JEE Mains Physics · System of Particles and Rotational Motion

Rolling on Inclines: Acceleration, Speed and Friction

A body rolling down a slope accelerates at g sin θ/(1 + k²/R²) and reaches the bottom with v² = 2gh/(1 + k²/R²); static friction supplies the spin and does no work.

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Why this matters

Twenty-two PYQs, nine of them asking for a number, and two from 2026. Thirteen use energy: the speed at the foot of a slope, the height a rolling body climbs, a race between shapes. Nine use forces: the acceleration down a slope, the friction rolling needs, a force applied at the top of a ball.

Concept 1 of 2: Speed and height of a body rolling on a slope

Rolling without slipping wastes no energy: the contact point does not slide, so friction does no work. The height lost becomes ½mv²(1 + k²/R²). A shape with more of its mass far out (larger k²/R²) puts more energy into spin, so it is slower at the bottom. The mass and the radius cancel.

Definition

Down a height h from rest: $mgh = \tfrac{1}{2}mv^{2}\left(1 + \dfrac{k^{2}}{R^{2}}\right)$ , so $v = \sqrt{\dfrac{2gh}{1 + k^{2}/R^{2}}}$ .
Race from rest down one slope: solid sphere first, then disc and solid cylinder (together), then hollow sphere, then ring, whatever their masses and radii.
Rolling UP a rough slope from speed v: $h = \dfrac{v^{2}}{2g}\left(1 + \dfrac{k^{2}}{R^{2}}\right)$ . Distance along the slope $= h/\sin\theta$ .
Up a SMOOTH slope there is no friction to stop the spin: only $\tfrac{1}{2}mv^{2}$ turns into height, $h = v^{2}/2g$ , and the body keeps spinning at the top.
A cylinder unwinding from strings (a yo-yo) obeys the same energy equation: $mgh = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}\left(\tfrac{1}{2}mR^{2}\right)\dfrac{v^{2}}{R^{2}} = \tfrac{3}{4}mv^{2}$ .
Connected bodies (a wheel pulling a block over a pulley): count every body's kinetic energy.

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)

Worked example

A solid sphere rolls without slipping down a slope from rest, losing 1.4 m of height. Find its speed at the bottom (g = 10 m/s²).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 July 2022 · Q5Moderate

Example 1 · System of Particles and Rotational Motion · Rolling on Inclines: Acceleration, Speed and Friction

A solid cylinder and a solid sphere, having same mass

M

and radius

R

, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be:

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.

Drill 12 more on speed and height of a body rolling on a slope

Concept 2 of 2: Acceleration and friction of a rolling body

Gravity's pull along the slope must both speed up the centre and spin the body up. Static friction at the contact point does the spinning: it acts up the slope, reducing the acceleration below g sin θ. The more of the mass that sits far out, the more friction is needed and the smaller the acceleration.

Definition

Down a slope: $mg\sin\theta - f = ma$ , $fR = I\dfrac{a}{R}$ , so $a = \dfrac{g\sin\theta}{1 + k^{2}/R^{2}}$ and $f = \dfrac{mg\sin\theta\,(k^{2}/R^{2})}{1 + k^{2}/R^{2}}$ .
It rolls without slipping only if $\mu \ge \dfrac{\tan\theta\,(k^{2}/R^{2})}{1 + k^{2}/R^{2}}$ . Otherwise it slips and kinetic friction acts.
Time down a slope of length l: $t = \sqrt{2l/a} \propto \sqrt{1 + k^{2}/R^{2}}$ . A disc takes $\sqrt{3/2}$ times as long rolling as sliding on a smooth slope.
A force F along the top of a body on rough level ground: $F + f = ma$ and $(F - f)R = I\dfrac{a}{R}$ , so $a = \dfrac{2F}{m(1 + k^{2}/R^{2})}$ and $f = F\dfrac{1 - k^{2}/R^{2}}{1 + k^{2}/R^{2}}$ , acting FORWARD. A solid sphere: $a = \dfrac{10F}{7m}$ .

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}}

Worked example

A solid cylinder rolls without slipping down a 30° slope (g = 10 m/s²). Find its acceleration and the least coefficient of friction that lets it roll.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q3Moderate

Example 2 · System of Particles and Rotational Motion · Rolling on Inclines: Acceleration, Speed and Friction

A solid sphere (A) of mass, 5 m and spherical shell (B) of mass

m

, both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of A and B , they start rolling without slipping with an acceleration of

a_{A}

and

a_{B}

, respectively. The ratio of

a_{A}

and

a_{B}

\_\_\_\_

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.

Drill 8 more on acceleration and friction of a rolling 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 (2)

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

Watch out for (6)

The slope's length is not the height
→ Speed and height of a body rolling on a slope
A smooth slope does not stop the spin
→ Speed and height of a body rolling on a slope
Mass and radius do not decide the race
→ Speed and height of a body rolling on a slope
A rolling body does not accelerate at g sin θ
→ Acceleration and friction of a rolling body
Friction pushes forward when the force is at the top
→ Acceleration and friction of a rolling body
Needed friction is not μN
→ Acceleration and friction of a rolling body

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