# Seesaw and torque

1. Mar 30, 2008

### compwiz3000

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?
$$\tau = I \cdot \alpha$$

$$\tau=F \cdot r$$

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

Last edited: Mar 30, 2008
2. Mar 30, 2008

### Staff: Mentor

That is correct.

3. Mar 30, 2008

### compwiz3000

Are you sure? Some people have told me that it's wrong, because I have to account for the mass of the seesaw.

4. Mar 30, 2008

### Staff: Mentor

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.

5. Mar 30, 2008

### compwiz3000

What if I cannot assume the masses are negligible? How would I derive that? And in that case, would angular acceleration increase?

6. Mar 31, 2008

### Staff: Mentor

Just include the rotational inertia of the seesaw as part of the total rotational inertia:
$$I_{total} = I_{seesaw} + mr^2 = 1/12 M L^2 + mr^2$$