PhysicsSystem of particles and rotational motionClass 11

Rotational dynamics and rolling: interactive simulation

A rolling body moves forward and spins at the same time, so its kinetic energy is shared between the two. How it is shared depends only on the shape factor k = I/MR² — which is why a solid sphere always beats a ring down a slope, whatever their masses.

Useful for: JEE Main, JEE Advanced, NEET, AP Physics 1, AP Physics C: Mechanics, IB Physics HL · 4 modes in one lab

What you can explore

  1. 1

    Rolling race

    Ring, disc, solid and hollow spheres and a frictionless block race down the same incline.
  2. 2

    Rolling vs slipping

    Forward and backward slipping, kinetic friction, and the moment pure rolling begins.
  3. 3

    Moment of inertia

    Rod, ring, disc, plate and spheres with the parallel and perpendicular axis theorems and radius of gyration.
  4. 4

    Toppling vs sliding

    Where the normal force acts, and whether a pushed block slides or tips over.

Try this

This lab has four parts. Race different bodies, give a ball a push and a spin and watch friction bring it to pure rolling, slide an axis to see the parallel axis theorem, and push a tall block to find out whether it slides or tips over.

  1. 1

    Race all five bodies on a 30° slope.

    What you should see

    Frictionless block 1.28 s, solid sphere 1.51 s, disc 1.56 s, hollow sphere 1.65 s, ring 1.81 s. Same order at every angle.

  2. 2

    In Rolling vs slipping, push a solid sphere at 6 m/s with no spin (μ = 0.3).

    What you should see

    It slips forward, friction slows it and spins it up, and after 0.58 s it rolls at 6 / 1.4 ≈ 4.29 m/s.

  3. 3

    Give the ball backspin larger than its speed.

    What you should see

    Backward slip: friction now acts forwards on the contact point and the ball can even reverse direction.

  4. 4

    In Moment of inertia, take a disc and shift the axis to its rim (d = R).

    What you should see

    I = MR²/2 + MR² = 3MR²/2.

  5. 5

    In Toppling, use m = 10 kg, b = 0.4 m, h = 1 m, push at 0.9 h with μs = 0.5.

    What you should see

    It topples at F ≈ 21.8 N, long before friction would let it slide (49 N), because μs > b/2H ≈ 0.22.

Formulas behind it

Acceleration when rolling down an incline

a=gsin⁡θ1+I/MR2a = \frac{g\sin\theta}{1 + I/MR^2}

Condition for pure rolling

vcm=ωRv_{cm} = \omega R

Final speed after slipping stops

vf=v0+k ω0R1+k,k=IMR2v_f = \frac{v_0 + k\,\omega_0 R}{1 + k},\quad k = \frac{I}{MR^2}

Parallel axis theorem

I=Icm+Md2I = I_{cm} + Md^2

Perpendicular axis theorem (flat bodies)

Iz=Ix+IyI_z = I_x + I_y

Block topples before sliding when

μs>b2H\mu_s > \frac{b}{2H}

Common mistakes

  • Using a = g sin θ for a rolling body. That is only for a frictionless slide.
  • Thinking friction does negative work in pure rolling. The contact point is at rest, so static friction does no work.
  • Adding I_cm to Md² when the axis is not parallel to the one through the centre of mass.
  • Using the perpendicular axis theorem for a sphere or any 3D body. It only works for flat (laminar) bodies.
Practise 280 questionsCheck your answer instantly, with a worked solution.Rotational Motion chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a rotational dynamics and rolling question?Snap it and get a whiteboard video solution.

FAQ

Why does a solid sphere roll down faster than a ring of the same mass and radius?

More of the ring’s mass is far from the axis, so a larger share of its energy goes into spinning. With k = 1 the ring gets a = g sin θ / 2; the solid sphere (k = 2/5) gets 5/7 g sin θ.

What is the direction of friction on a body rolling down an incline?

Up the slope. It is the only force with a torque about the centre, so it is what makes the body spin faster as it speeds up.

What is the kinetic energy of a rolling body?

KE = ½Mv² + ½Iω² = ½Mv²(1 + k). For a solid sphere that is 7/10 Mv².

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