Moment of Inertia: Disk vs Hoop vs Sphere

The moment of inertia $I$ measures how hard it is to angularly accelerate an object — the rotational analog of mass. The bigger $I$, the harder it is to spin up. For three common shapes with the same mass $m$ and radius $R$:

• Solid sphere: $I = \tfrac{2}{5}\,mR^{2}$
• Solid disk (cylinder): $I = \tfrac{1}{2}\,mR^{2}$
• Thin hoop (ring): $I = mR^{2}$

The hoop has the largest moment of inertia because all its mass is at the maximum distance from the center. The sphere has the smallest because its mass is distributed throughout the volume, much of it close to the center.

Rolling Down a Ramp

When an object rolls without slipping down an incline of angle $\theta$, both the linear and rotational motion must be solved together. The result is a clean formula for the acceleration:

$a = \dfrac{g\sin\theta}{1 + I/(mR^{2})}$

The factor $I/(mR^{2})$ is dimensionless and depends only on the object's shape, not its actual mass or radius:

• Sphere: $I/(mR^{2}) = 2/5$ → $a = (5/7)\,g\sin\theta$
• Disk: $I/(mR^{2}) = 1/2$ → $a = (2/3)\,g\sin\theta$
• Hoop: $I/(mR^{2}) = 1$ → $a = (1/2)\,g\sin\theta$

The sphere is fastest, the hoop is slowest. It's a race that always has the same winner, regardless of mass or size.

Step-by-Step Solution

Given: Three rolling objects (sphere, disk, hoop) with equal $m$ and $R$, on a 30° incline, $g = 9.81\;\text{m/s}^{2}$.

Find: The acceleration of each, and the order they reach the bottom.

---

Step 1 — Compute each $I/(mR^{2})$.

• Sphere: $\dfrac{(2/5)mR^{2}}{mR^{2}} = \dfrac{2}{5} = 0.400$
• Disk: $\dfrac{(1/2)mR^{2}}{mR^{2}} = \dfrac{1}{2} = 0.500$
• Hoop: $\dfrac{mR^{2}}{mR^{2}} = 1.000$

Step 2 — Compute $g\sin\theta$ for $\theta = 30°$.

$g\sin 30° = 9.81 \times 0.5 = 4.905\;\text{m/s}^{2}$

Step 3 — Compute the acceleration of each shape.

• Sphere: $a = \dfrac{4.905}{1 + 0.4} = \dfrac{4.905}{1.4} \approx 3.504\;\text{m/s}^{2}$
• Disk: $a = \dfrac{4.905}{1 + 0.5} = \dfrac{4.905}{1.5} \approx 3.270\;\text{m/s}^{2}$
• Hoop: $a = \dfrac{4.905}{1 + 1} = \dfrac{4.905}{2} \approx 2.453\;\text{m/s}^{2}$

Step 4 — Time to roll a 5 m incline (using $d = \tfrac{1}{2}at^{2}$).

For each shape:

• Sphere: $t = \sqrt{2(5)/3.504} = \sqrt{2.854} \approx 1.689\;\text{s}$
• Disk: $t = \sqrt{2(5)/3.27} = \sqrt{3.058} \approx 1.749\;\text{s}$
• Hoop: $t = \sqrt{2(5)/2.453} = \sqrt{4.077} \approx 2.019\;\text{s}$

Step 5 — Race results.

The sphere wins in $\approx 1.69$ s. The disk comes second at $\approx 1.75$ s. The hoop finishes last at $\approx 2.02$ s — about 20% slower than the sphere.

Why? Energy conservation says all three convert the same amount of PE into combined translational and rotational KE. But the hoop has to "spend" more of that energy on rotation (because of its larger $I$) and less on translation, so it ends up moving more slowly.

---

Answer:

For a 30° ramp with equal $m$ and $R$:

• Sphere: $a = (5/7)g\sin\theta \approx 3.504\;\text{m/s}^{2}$ ← fastest
• Disk: $a = (2/3)g\sin\theta \approx 3.270\;\text{m/s}^{2}$
• Hoop: $a = (1/2)g\sin\theta \approx 2.453\;\text{m/s}^{2}$ ← slowest

The order is sphere → disk → hoop, and it's the same for any mass, any radius, and any incline angle. Mass and radius cancel out — only the shape matters.

Try It

• Adjust the slope angle — all three objects accelerate proportionally, but the order never changes.
• Watch the race: the sphere always wins, the disk is always second, the hoop is always last.
• The HUD reports the live speed of each.