JEE Mains Physics · System of Particles and Rotational Motion

Rolling: Velocities and Kinetic Energy

A body rolls without slipping when v = ωR: its contact point is momentarily at rest, its top moves at 2v, and its kinetic energy is ½mv²(1 + k²/R²), split between translation and rotation in a ratio fixed by its shape.

All System of Particles and Rotational Motion notes JEE Mains Physics strategy

Why this matters

Thirteen PYQs, six of them asking for a number, and one from 2026. Five use the rolling condition v = ωR: the speed of a point on the rim, or the moment a slipping body starts to roll. Eight split a rolling body's kinetic energy between translation and rotation, or find its speed from that energy.

Concept 1 of 2: The rolling condition and the speeds of points on a rolling body

Rolling is a translation of the centre plus a spin about it. Without slipping, the two cancel exactly at the ground: the contact point is at rest for an instant. So the whole body is turning about the contact point at that instant, and every point's speed is ω times its distance from the contact point.

Definition

Rolling without slipping: $v_{cm} = \omega R$ .
Speed of any point $= \omega \times$ its distance from the contact point. Contact point 0; centre v; top 2v; a rim point level with the centre $\sqrt{2}\,v$ .
The top point after half a turn: it has moved $\pi R$ forward and $2R$ down, a displacement of $R\sqrt{\pi^{2} + 4}$ .
Slipping to rolling on a rough floor: kinetic friction $\mu mg$ slows the centre, $v = v_0 - \mu gt$ , and changes the spin, $\alpha = \mu mgR/I$ . Rolling begins when $v = \omega R$ ; friction then stops acting.
A disc launched sliding with no spin: $v_0 - \mu gt = 2\mu gt$ , so $t = \dfrac{v_0}{3\mu g}$ and it rolls on at $\tfrac{2}{3}v_0$ . A ring rolls on at $\tfrac{1}{2}v_0$ .

Rolling without slipping

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

Worked example

A wheel of radius 0.4 m rolls without slipping with its centre moving at 6 m/s. Find its angular speed and the speeds of the top point, the contact point, and a rim point level with the centre.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q7Moderate

Example 1 · System of Particles and Rotational Motion · Rolling: Velocities and Kinetic Energy

A solid cylinder having radius R and length L is slipping on a rough horizontal plane. At time

t = 0

the cylinder has a translational velocity

v_{0}= 49\text{ }m/s

, perpendicular to its axis and a rotational velocity

v_{0}/4R

about the centre. The time taken by the cylinder to start rolling is

\_\_\_\_

seconds. (coefficient of kinetic friction

\mu_{K}= 0.25

and

g = 9.8\text{ }m/s^{2}

)

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.

Drill 4 more on the rolling condition and the speeds of points on a rolling body

Concept 2 of 2: Kinetic energy of rolling bodies

A rolling body has two kinetic energies: ½mv² for moving and ½Iω² for spinning. With v = ωR the total is ½mv²(1 + k²/R²). The number k²/R² depends only on the shape, so each shape splits its energy in a fixed ratio, whatever its mass, radius or speed.

Definition

$K = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2} = \tfrac{1}{2}mv^{2}\left(1 + \dfrac{k^{2}}{R^{2}}\right)$ .
Rotational share $= \dfrac{k^{2}/R^{2}}{1 + k^{2}/R^{2}}$ ; translational : rotational $= 1 : k^{2}/R^{2}$ .
Given the total kinetic energy, $v^{2} = \dfrac{2K}{m(1 + k^{2}/R^{2})}$ .
Angular momentum of a rolling body about a point on the ground below its path: $mvR + I\omega$ . For a thin spherical shell, $mR^{2}\omega + \tfrac{2}{3}mR^{2}\omega = \tfrac{5}{3}mR^{2}\omega$ .

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.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q79Moderate

Example 2 · System of Particles and Rotational Motion · Rolling: Velocities and Kinetic Energy

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

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.

Drill 7 more on kinetic energy of rolling bodies

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 (1)

The rolling condition and the speeds of points on a rolling body
Rolling without slipping
$v_{cm} = \omega R \qquad v_P = \omega\, r_P \quad (r_P = \text{distance of P from the contact point})$

Reference tables (1)

Kinetic energy of rolling bodies4 rows

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.

Watch out for (4)

The top of a rolling wheel moves at 2v
→ The rolling condition and the speeds of points on a rolling body
Friction acts only while the body slips
→ The rolling condition and the speeds of points on a rolling body
½mv² alone for a rolling body
→ Kinetic energy of rolling bodies
Share of the total, or ratio of the parts
→ Kinetic energy of rolling bodies

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.

Take the System of Particles and Rotational Motion chapter test Drill all 13 questions