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44.2: Acceleration

  • Page ID
    92302
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    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.


    44.2: Acceleration is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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