A seesaw can be modeled as a rigid board rotating about a fixed pivot. Two people sitting at different distances
from the pivot create gravitational torques. The same system can also have angular momentum if it is rotating.
\[
\tau_{\mathrm{net}}=I\alpha,
\qquad
L=I\omega.
\]
1. Coordinate and sign convention
Take the pivot as the origin. Let the left rider sit at distance \(r_1\) from the pivot and the right rider sit at
distance \(r_2\). With counterclockwise torque treated as positive:
\[
\tau_{\mathrm{left}}=m_1gr_1.
\]
\[
\tau_{\mathrm{right}}=-m_2gr_2.
\]
If the board center is not exactly at the pivot, let \(x_p\) be the pivot offset from the board center. In the
convention used by the calculator, the board contributes
\[
\tau_{\mathrm{board}}=Mgx_p.
\]
2. Balance condition
The seesaw is balanced when the net torque about the pivot is zero:
\[
\tau_{\mathrm{net}}=0.
\]
Therefore,
\[
m_1gr_1-m_2gr_2+Mgx_p=0.
\]
Since every term contains \(g\), the balance condition can be written as
\[
m_1r_1+Mx_p=m_2r_2.
\]
For a board whose center is exactly at the pivot, \(x_p=0\), so the familiar balance rule becomes
\[
m_1r_1=m_2r_2.
\]
3. Moment of inertia of the seesaw system
Moment of inertia measures resistance to angular acceleration. For a uniform board of mass \(M\) and length
\(\ell\), about an axis through its center:
\[
I_{\mathrm{cm}}=\frac{1}{12}M\ell^2.
\]
If the pivot is offset from the board center, use the parallel-axis theorem:
\[
I_{\mathrm{board}}=\frac{1}{12}M\ell^2+Mx_p^2.
\]
Each rider is modeled as a point mass:
\[
I_1=m_1r_1^2,
\qquad
I_2=m_2r_2^2.
\]
The total moment of inertia is
\[
I=I_{\mathrm{board}}+I_1+I_2.
\]
4. Angular acceleration
Newton's second law for rotation is
\[
\tau_{\mathrm{net}}=I\alpha.
\]
Therefore,
\[
\alpha=\frac{\tau_{\mathrm{net}}}{I}.
\]
If \(\tau_{\mathrm{net}}>0\), the left side tends to move downward in the calculator's default sign convention. If
\(\tau_{\mathrm{net}}<0\), the right side tends to move downward.
5. Angular momentum
If the seesaw rotates with angular speed \(\omega\), its angular momentum about the pivot is
\[
L=I\omega.
\]
The direction of angular momentum follows the right-hand rule. In a side-view diagram, positive rotation can be
represented as angular momentum coming out of the page; negative rotation can be represented as angular momentum
going into the page.
6. Rotational kinetic energy
The rotational kinetic energy is
\[
K_{\mathrm{rot}}=\frac12I\omega^2.
\]
A larger moment of inertia or a larger angular speed increases rotational kinetic energy.
7. Worked example
Suppose the board has
\[
M=20\ \mathrm{kg},
\qquad
\ell=4\ \mathrm{m},
\qquad
x_p=0.
\]
The left rider has
\[
m_1=60\ \mathrm{kg},
\qquad
r_1=1.2\ \mathrm{m}.
\]
The right rider has
\[
m_2=36\ \mathrm{kg},
\qquad
r_2=2.0\ \mathrm{m}.
\]
The torque balance is
\[
\tau_{\mathrm{left}}=(60)(9.81)(1.2)=706.32\ \mathrm{N\,m}.
\]
\[
\tau_{\mathrm{right}}=-(36)(9.81)(2.0)=-706.32\ \mathrm{N\,m}.
\]
\[
\tau_{\mathrm{net}}=0.
\]
So the seesaw is balanced.
The board inertia is
\[
I_{\mathrm{board}}=\frac{1}{12}(20)(4)^2=26.67\ \mathrm{kg\,m^2}.
\]
Rider contributions:
\[
I_1=(60)(1.2)^2=86.4\ \mathrm{kg\,m^2}.
\]
\[
I_2=(36)(2.0)^2=144\ \mathrm{kg\,m^2}.
\]
Total inertia:
\[
I=26.67+86.4+144=257.07\ \mathrm{kg\,m^2}.
\]
If \(\omega=0.8\ \mathrm{rad/s}\), then
\[
L=I\omega=(257.07)(0.8)=205.65\ \mathrm{kg\,m^2/s}.
\]
Key idea: balance depends on torque \(mgr\), while angular momentum depends on rotational inertia \(mr^2\) and
angular speed \(\omega\).
