44.2: Acceleration

Now let's find the (translational) acceleration of the body down the incline. If the distance down the incline is $x$, then the velocity $v$ at the bottom of the incline is related to $x$ by

$$v^{2}=2 a x$$

By geometry, $\sin \theta=h / x$, and so $x=h / \sin \theta$; using this to substitute for $x$, we have

$$v^{2}=2 a \frac{h}{\sin \theta},$$

or, solving for the acceleration $a$,

$$a=\frac{v^{2} \sin \theta}{2 h}$$

Now let's use Eq. (41.1.12) to substitute for $v$; the result is an expression for the acceleration of a body rolling down an incline,

$$a=\frac{g \sin \theta}{\beta+1}$$

Table 44.3.1 shows values of $\beta$ and $a$ for several common geometries.

Equation $\PageIndex{4}$ has some interesting consequences. For example, if you start a solid sphere and a cylindrical shell at the top of a incline and release them at the same time, which one will reach the bottom first? From Table 44.3.1, you can see that the solid sphere will will win: its acceleration (5/7) $g \sin \theta$ is greater than the cylindrical shell's acceleration of $(1 / 2) g \sin \theta$. What's surprising about this is that all solid spheres will beat all cylindrical shells, regardless of mass or radius. In general, the object with the smaller $\beta$ will win such a race, since that will give the smallest denominator in Eq. $\PageIndex{4}$ and therefore the larger acceleration.