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Rolling Race & Moment of Inertia

Put up to four bodies on a ramp and let them go together. Choose each one's shape, mass and size, tilt the ramp and set how much it grips. The table under the plots works out every body's acceleration, time and where its energy went.

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Lesson

The theory — Rolling Race & Moment of Inertia

The moment of inertia I is how hard a body is to set spinning, as mass is how hard it is to push; for a body of mass m and radius r it is written I = k m r², where the shape factor k depends only on how the mass is spread: 2/5 for a solid sphere, 1/2 for a solid cylinder, 2/3 for a thin hollow sphere, 1 for a hoop. A body that rolls without slipping unrolls exactly as much of its circumference as the distance it travels, v = ωr, and down a ramp of angle θ it accelerates at a = g sin θ ÷ (1 + k).

What each symbol means

k
the shape factor, I ÷ mr². The only property of the body that decides its time while it rolls; once it slips, k no longer enters
I
the moment of inertia, in kg·m². The rotational counterpart of mass
a
the acceleration down the ramp. g sin θ for a frictionless slide, less for anything that rolls
μ_s
static friction, what the contact can hold without sliding. Rolling needs μs ≥ tan θ · k ÷ (1 + k)
μ_k
kinetic friction, what the contact gives once it slides. Usually lower than μs

Mass and size cancel, and the shape decides the race 🖖

Press Heavy vs light cylinder. One cylinder weighs 5 kg and the other 0.2 kg, one is five times the radius of the other, and they cross the line together at 1.537 s. A heavier body is pulled harder and is harder to speed up in exactly the same proportion, and a bigger one of the same shape has r² times the moment of inertia but spins 1/r times as fast, so its spin energy, ½Iω² = ½kmv², comes out in proportion to its mass all the same. What is left is one number for the shape, k = I ÷ mr², and the acceleration is g sin θ ÷ (1 + k). Press Classic four-body race: the solid sphere (k = 0.4) beats the solid cylinder (0.5), which beats the hollow sphere (0.667), which beats the hoop (1). Any solid sphere that rolls without slipping takes 1.485 s on this ramp.

A hoop puts half its energy into spinning 🖖

The energy bars show where each body's lost height went. Everything that rolls has to share it between moving down the ramp and turning, and the share that goes into turning is k ÷ (1 + k): 28.6% for the solid sphere and exactly 50% for the hoop, whose mass is all out at the rim. Energy spent on spin is not spent on speed, which is the whole reason the hoop loses. Press Soup cans. The can of liquid wins, in 1.316 s, because in the limit the page models the liquid does not turn with the tin: only the tin spins, and 9.1% of the energy goes into rotation. Set the same can's contents solid and it takes 1.563 s. The empty can, open-ended in this model and so a hoop, comes last.

On ice the slowest roller slips first, and slipping can win 🖖

Rolling needs grip, and the hoop needs the most: it can roll only while the ramp is shallower than its critical angle, 11.31° at a friction of 0.1 against the solid sphere's 19.29°. Press Icy ramp. At 15° the hoop and the hollow sphere slip, put less of their energy into spin, and speed up, yet they still finish behind the two that roll. With a single friction coefficient that is always so: a body that slips is at best as fast as one that rolls and never overtakes it. Most real surfaces grip harder standing still than sliding. Press Slipping hoop overtakes, where static friction is 0.3 and sliding friction 0.1, and the slipping hoop finishes in 0.849 s against the rolling ball's 0.942.

Problem solved in full

  1. A ball and a hoop down the same ramp 4 steps

    A solid ball and a hoop are released together from rest at the top of a ramp 2 m long, tilted at 15°. Both roll without slipping. Find how long each takes to reach the bottom, and how far behind the hoop is when the ball finishes. Take g = 9.81 m/s². Press Classic four-body race to check.

    Open this problem in the tool

    1. The part of gravity along the ramp is the same for both:

    2. Each body's acceleration divides it by one plus its shape factor:

    3. From rest, s = ½at², so the time for 2 m is:

    4. When the ball finishes, the hoop has covered only ½ a_hoop t_ball²:

    Answer

    The ball takes 1.485 s and the hoop 1.775 s. When the ball crosses the line the hoop is 0.600 m short, nearly a third of the ramp. Neither mass nor radius appeared anywhere in the working: the same answer holds for any uniform solid ball, marble or bowling ball.

· Reviewed against published sources. Corrections welcome.