Is System of Particles and Rotational Motion part of the EAPCET Physics syllabus?

Yes. System of Particles and Rotational Motion is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in System of Particles and Rotational Motion?

Applying the parallel axis theorem from an axis that does not pass through the centre of mass — I_cm must be the starting point. Using the perpendicular axis theorem for a solid sphere or cylinder — it holds only for plane laminae. Assuming rotational kinetic energy is conserved when angular momentum is conserved — when I decreases, KE = L²/2I actually increases. Mixing up standard results, such as using ½MR² for a ring (correct is MR²) or ML²/3 for a rod about its centre (correct is ML²/12).

EAPCET Physics · Intermediate first year

System of Particles and Rotational Motion

This chapter extends mechanics to extended bodies: centre of mass, torque, moment of inertia, angular momentum and rolling motion.

Subject: Physics
Taught in: Intermediate first year
Physics in the paper: 40 of 160 questions

Key formulas and results

$\vec{r}_{\text{cm}} = \frac{\sum m_{i}\vec{r}_{i}}{M}$
$\vec{\tau} = \vec{r} \times \vec{F} = I\alpha$ and $L = I\omega$
Angular momentum is conserved when the net external torque is zero
Parallel axis: $I = I_{\text{cm}} + Md^{2}$ ; perpendicular axis (laminae): $I_{z} = I_{x} + I_{y}$
Moments of inertia: ring $MR^{2}$ , disc $\tfrac{1}{2}MR^{2}$ , solid sphere $\tfrac{2}{5}MR^{2}$ , rod about its centre $\frac{ML^{2}}{12}$
Rolling without slipping: $v = R\omega$ ; $\text{KE} = \tfrac{1}{2}mv^{2}\left(1 + \frac{k^{2}}{R^{2}}\right)$
Velocity of the centre of mass: $v_{\text{cm}} = \frac{\sum m_{i}v_{i}}{M}$ ; total momentum $= Mv_{\text{cm}}$
Centre of mass of a body with a hole: $x_{\text{cm}} = \frac{A_{1}x_{1} - A_{2}x_{2}}{A_{1} - A_{2}}$
Centre of mass from the centre: semicircular ring $\frac{2R}{\pi}$ , semicircular disc $\frac{4R}{3\pi}$ , hollow hemisphere $\frac{R}{2}$ , solid hemisphere $\frac{3R}{8}$
Centre of mass of a cone measured from its vertex: solid $\frac{3H}{4}$ , hollow $\frac{2H}{3}$
Moments of inertia about the symmetry axis: hollow sphere $\tfrac{2}{3}MR^{2}$ , hollow cylinder $MR^{2}$ , solid cone $\frac{3MR^{2}}{10}$ , hollow cone $\tfrac{1}{2}MR^{2}$ ; rod about one end $\frac{ML^{2}}{3}$
Rectangular plate about a perpendicular axis through its centre: $I = \frac{M(l^{2} + b^{2})}{12}$ ; square plate $\frac{Ml^{2}}{6}$
Radius of gyration: $I = Mk^{2}$ , so $k = \sqrt{I/M}$
Rocket: thrust $F = v_{r}\frac{dm}{dt}$ ; speed after burning (no gravity) $v = v_{r}\ln(M_{0}/M)$

What this chapter covers

Centre of mass
Torque and angular momentum
Moment of inertia
Parallel and perpendicular axis theorems
Conservation of angular momentum
Rolling motion

How it is usually asked

Find a centre of mass
Find a moment of inertia using the axis theorems
Conservation of angular momentum
Bodies rolling down an incline

Understanding System of Particles and Rotational Motion

For an extended body, the centre of mass is the point that moves as if all the mass and all the external forces were concentrated there. Internal forces cannot change its motion. Rotation then has its own versions of the linear quantities: torque plays the role of force, moment of inertia plays the role of mass, and angular momentum L = Iω plays the role of linear momentum.

Moment of inertia depends on how mass is distributed about the chosen axis, not just on the total mass. The parallel axis theorem shifts an axis from the centre of mass; the perpendicular axis theorem works only for flat laminae. When no external torque acts, L is conserved, so reducing I increases ω, as with a spinning skater pulling in her arms.

Rolling without slipping combines translation of the centre of mass with rotation about it, linked by v = Rω. The total kinetic energy is ½mv²(1 + k²/R²), so a body with a smaller k²/R², such as a solid sphere rather than a ring, gains more speed rolling down the same incline. EAPCET often asks for this energy split, speeds at the bottom of inclines, or moments of inertia about shifted axes.

Worked example

EAPCET-style question

A solid sphere rolls without slipping on a horizontal surface. What fraction of its total kinetic energy is rotational?

(1) $\frac{2}{7}$
(2) $\frac{2}{5}$
(3) $\frac{5}{7}$
(4) $\frac{1}{2}$

Solution

For a solid sphere, $I = \tfrac{2}{5}MR^{2}$ , so $\frac{k^{2}}{R^{2}} = \frac{2}{5}$ .
Translational KE $= \tfrac{1}{2}Mv^{2}$ ; rotational KE $= \tfrac{1}{2}I\omega^{2} = \tfrac{1}{2}\left(\tfrac{2}{5}MR^{2}\right)\left(\frac{v}{R}\right)^{2} = \tfrac{1}{5}Mv^{2}$ .
Total KE $= \tfrac{1}{2}Mv^{2} + \tfrac{1}{5}Mv^{2} = \tfrac{7}{10}Mv^{2}$ .
Fraction rotational $= \dfrac{1/5}{7/10} = \frac{2}{7}$ .

Answer: $\frac{2}{7}$

Shortcuts and results to remember

A man of mass $m$ walking a length $L$ on a plank of mass $M$ on a smooth floor moves the plank back by $\frac{mL}{m + M}$ , because the centre of mass stays fixed.
A block of mass $m$ sliding down a smooth wedge (mass $M$ , height $h$ , angle $\theta$ ) on a smooth floor shifts the wedge by $\frac{mh\cot\theta}{m + M}$ .
An internal explosion does not change the path of the centre of mass: if a projectile of range $R$ splits into three equal parts at the top, one retracing its path and one dropping from rest, the third lands at $5R/2$ .
The centre of mass moves in a straight line if $a_{\text{cm}} = 0$ or $a_{\text{cm}}$ is parallel to $v_{\text{cm}}$ ; otherwise (constant $a_{\text{cm}}$ ) it follows a parabola.

Common mistakes to avoid

Applying the parallel axis theorem from an axis that does not pass through the centre of mass — I_cm must be the starting point.
Using the perpendicular axis theorem for a solid sphere or cylinder — it holds only for plane laminae.
Assuming rotational kinetic energy is conserved when angular momentum is conserved — when I decreases, KE = L²/2I actually increases.
Mixing up standard results, such as using ½MR² for a ring (correct is MR²) or ML²/3 for a rod about its centre (correct is ML²/12).

How to prepare it

Memorise the standard moments of inertia; most questions are one theorem away from them.

Once the formulas feel familiar, test them the way the exam does: practise System of Particles and Rotational Motion questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise System of Particles and Rotational Motion chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is System of Particles and Rotational Motion part of the EAPCET Physics syllabus?: Yes. System of Particles and Rotational Motion is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare System of Particles and Rotational Motion for EAPCET?: Memorise the standard moments of inertia; most questions are one theorem away from them.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in System of Particles and Rotational Motion?: Applying the parallel axis theorem from an axis that does not pass through the centre of mass — I_cm must be the starting point. Using the perpendicular axis theorem for a solid sphere or cylinder — it holds only for plane laminae. Assuming rotational kinetic energy is conserved when angular momentum is conserved — when I decreases, KE = L²/2I actually increases. Mixing up standard results, such as using ½MR² for a ring (correct is MR²) or ML²/3 for a rod about its centre (correct is ML²/12).