1. Physics
  2. Rotational Motion
  3. Rolling Motion

Rotational Motion · JEE & NEET Physics

Rolling Motion: notes and previous year questions

Moving along and turning at once: v = Rω, the speeds of points on a wheel, rolling energy, racing down slopes, friction, sliding into rolling, and rolling uphill.

Rolling Motion in short

  • Rolling is moving along plus turning about the centre.
  • In pure rolling v = Rω and the contact point is at rest; the top moves at 2v.
  • At each instant the wheel turns about the contact point.
  • Rolling KE = ½Mv²(1 + I/MR²); the turning share grows with I/MR².

1What is rolling?

A rolling wheel does two things at once: its centre moves along in a straight line, and the wheel turns about its centre. A point on the rim traces loops called a cycloid.

  • Pure rolling (no slipping): it turns exactly as fast as it moves, v=Rωv = R\omega.
  • Sliding: it moves without turning enough; the bottom skids forward.
  • Spinning in place: it turns faster than it moves; the bottom skids backward.

2The rolling rule, v = Rω

Add the two motions. Moving along gives every point vv forward. Turning gives every rim point RωR\omega along the rim. At the top both point forward: v+Rωv + R\omega. At the bottom they are opposite: v−Rωv - R\omega. In pure rolling the bottom does not slide, so

vcm=Rω,acm=Rαv_{cm} = R\omega, \qquad a_{cm} = R\alpha

Then the top moves at 2v2v, the centre at vv and the contact point not at all. In one turn the wheel rolls its circumference, 2πR2\pi R.

3Turning about the contact point

At each instant the contact point P is at rest, so the wheel is simply turning about P at the same ω\omega. Each point moves square to its line to P, at ω\omega times its distance from P.

vθ=v2(1+cos⁡θ)=2vcos⁡(θ/2)v_\theta = v\sqrt{2(1 + \cos\theta)} = 2v\cos(\theta/2)θ measured from the top: 2v at the top, √2 v at the side, 0 at the bottom.

A disc rolling at 6 m/s: the rim point level with the centre moves at 62≈8.56\sqrt2 \approx 8.5 m/s, at 45° to the ground; the top moves at 12 m/s.

4Kinetic energy of rolling

KE=12Mv2+12Iω2=12Mv2(1+IMR2)KE = \tfrac12 Mv^2 + \tfrac12 I\omega^2 = \tfrac12 Mv^2\left(1 + \frac{I}{MR^2}\right)Moving plus turning, with ω = v/R. Equally ½Mv²(1 + k²/R²).
BodyI/MR²Turning : movingTurning share
Ring11 : 150%
Hollow sphere2/32 : 340%
Disc, solid cylinder1/21 : 233%
Solid sphere2/52 : 529%

5Rolling down a slope

Along the slope: Mgsin⁡θ−f=MaMg\sin\theta - f = Ma. Friction turns the body: fR=IαfR = I\alpha with a=Rαa = R\alpha, so f=IR2af = \frac{I}{R^2}a. Together:

a=gsin⁡θ1+I/MR2a = \frac{g\sin\theta}{1 + I/MR^2}
v=2gh1+I/MR2v = \sqrt{\frac{2gh}{1 + I/MR^2}}From energy: Mgh = ½Mv²(1 + I/MR²).
t=2L(1+I/MR2)gsin⁡θt = \sqrt{\frac{2L(1 + I/MR^2)}{g\sin\theta}}Time down a slope of length L.
BodyI/MR²av after a drop h
Solid sphere2/557gsin⁡θ\frac57 g\sin\theta10gh/7\sqrt{10gh/7}
Disc, solid cylinder1/223gsin⁡θ\frac23 g\sin\theta4gh/3\sqrt{4gh/3}
Hollow sphere2/335gsin⁡θ\frac35 g\sin\theta6gh/5\sqrt{6gh/5}
Ring, hollow cylinder112gsin⁡θ\frac12 g\sin\thetagh\sqrt{gh}
Block sliding, no friction0gsin⁡θg\sin\theta2gh\sqrt{2gh}

Only the shape matters, not the mass or radius. The race order: solid sphere, disc, hollow sphere, ring.

6Friction in rolling

In pure rolling the contact point does not slide, so the friction is static and does no work: the point it acts on does not move. Mechanical energy is conserved. Friction is needed only when the speed changes; steady rolling on level ground needs none. On a frictionless (icy) slope, there is no torque: the body slides down without turning, at gsin⁡θg\sin\theta.

f=I/MR21+I/MR2 Mgsin⁡θf = \frac{I/MR^2}{1 + I/MR^2}\,Mg\sin\thetaUp the slope. Solid sphere of 2 kg on 30°: f = (2/7) × 10 ≈ 2.86 N.
μ≥tan⁡θ1+MR2/I\mu \ge \frac{\tan\theta}{1 + MR^2/I}Ring or hollow cylinder: tan θ/2; solid cylinder: tan θ/3; solid sphere: (2/7) tan θ.

7From sliding to rolling

A solid cylinder is sent sliding at v0v_0 without turning. Kinetic friction μMg\mu Mg acts backward: it slows the centre, v=v0−μgtv = v_0 - \mu gt, and its torque spins the cylinder up, α=μMgRMR2/2=2μgR\alpha = \frac{\mu MgR}{MR^2/2} = \frac{2\mu g}{R}, so Rω=2μgtR\omega = 2\mu gt. Skidding stops when they meet:

t=v03μg,v=2v03,s=5v0218μgt = \frac{v_0}{3\mu g}, \qquad v = \frac{2v_0}{3}, \qquad s = \frac{5v_0^2}{18\mu g}

With v0=6v_0 = 6 m/s, rolling starts after 1 s if μ=0.2\mu = 0.2; it has skidded 5 m and moves at 4 m/s. Afterwards it rolls on steadily and friction stops acting.

8Rolling uphill

All the kinetic energy, moving and turning, becomes height: 12Mv2(1+IMR2)=Mgh\frac12Mv^2\left(1 + \frac{I}{MR^2}\right) = Mgh.

  • A hollow sphere at 4 m/s: h=16×5/320≈1.33h = \frac{16 \times 5/3}{20} \approx 1.33 m, which is 2.67 m along a 30° slope.
  • Same speed: the ring climbs highest (most turning energy).
  • Same kinetic energy: all shapes climb to the same height, h=KE/Mgh = KE/Mg.
  • Same height, different slopes: the same speed at the bottom, but the steeper slope is quicker.

Summary

Key ideas

  • Rolling is moving along plus turning about the centre.
  • In pure rolling v = Rω and the contact point is at rest; the top moves at 2v.
  • At each instant the wheel turns about the contact point.
  • Rolling KE = ½Mv²(1 + I/MR²); the turning share grows with I/MR².
  • Down a slope a = g sin θ/(1 + I/MR²): sphere beats disc beats hollow sphere beats ring.
  • The result depends on shape, not on mass or radius.
  • Static friction makes the body turn but does no work.
  • Rolling needs μ ≥ tan θ/(1 + MR²/I); with less it slips.
  • A sliding cylinder starts rolling at t = v₀/3μg, moving at 2v₀/3.
  • Going uphill with the same speed, the ring climbs highest; with the same energy, all climb equally.

Every equation

Pure rolling
v=Rω, a=Rαv = R\omega,\ a = R\alpha
Top point
vtop=2vv_{top} = 2v
Rim point at angle θ
vθ=2vcos⁡(θ/2)v_\theta = 2v\cos(\theta/2)
Rolling KE
KE=12Mv2(1+I/MR2)KE = \tfrac12Mv^2(1 + I/MR^2)
Down a slope
a=gsin⁡θ1+I/MR2a = \frac{g\sin\theta}{1 + I/MR^2}
Speed after a drop h
v=2gh1+I/MR2v = \sqrt{\frac{2gh}{1 + I/MR^2}}
Time down length L
t=2L(1+I/MR2)gsin⁡θt = \sqrt{\frac{2L(1 + I/MR^2)}{g\sin\theta}}
Friction on a slope
f=I/MR21+I/MR2Mgsin⁡θf = \frac{I/MR^2}{1 + I/MR^2}Mg\sin\theta
Least μ to roll
μ≥tan⁡θ1+MR2/I\mu \ge \frac{\tan\theta}{1 + MR^2/I}
Cylinder: slide → roll time
t=v0/3μgt = v_0/3\mu g
Speed when rolling begins
v=2v0/3v = 2v_0/3
Skid distance
s=5v02/18μgs = 5v_0^2/18\mu g

Previous year questions with solutions

Real JEE and NEET questions on rolling motion. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A solid cylinder having radius RR and length LL is slipping on a rough horizontal plane. At time t=0t=0 the cylinder has a translational velocity vo=49m/sv_{o}=49 m/s, perpendicular to its axis and a rotational velocity vo/4Rv_{o}/4R about the centre. The time taken by the cylinder to start rolling is ____\_\_\_\_ seconds. (coefficient of kinetic friction μK=0.25{\mu}_{K}=0.25 and g=9.8m/s2g=9.8 m/s^{2} )

  1. A15
  2. B5
  3. C10
  4. D7.5
Show answer and solution

Answer: Option B

Here the cylinder starts with some forward spin, but not enough: ω0R=v04\omega_{0}R = \dfrac{v_{0}}{4}, less than v0v_{0}, so its bottom point still slides forward and friction μkmg\mu_{k}mg acts backward. The centre: v=v0−μkgtv = v_{0} - \mu_{k}gt. The spin: for a solid cylinder α=μkmgR12mR2=2μkgR\alpha = \dfrac{\mu_{k}mgR}{\frac{1}{2}mR^{2}} = \dfrac{2\mu_{k}g}{R}, so ωR=v04+2μkgt\omega R = \dfrac{v_{0}}{4} + 2\mu_{k}gt. Rolling starts when they are equal: 3v04=3μkgt\dfrac{3v_{0}}{4} = 3\mu_{k}gt, so t=v04μkg=494×0.25×9.8=5 st = \dfrac{v_{0}}{4\mu_{k}g} = \dfrac{49}{4 \times 0.25 \times 9.8} = 5\ \mathrm{s}.

Check with the angular momentum about the ground: mv0R+12mR2v04R=mvR+12mRvmv_{0}R + \dfrac{1}{2}mR^{2}\dfrac{v_{0}}{4R} = mvR + \dfrac{1}{2}mRv gives v=34v0v = \dfrac{3}{4}v_{0}, and v0−vμkg=12.252.45=5 s\dfrac{v_{0} - v}{\mu_{k}g} = \dfrac{12.25}{2.45} = 5\ \mathrm{s}. The trap is A, 15 s15\ \mathrm{s}, which lets the centre slow all the way to v04\dfrac{v_{0}}{4} as if the spin never grew. The length LL plays no part.

Q2NEET 2018One correct option

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (KtK_{t}) as well as rotational kinetic energy (KrK_{r}) simultaneously. The ratio Kt:(Kt+Kr)K_{t} : (K_{t} + K_{r}) for the sphere is

  1. A7 : 10
  2. B5 : 7
  3. C10 : 7
  4. D2 : 5
Show answer and solution

Answer: Option B

For a solid sphere k2R2=25\dfrac{k^{2}}{R^{2}} = \dfrac{2}{5}, so Kr=25KtK_{r} = \dfrac{2}{5}K_{t} and Kt+Kr=75KtK_{t} + K_{r} = \dfrac{7}{5}K_{t}. Then Kt:(Kt+Kr)=1:75=5:7K_{t} : (K_{t} + K_{r}) = 1 : \dfrac{7}{5} = 5 : 7.

The trap is D, 2:52 : 5, which is Kr:KtK_{r} : K_{t}, a different pair from the one asked for. A, 7:107 : 10, is the factor in K=710Mv2K = \dfrac{7}{10}Mv^{2}, and C, 10:710 : 7, is B turned upside down: a part cannot be larger than the whole. Check: the rotational share is 27\dfrac{2}{7}, and 57+27=1\dfrac{5}{7} + \dfrac{2}{7} = 1.

Q3JEE Advanced 2018Numerical answer

A ring and disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60∘{60}^{\circ } with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2−3)/10s,(2-\sqrt{3})/\sqrt{10} s, then the height of the top of the inclined plane, in metres is ______________ . Take g=10ms−2.g=10 ms^{-2}.

Show answer and solution

Answer: 0.75 m

Along the slope the length is s=hsin⁡θs = \dfrac{h}{\sin\theta}, and s=12at2s = \dfrac{1}{2}at^{2} with a=gsin⁡θ1+k2/R2a = \dfrac{g\sin\theta}{1 + k^{2}/R^{2}} gives t=2h(1+k2/R2)gsin⁡2θt = \sqrt{\dfrac{2h(1 + k^{2}/R^{2})}{g\sin^{2}\theta}}. With gsin⁡260∘=10×34=7.5g\sin^{2}60^{\circ} = 10 \times \dfrac{3}{4} = 7.5:

ring (k2R2=1\dfrac{k^{2}}{R^{2}} = 1): t=4h7.5=4h30t = \sqrt{\dfrac{4h}{7.5}} = \dfrac{4\sqrt{h}}{\sqrt{30}}; disc (12\dfrac{1}{2}): t=3h7.5=23h30t = \sqrt{\dfrac{3h}{7.5}} = \dfrac{2\sqrt{3}\sqrt{h}}{\sqrt{30}}.

The difference is 2(2−3)h30=2−310\dfrac{2(2 - \sqrt{3})\sqrt{h}}{\sqrt{30}} = \dfrac{2 - \sqrt{3}}{\sqrt{10}}, so h=30210=32\sqrt{h} = \dfrac{\sqrt{30}}{2\sqrt{10}} = \dfrac{\sqrt{3}}{2} and h=0.75 mh = 0.75\ \mathrm{m}.

The trap is using hh itself as the distance travelled: the bodies roll along the slope, hsin⁡θ\dfrac{h}{\sin\theta}, which is why sin⁡2θ\sin^{2}\theta appears. The ring, with the larger ratio, is the one that arrives later.

Practice questions, easy to hard

Three questions from the rolling motion practice ladder: one easy, one medium, one hard.

Q4Numerical answer

A block sliding down a smooth slope has no spin to feed, so all the energy it gains goes into moving forward and a=gsin⁡θa = g\sin\theta, as if k=0k = 0. It beats every rolling body. From rest, s=12at2s = \dfrac{1}{2}at^{2}, so over the same length the time goes as 1a\dfrac{1}{\sqrt{a}}.

A ring rolls without slipping down a rough slope, and a small block slides down a smooth slope of the same angle and the same length. Both start from rest. What is the ratio of their times, tringtblock\dfrac{t_{\mathrm{ring}}}{t_{\mathrm{block}}}? Give it to two decimal places.

Show answer and solution

Answer: 1.41

The ring has a=gsin⁡θ1+1=12gsin⁡θa = \dfrac{g\sin\theta}{1 + 1} = \dfrac{1}{2}g\sin\theta, half the block's. Time goes as 1a\dfrac{1}{\sqrt{a}}, so tringtblock=2≈1.41\dfrac{t_{\mathrm{ring}}}{t_{\mathrm{block}}} = \sqrt{2} \approx 1.41.

The trap is 22: halving the acceleration does not double the time, because the distance grows as t2t^{2}. The answer does not depend on the angle or the length, since both are the same for the two.

Q5Numerical answer

While the ball slides, both motions change steadily. The centre: v=v0−μgtv = v_{0} - \mu gt. The spin, from τ=Iα\tau = I\alpha with τ=μMgR\tau = \mu MgR: for a solid sphere α=μMgR25MR2=5μg2R\alpha = \dfrac{\mu MgR}{\frac{2}{5}MR^{2}} = \dfrac{5\mu g}{2R}, so starting from no spin, ωR=52μgt\omega R = \dfrac{5}{2}\mu gt. Sliding stops when v=ωRv = \omega R.

A bowling ball, a uniform solid sphere, is launched at 7 m/s7\ \mathrm{m/s} with no spin on a lane with μ=0.2\mu = 0.2. Taking g=10 m/s2g = 10\ \mathrm{m/s^{2}}, at what speed, in m/s\mathrm{m/s}, does it start rolling without slipping?

Show answer and solution

Answer: 5 m/s

7−2t=5t7 - 2t = 5t gives t=1 st = 1\ \mathrm{s}, so v=7−2=5 m/sv = 7 - 2 = 5\ \mathrm{m/s}. Check: ωR=5×1=5 m/s\omega R = 5 \times 1 = 5\ \mathrm{m/s}, the same.

The answer is 57v0\dfrac{5}{7}v_{0}, and μ\mu does not appear in it: a rougher lane makes the change quicker but ends at the same speed. The trap is 7 m/s7\ \mathrm{m/s}, thinking the ball keeps its speed and simply starts to spin; the spin's energy has to come from somewhere.