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I'm looking at an explanation of pendulums, and the following is said:

enter image description here

What I don't understand is where it says "if restoring force is given by $mg\theta$..." - conceptually, what is a force times an angle? Never being one to simply commit something to memory for the sake of it, I really want to understand this...

Any help greatly appreciated!

Tim.

EDIT:

Follow up question here.

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    $\begingroup$ Because an angle is dimensionless, its force. The exact value is mgsin(theta), but for small angles sin is identical to the angle (in radians). For that mgsin(theta) imagine the forces on that pendulum at some point. $\endgroup$ – Georg Nov 9 '11 at 11:00
  • $\begingroup$ Your edit should be a second question. $\endgroup$ – Colin K Nov 10 '11 at 2:46
  • $\begingroup$ Colin, you're right - have fixed that up. Cheers. $\endgroup$ – Hanshan Nov 10 '11 at 3:13
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As @Georg puts it, the force is $mg\sin{\theta}$, but then for small $\theta$ one can assume $\sin{\theta} = \theta$.

So, the force becomes, $mg\theta$.

And regarding the dimensions, $\sin{\theta}$ and $\theta$ are dimension less quantities.

So dimensionally, $mg$ and $mg\sin{\theta}$ are same $\left[ kg \cdot m \cdot s^{-2} \right]$.

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  • $\begingroup$ Thanks guys. So essentially it is just a way to simplify the equations/maths? $\endgroup$ – Hanshan Nov 9 '11 at 21:37
  • $\begingroup$ @nulliusinverba: Right. It works as long as the angle is small. $\endgroup$ – Mike Dunlavey Nov 10 '11 at 2:15

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