system continues on the original parabolic path — internal forces don't change CM motion. This is a very common NEET concept question.

2. Angular Kinematics

Rotational motion uses angular analogues of linear kinematics. The angular variables are:

Linear	Formula	Angular	Formula
Displacement $s$	—	Angular displacement $heta$	$heta$ (radians)
Velocity $v$	$ds/dt$	Angular velocity $omega$	$d heta/dt$
Acceleration $a$	$dv/dt$	Angular acceleration $alpha$	$domega/dt$
$v = u + at$	—	$omega = omega_0 + alpha t$	—
$s = ut + rac{1}{2}at^2$	—	$heta = omega_0 t + rac{1}{2}alpha t^2$	—
$v^2 = u^2 + 2as$	—	$omega^2 = omega_0^2 + 2alpha heta$	—

Relation between linear and angular quantities (for a point at radius $r$ from axis):

v = romega, quad a_t = ralpha, quad a_c = omega^2 r = rac{v^2}{r}

$a_t$ = tangential acceleration (changes speed), $a_c$ = centripetal acceleration (changes direction).

3. Moment of Inertia

The Moment of Inertia (I) is the rotational analogue of mass. It measures the resistance to angular acceleration:

I = sum m_i r_i^2 = int r^2, dm

SI unit: kg·m². $r$ is the perpendicular distance from the axis of rotation.

Standard moments of inertia (about axis through CM):

Body	I (about CM axis)
Thin ring (radius R) — about diameter	$rac{1}{2}MR^2$
Thin ring — about central axis (⊥ plane)	$MR^2$
Solid disc / cylinder — about central axis	$rac{1}{2}MR^2$
Solid disc — about diameter	$rac{1}{4}MR^2$
Solid sphere — about diameter	$rac{2}{5}MR^2$
Hollow sphere (thin shell) — about diameter	$rac{2}{3}MR^2$
Thin rod — about centre (⊥ rod)	$rac{1}{12}ML^2$
Thin rod — about end (⊥ rod)	$rac{1}{3}ML^2$

Parallel Axis Theorem: $I = I_{cm} + Md^2$ where $d$ is the distance from CM axis to new axis.

Perpendicular Axis Theorem (only for planar laminas): $I_z = I_x + I_y$ where $z$ is perpendicular to the plane.

Radius of Gyration (K): $I = MK^2 implies K = sqrt{I/M}$ . It is the distance from axis at which whole mass can be assumed to be concentrated.

4. Torque, Angular Momentum, and Newton's 2nd Law for Rotation

Torque ( $au$ ) is the rotational analogue of force. It is the turning effect of a force:

ec{ au} = ec{r} imes ec{F} quad Rightarrow quad au = rFsin heta = F cdot d

where $d = rsin heta$ is the perpendicular distance (moment arm) from the axis to the line of action of force.

Newton's 2nd Law for Rotation:

ec{ au}_{net} = I ec{alpha}

Angular Momentum (L): Rotational analogue of linear momentum:

ec{L} = I ec{omega} quad ext{(for rigid body)} quad ext{and} quad ec{L} = ec{r} imes ec{p} quad ext{(for particle)}

ec{ au}_{net} = rac{d ec{L}}{dt}

Conservation of Angular Momentum: If net external torque = 0, then $L = Iomega = ext{constant}$ .

Classic examples: skater pulling arms in (I decreases → ω increases), diver tucking (same principle), planet in elliptical orbit (Kepler's 2nd law).

NEET tip: The most tested aspect is conservation of angular momentum. When a person walks toward the centre of a rotating platform, the system's I decreases so ω increases, keeping L constant. This is also why pulsars spin faster as they collapse.

5. Rolling Motion Without Slipping

When a body rolls without slipping, the contact point has zero instantaneous velocity. The condition is:

v_{cm} = Romega quad ext{(rolling condition)}

Any point on the rolling body has velocity = $v_{cm}$ (translation) + $Romega$ (rotation). At the contact point, these cancel (v = 0). At the top, they add (v = 2v_{cm}).

Total KE of rolling body:

KE_{rolling} = rac{1}{2}mv_{cm}^2 + rac{1}{2}I_{cm}omega^2 = rac{1}{2}mv_{cm}^2left(1 + rac{K^2}{R^2} ight)

where $K$ = radius of gyration. For solid sphere: $KE = rac{7}{10}mv^2$ . For hollow sphere: $KE = rac{5}{6}mv^2$ . For disc: $KE = rac{3}{4}mv^2$ . For ring: $KE = mv^2$ .

Speed of rolling body at bottom of incline (height $h$ , starting from rest):

v = sqrt{ rac{2gh}{1 + K^2/R^2}}

Body with smaller $K^2/R^2$ reaches bottom faster. Ranking (fastest first): solid sphere ( $rac{2}{5}$ ) > solid disc ( $rac{1}{2}$ ) > hollow sphere ( $rac{2}{3}$ ) > ring ( $1$ ).

Pro tip: Static friction provides the torque for rolling — it does NOT dissipate energy in pure rolling. Work done by static friction in rolling without slipping = 0 (since contact point has zero velocity). Only kinetic friction (slipping) dissipates energy.

6. Equilibrium of Rigid Bodies

A rigid body is in mechanical equilibrium when both translational and rotational equilibrium conditions are met simultaneously:

sum ec{F} = 0 quad ext{(translational equilibrium)}

sum ec{ au} = 0 quad ext{(rotational equilibrium — about ANY axis)}

The second condition can be applied about any convenient axis — choose the axis to eliminate unknown forces from the torque equation.

Principle of Moments: For a lever in equilibrium: $F_1 cdot d_1 = F_2 cdot d_2$ (clockwise torque = anticlockwise torque).

Example — beam supported at two points: A uniform beam of mass $M$ and length $L$ is supported at distances $a$ and $b$ from each end. Taking torque about one support eliminates its reaction from the equation, giving the other reaction directly.

Caution: In equilibrium problems,

sum F = 0

gives two equations (x and y components) and

sum au = 0

gives one more — three equations for three unknowns. If you have more unknowns, the problem is statically indeterminate (not in NEET scope).

NEET tip: Always take torque about the point where the most unknowns act — usually an endpoint or support. The weight of a uniform body acts at its CM (midpoint for a uniform rod).

7. NEET Traps & Formula Summary

Trap 1 — Moment of inertia depends on axis: The same body has different I values about different axes. Memorise the standard ones and use parallel/perpendicular axis theorems.

Trap 2 — Rolling vs sliding: A body slides (frictionless) when $v_{cm} eq Romega$. In rolling without slipping, static friction acts — it does NO work and does NOT dissipate energy.

Trap 3 — Angular momentum vs angular velocity:

L = Iomega

. When a skater pulls arms in,

I

decreases and

omega

increases to keep

L

constant. Neither momentum is conserved here — only angular momentum.

Trap 4 — Direction of torque: Torque is a vector (axial vector = pseudovector). Use the right-hand rule: curl fingers from

ec{r}

ec{F}

, thumb points in direction of

ec{ au}

Formula Sheet:

Torque	$au = r Fsin heta = Ialpha$
Angular momentum	$L = Iomega = mvrsin heta$
Rolling KE	$rac{1}{2}mv^2(1 + K^2/R^2)$
Rolling speed (incline)	$v=sqrt{2gh/(1+K^2/R^2)}$
Solid sphere I	$rac{2}{5}MR^2$
Hollow sphere I	$rac{2}{3}MR^2$
Solid disc I (axis)	$rac{1}{2}MR^2$
Ring I (axis)	$MR^2$
Parallel axis theorem	$I = I_{cm} + Md^2$
Perp. axis theorem	$I_z = I_x + I_y$ (lamina)

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