Seesaw and torque

Main Question or Discussion Point

When I do some physics derivation, I find that on a seesaw, if the object is farther away from the fulcrum, the angular acceleration decreases. Is this true? If not, where did I go wrong?
[tex]\tau = I \cdot \alpha[/tex]

[tex]\tau=F \cdot r[/tex]

Then, [tex]\alpha = \frac{F \cdot r}{I} = \frac{F_g \cdot r}{mr^2}=\frac{g \cdot m \cdot r}{mr^2}=\frac{g}{r}[/tex], so if the distance "r" increases, angular acceleration decreases...did I do something wrong?
 
Last edited:

Answers and Replies

28,493
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That is correct.
 
Are you sure? Some people have told me that it's wrong, because I have to account for the mass of the seesaw.
 
28,493
4,837
It's not that it is wrong, it simply makes some assumptions. The formula you derived is for a point mass on a massless seesaw. If your seesaw is not light enough to approximate as massless or if your load is too large to be considered a point then your formula doesn't apply.
 
What if I cannot assume the masses are negligible? How would I derive that? And in that case, would angular acceleration increase?
 
Doc Al
Mentor
44,827
1,083
Just include the rotational inertia of the seesaw as part of the total rotational inertia:
[tex]I_{total} = I_{seesaw} + mr^2 = 1/12 M L^2 + mr^2[/tex]
 

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