Let ${\displaystyle r}$ be the distance between the center of mass and the point of contact; when the surface is flat, this is the radius of the object around its widest cross section. Since the center of mass has an immediate velocity as if it was rotating around the point of contact, its velocity is ${\displaystyle v_{\text{c.o.m.))=r\omega }$. Due to symmetry, the object center of mass is a point in its axis. Let ${\displaystyle I_{\text{rotation))}$ be inertia of pure rotation around the axis of symmetry, then according to the parallel axis theorem, the rotational inertia associated with rolling is ${\displaystyle I_{\text{rolling))=mr^{2}+I_{\text{rotation))}$ (same as the rotational inertia of pure rotation around the point of contact). Using the general formula for kinetic energy of rotation, we have:

${\displaystyle {\begin{aligned}K_{\text{rolling))&={\frac {1}{2))I_{\text{rolling))\omega ^{2}\\&={\frac {1}{2))mr^{2}\omega ^{2}+{\frac {1}{2))I_{\text{rotation))\omega ^{2}\\&={\frac {1}{2))m(r\omega )^{2}+{\frac {1}{2))I_{\text{rotation))\omega ^{2}\\&={\frac {1}{2))mv_{\text{c.o.m.))^{2}+{\frac {1}{2))I_{\text{rotation))\omega ^{2}\\&=K_{\text{translation))+K_{\text{rotation))\\\end{aligned))}$

Assume that the object experiences an external force ${\displaystyle F_{\text{external))}$ which exerts no torque (it has 0 moment arm), static friction at the point of contact (${\displaystyle F_{\text{friction))}$) provides the torque and the other forces involved cancel. ${\displaystyle F_{\text{friction))}$ is tangential to the object and surface at the point of contact and opposite in direction to ${\displaystyle F_{\text{external))}$. Using the sign convention by which this force is positive, the net force is:

${\displaystyle F_{\text{net))=F_{\text{external))-F_{\text{friction))}$

${\displaystyle (1)}$

Because there is no slip, ${\displaystyle r\alpha =a}$ holds. Substituting ${\displaystyle \alpha }$ and ${\displaystyle a}$ for the linear and rotational version of Newton's second law, then solving for ${\displaystyle F_{\text{friction))}$:

${\displaystyle {\begin{aligned}r{\frac {\tau }{I))&={\frac {F_{\text{net))}{m))\\r{\frac {rF_{\text{friction))}{I))&={\frac {F_{\text{net))}{m))\\F_{\text{friction))&={\frac {IF_{\text{net))}{mr^{2))}\\\end{aligned))}$

Expanding ${\displaystyle F_{\text{friction))}$ in ${\displaystyle (1)}$:

${\displaystyle {\begin{aligned}F_{\text{net))&=F_{\text{external))-{\frac {IF_{\text{net))}{mr^{2))}\\&=F_{\text{external))-{\frac {I}{mr^{2))}F_{\text{net))\\&={\frac {F_{\text{external))}{1+{\frac {I}{mr^{2))))}\\\end{aligned))}$

The last equality is the first formula for ${\displaystyle F_{\text{net))}$; using it together with Newton's second law, then reducing, the formula for ${\displaystyle a}$ is obtained:

${\displaystyle {\begin{aligned}a&={\frac {F_{\text{net))}{m))\\&={\frac {\left({\frac {F_{\text{external))}{1+{\frac {I}{mr^{2))))}\right)}{m))\\&={\frac {F_{\text{external))}{m\left(1+{\frac {I}{mr^{2))}\right)))\\&={\frac {F_{\text{external))}{m+{\frac {I}{r^{2))))}\\\end{aligned))}$

The radius of gyration can be incorporated in the first formula for ${\displaystyle F_{\text{net))}$ as follows:

${\displaystyle {\begin{aligned}r_{\text{gyr.))&={\sqrt {\frac {I}{m))}\\r_{\text{gyr.))^{2}&={\frac {I}{m))\\{\frac {I}{mr^{2))}&={\frac {\left({\frac {I}{m))\right)}{r^{2))}\\&={\frac {r_{\text{gyr.))^{2)){r^{2))}\\&=\left({\frac {r_{\text{gyr.))}{r))\right)^{2}\\\end{aligned))}$

Substituting the latest equality above in the first formula for ${\displaystyle F_{\text{net))}$ the second formula for it:

${\displaystyle F_{\text{net))={\frac {F_{\text{external))}{1+\left({\frac {r_{\text{gyr.))}{r))\right)^{2))))$

Assuming that the object is placed so as to roll downward in the direction of the inclined plane (and not partially sideways), the weight can be decomposed in a component in the direction of rolling and a component perpendicular to the inclined plane. Only the first force component makes the object roll, the second is balanced by the contact force, but it does not form an action‐reaction pair with it (just as an object in rest on a table). Therefore, for this analysis, only the first component is considered, thus:

${\displaystyle F_{\text{external))=gm\sin \left(\theta \right)}$

${\displaystyle {\begin{aligned}a&={\frac {F_{\text{net))}{m))\\&={\frac {\left({\frac {F_{\text{external))}{1+\left({\frac {r_{\text{gyr.))}{r))\right)^{2))}\right)}{m))\\&={\frac {\left({\frac {gm\sin \left(\theta \right)}{1+\left({\frac {r_{\text{gyr.))}{r))\right)^{2))}\right)}{m))\\&={\frac {gm\sin \left(\theta \right)}{m\left(1+\left({\frac {r_{\text{gyr.))}{r))\right)^{2}\right)))\\&={\frac {g\sin \left(\theta \right)}{1+\left({\frac {r_{\text{gyr.))}{r))\right)^{2))}\\\end{aligned))}$

In the last equality the denominator is the same as in the formula for force, but the factor ${\displaystyle m}$ disappears because its instance in the force of gravity cancels with its instance due to Newton's third law.