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Answer 1 · 2017-02-20T15:25:36

The acceleration is $a = g sin θ$ $a=gsintheta$

The forces acting on the pendulum will be ${\to F}_{g}$ $vecF_g$ and $\to T$ $vecT$ , this second being the tension in the cord.

![webassign.net/labsgraceperiod/ncsulcpmech2/lab_7/http://manual.html](https://useruploads.socratic.org/cbM57UkGTW6nUhRxSv6w_figure8.jpg)

As the diagram shows, we can resolve ${\to F}_{g}$ $vecF_g$ into two components, namely $m g sin θ$ $mgsintheta$ and $m g cos θ$ $mgcostheta$ .

Because the rope does not stretch, it must be that $T = m g cos θ$ $T=mgcostheta$ , and these forces cancel.

Therefore, the net force acting on the pendulum is the "restoring force", $m g sin θ$ $mgsintheta$

$F_{net} = m g sin θ$ $F_"net" = mgsintheta$

$m a = m g sin θ$ $ma = mgsintheta$

and finally,

$a = g sin θ$ $a=gsintheta$

If you prefer, you can express the acceleration as an angular acceleration:

$α = \frac{a}{L}$ $alpha=a/L$

in which case the angular acceleration of the pendulum bob, moving in a circular arc, is

$α = \frac{g}{L} sin θ$ $alpha=g/L sintheta$