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For the complete uniform disc, the surface density is $\sigma =M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, so its contribution to the axial moment of inertia is $r^{2}dm$. HenceThis derives the moment of inertia of a uniform disc rather than assuming it.

Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\frac{1}{2}(M/4)(a/2{)}^2=M{a}^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ andThe moment of inertia of a uniform annulus gives the same value directly.

Let $A_c$ be the downhill centre-of-mass acceleration and let $F$ denote the magnitude of the uphill static friction. Translation, rotation about the centre, and rolling without slipping giveThus $F=I_A{A}_c/a^2=(5/8)m_A{A}_c$. Eliminating $F$ gives the rolling acceleration with rotational inertiaSubstitution gives the contact forcesThe normal force balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the torque that increases its rolling angular speed. For a plane with $0\le \alpha <\pi /2$, the assumed no-slip motion requires a coefficient of static friction satisfying ${\mu }_s\ge F/N=(5/13)\tan \alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational kinetic energy.

For the moment of inertia of a uniform solid sphere, write the mass density as $\rho =3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2{\sin }^2\vartheta$. The volume element is $R^2\sin \vartheta dRd\vartheta d\phi$, so

Take $0<\alpha <\pi /2$ and $\ell >0$, with a fixed plane. The normal force has magnitude $N=mg\cos \alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the normal force have zero torque about the marble's centre, so a marble released from rest does not rotate. The component of Newton's second law down the plane is $m{A}_1=mg\sin \alpha$, henceThe normal force does no work because the velocity is tangent to the plane. Gravity is conservative, so mechanical energy, the sum of translational kinetic energy and gravitational potential energy, is conserved.

For rolling without slipping, let the static friction force have magnitude $f$, pointing up the plane. It supplies the torque that spins the marble while reducing its translational acceleration. If $\Omega$ is the angular velocity and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega =\dot{s}$, hence $a\dot{\Omega }=A_2$. The translation and rotation equations areThus $f=I{A}_2/a^2$ andThese are the equations for rolling acceleration with rotational inertia; the plane must be rough enough to supply this static friction.

The point of contact is instantaneously at rest relative to the fixed plane, so static friction does no total work on the rolling marble. Equivalently, its translational power is $-f\dot{s}$ and its rotational power is $fa\Omega =f\dot{s}$, which cancel. The normal force also does no work. Consequently mechanical energy is conserved here too, now with rotational kinetic energy included:where $H$ is the centre's height. No energy loss is implied by the presence of ideal static friction. Constant acceleration from rest gives

For the hollow marble, put $I=\lambda m{a}^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin \alpha /(1+\lambda )$. ThereforeThis again assumes rolling without slipping and enough static friction; the ratio depends on the mass distribution through its moment of inertia.

Using spherical shells and averaging the squared distance from the axis givesSliding without friction has acceleration $g\sin \alpha$, soFor rolling, $M{a}_{\mathrm{cm}}=Mg\sin \alpha -F$ and $I(a_{\mathrm{cm}}/a)=Fa$, whenceThe rolling acceleration with rotational inertia gives $5g\sin \alpha /7$ for the uniform sphere, and thereforeMechanical energy is conserved in both idealizations: there is no friction in sliding, while static friction does no work at the instantaneous contact point in pure rolling.

For $I=\gamma M{a}^2$, the same calculation gives

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's $1\mathrm{cm}$ pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at $14^{\circ }$ and reducing by $2^{\circ }$, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against $\sin \theta$, testing the ideal slope $g$. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass $M$, wheel radii $r_i$ and axial moments of inertia $I_i$, rigid wheels with fixed axles and no losses giveTo derive it, combine translational kinetic energy $M{v}^2/2$ with wheel rotational kinetic energy ${\sum }_i{I}_i(v/r_i{)}^2/2$, then equate their rate of change to the release of gravitational potential energy $Mg\sin \theta v$. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields $a\simeq g\sin \theta$; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.