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JEE Mains Physics · System of Particles and Rotational Motion

Rotational Kinematics, Torque and Equilibrium

Angles, angular speeds and angular accelerations follow the same equations as straight-line motion; torque, r × F, is what turns a body, and a body in equilibrium has zero net force and zero net torque.

All System of Particles and Rotational Motion notesJEE Mains Physics strategy

Why this matters

Seventeen PYQs, fourteen of them multiple choice, and three from 2026. Five are rotational kinematics, with a constant or a time-varying angular acceleration. Eight find a torque from r × F or pick out the correct expressions for it. Four balance a rod, a metre scale or a plate.

Concept 1 of 3: Rotational kinematics with constant and varying angular acceleration

Rotation about a fixed axis is one-dimensional motion in disguise: angle θ plays the part of distance, ω of speed and α of acceleration. With α constant, the three equations of motion carry over letter for letter. With α changing in time, differentiate or integrate instead.

Definition

  • Constant α: ω=ω0+αt\omega = \omega_0 + \alpha t, θ=ω0t+12αt2\theta = \omega_0t + \tfrac{1}{2}\alpha t^{2}, ω2=ω02+2αθ\omega^{2} = \omega_0^{2} + 2\alpha\theta, and θ=12(ω0+ω)t\theta = \tfrac{1}{2}(\omega_0 + \omega)t.
  • From rest, the angles turned in successive equal time intervals are in the ratio 1 : 3 : 5 : 7.
  • Varying α: ω=dθ/dt\omega = d\theta/dt, α=dω/dt\alpha = d\omega/dt. Given α(t)\alpha(t), integrate twice, using the starting ω and θ as the constants.
  • Units: revolutions =θ/2π= \theta/2\pi; 1 rpm=2π601\ \text{rpm} = \dfrac{2\pi}{60} rad/s.

Equations of rotation with constant angular acceleration

ω=ω0+αtθ=ω0t+12αt2ω2=ω02+2αθ\omega = \omega_0 + \alpha t \qquad \theta = \omega_0t + \tfrac{1}{2}\alpha t^{2} \qquad \omega^{2} = \omega_0^{2} + 2\alpha\theta

Worked example

A disc spinning at 600 rpm speeds up uniformly to 1200 rpm in 5 s. Find its angular acceleration and the number of revolutions it makes in those 5 s.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q5Moderate

Example 1 · System of Particles and Rotational Motion · Rotational Kinematics, Torque and Equilibrium

A wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first 2 s it rotates through an angle θ1\theta_{1} and in the next 2 s it rotates through an angle θ2\theta_{2}. The ratio θ2θ1\frac{\theta_{2}}{\theta_{1}} is ____\_\_\_\_ .

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.
Drill 4 more on rotational kinematics with constant and varying angular acceleration

Concept 2 of 3: Torque as the cross product of position and force

A force turns a body more when it is applied further from the axis and more nearly at right angles to the line from the axis. The cross product r × F captures both. A force whose line passes through the point gives no torque about it.

Definition

  • τ⃗=r⃗×F⃗\vec\tau = \vec r \times \vec F, with r⃗\vec r from the chosen point to where the force acts; ∣τ⃗∣=rFsin⁡θ|\vec\tau| = rF\sin\theta.
  • About a point P: r⃗=r⃗force−r⃗P\vec r = \vec r_{\text{force}} - \vec r_P.
  • In a plane: τz=xFy−yFx\tau_z = xF_y - yF_x.
  • τ⃗\vec\tau is perpendicular to both r⃗\vec r and F⃗\vec F.
  • Correct expressions for torque: r⃗×F⃗\vec r \times \vec F, dL⃗dt\dfrac{d\vec L}{dt}, r⃗×dp⃗dt\vec r \times \dfrac{d\vec p}{dt}, and Iα for a fixed axis (I is the moment of inertia, met on the next pages). r⃗×L⃗\vec r \times \vec L is NOT a torque.

Torque

τ⃗=r⃗×F⃗=∣i^j^k^xyzFxFyFz∣\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}

Worked example

A force F⃗=(2i^+3j^−k^)\vec F = (2\hat i + 3\hat j - \hat k) N acts at the point r⃗=(i^−j^+2k^)\vec r = (\hat i - \hat j + 2\hat k) m. Find the torque about the origin.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q4Moderate

Example 2 · System of Particles and Rotational Motion · Rotational Kinematics, Torque and Equilibrium

The position of an object having mass 0.1 kg as a function of time tt is given as r→=(10t2i^+5t3j^)m\overrightarrow{r}=\left( 10t^{2}\widehat{i}+ 5t^{3}\widehat{j} \right)m. At t=1 st = 1\text{ }s, which of the following statements are correct? (A) The linear momentum p→=(2i^+1.5j^)kg⋅m/s\overrightarrow{p} = (2\widehat{i} + 1.5\widehat{j})kg \cdot m/s. (B) The force acting on the object F→=(2i^+3j^)N\overrightarrow{F} = (2\widehat{i} + 3\widehat{j})N. (C) The angular momentum of the object about its origin L→=15k^Js\overrightarrow{L} = 15\widehat{k}Js. (D) The torque acting on the object about its origin τ→=20k^Nm\overrightarrow{\tau} = 20\widehat{k}Nm. Choose the correct answer from the option given below.

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.
Drill 7 more on torque as the cross product of position and force

Concept 3 of 3: Equilibrium of a rigid body

A body stays at rest only if the forces balance AND the turning effects balance. Torques can be taken about any point, so pick the point where an unknown force acts: that force then drops out. A uniform rod's own weight acts at its middle and must be counted.

Definition

  • ∑F⃗=0\sum\vec F = 0 and ∑τ⃗=0\sum\vec\tau = 0 about any point.
  • Take torques about the point where an unknown force acts, such as a pivot or a support.
  • A uniform rod or metre scale: its weight acts at its centre (the 50 cm mark of a metre scale).
  • A uniform bar resting on the ground at one end and held up by a vertical force at the other: torques about the ground end give the holder W/2, at any angle.
  • A net force of zero with a nonzero torque (a couple) still turns the body.

Conditions for equilibrium

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

Worked example

A uniform metre rule of mass 120 g is pivoted at its 30 cm mark. Where must a 300 g mass hang to keep it level?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q92Moderate

Example 3 · System of Particles and Rotational Motion · Rotational Kinematics, Torque and Equilibrium

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

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.
Drill 3 more on equilibrium of a rigid body

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Rotational kinematics with constant and varying angular acceleration

    Equations of rotation with constant angular acceleration

    ω=ω0+αtθ=ω0t+12αt2ω2=ω02+2αθ\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

    τ⃗=r⃗×F⃗=∣i^j^k^xyzFxFyFz∣\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

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

Watch out for (7)

Test yourself on System of Particles and Rotational Motion

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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