Rolling on Inclined Plane

Consider a round object of mass $m$ and radius $R$ rolling without slipping down an incline of angle $\theta$. Two forces act on it along the direction of the incline: the driving component of gravity, $mg\sin\theta$, and an opposing static frictional force $f$ (the perpendicular component of gravity, $mg\cos\theta$, is simply balanced by the normal force $N$ from the incline and does no work).

Translational equation of motion, along the incline:

Rotational equation of motion, taking torques about the object's own center: the gravity component $mg\sin\theta$ produces no torque here (it passes right through the center), so only the friction $f$, acting at the rim, contributes: $Rf=I\alpha$. Using $a=R\alpha$ and $I=mK^2$, this gives $Rf=mK^2\left(\dfrac{a}{R}\right)$, so $f=\dfrac{ma K^2}{R^2}$.

Combining the two equations, substituting this expression for $f$ into (5.61):

Final speed. Using $v^2=u^2+2as$ with $u=0$ (released from rest) and incline length $s=\dfrac{h}{\sin\theta}$ (for a vertical drop $h$):

Time to reach the bottom. Using $v=u+at$ with $u=0$:

$t=\frac{v}{a}=\sqrt{\dfrac{2h\left(1+K^2/R^2\right)}{g\sin^2\theta}}.\qquad(5.64)$ …