Why Does My Physics Book Use L/2 for Calculating Linear Acceleration?

AI Thread Summary
The discussion centers on understanding the use of L/2 in calculating linear acceleration for a rod in a physics problem. The acceleration of the rod's center, stated as 3g/4, is derived from the angular acceleration formula, where linear acceleration is calculated using a = α × (L/2). L/2 is significant because it represents the distance from the pivot to the center of mass, which is crucial for torque calculations. The varying linear acceleration along the rod is explained by the relationship a = αR, where R is the distance from the pivot point. This highlights the importance of understanding both angular and linear dynamics in physics problems.
UrbanXrisis
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The sample problem in my physics book that I do not understand is http://home.earthlink.net/~urban-xrisis/clip001.jpg

The only thing that I do not understand is the LAST sentance...
For example,the middle of the rod has an acceleration of 3g/4

I'm pretty sure that they are talking about the centripetal acceleration but they do not show how they got it. The book stops there. Any ideas of how they got 3g/4?
 
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Yes,\alpha=\frac{3g}{2L}

.So for the middle of the rod,the linear acceleration is:

a=\alpha\times\frac{L}{2}

which gives the answer you asked about.

Daniel.
 
why was (L/2) used for the original equation that asked for the linear acceleration of the whole rod?
 
UrbanXrisis said:
why was (L/2) used for the original equation that asked for the linear acceleration of the whole rod?
What equation are you referring to? While the rod has a single angular acceleration, the linear acceleration varies along the length (as Daniel illustrated): a = \alpha R, where R is the distance from the pivot.

L/2 is used in determining the torque due to the weight of the rod, which acts at the center of mass.
 

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