A merry-go-round platform problem is a classic example of angular momentum conservation. A circular platform rotates
freely about a vertical axle while a person walks radially inward or outward.
\[
L_i=L_f.
\]
This is true if the net external torque about the axle is negligible.
1. Physical model
The system includes:
- a uniform circular platform of mass \(M\) and radius \(R\),
- a person of mass \(m\), modeled as a point mass,
- a vertical rotation axis through the center of the platform.
The person begins at radius \(r_i\) and moves to radius \(r_f\). The angular velocity changes from
\(\omega_i\) to \(\omega_f\).
2. Moment of inertia
A uniform circular platform is modeled as a solid disk:
\[
I_{\mathrm{platform}}=\frac12MR^2.
\]
The person is modeled as a point mass:
\[
I_{\mathrm{person}}=mr^2.
\]
Therefore, when the person is at radius \(r\), the total moment of inertia is
\[
I(r)=\frac12MR^2+mr^2.
\]
Initially,
\[
I_i=\frac12MR^2+mr_i^2.
\]
Finally,
\[
I_f=\frac12MR^2+mr_f^2.
\]
3. Conservation of angular momentum
Angular momentum for this rotating system is
\[
L=I\omega.
\]
If no external torque acts about the axle, angular momentum is conserved:
\[
I_i\omega_i=I_f\omega_f.
\]
Solving for final angular velocity:
\[
\omega_f=\frac{I_i}{I_f}\omega_i.
\]
Substituting the inertia formulas:
\[
\omega_f=
\frac{\frac12MR^2+mr_i^2}{\frac12MR^2+mr_f^2}\omega_i.
\]
4. Why moving inward speeds up the platform
The person contributes \(mr^2\) to the total moment of inertia. When the person walks inward, \(r\) decreases, so
the total moment of inertia decreases.
\[
r_f
Since
\[
I_i\omega_i=I_f\omega_f,
\]
a smaller \(I_f\) requires a larger \(\omega_f\). Therefore, the platform spins faster.
5. Moving outward slows the platform
If the person walks outward, the opposite happens:
\[
r_f>r_i
\quad\Longrightarrow\quad
I_f>I_i.
\]
A larger final moment of inertia means the final angular velocity is smaller:
\[
\omega_f<\omega_i.
\]
6. Rotational kinetic energy
Rotational kinetic energy is
\[
K=\frac12I\omega^2.
\]
Initial kinetic energy:
\[
K_i=\frac12I_i\omega_i^2.
\]
Final kinetic energy:
\[
K_f=\frac12I_f\omega_f^2.
\]
Even though angular momentum is conserved, rotational kinetic energy is not necessarily conserved. When the person
walks inward, \(K_f\) often increases. The extra energy comes from the work done by the person while walking.
7. Solving inverse problems
The same conservation equation can solve for other unknowns.
To solve for the final radius:
\[
I_f=\frac{I_i\omega_i}{\omega_f}.
\]
\[
r_f=\sqrt{\frac{I_f-I_{\mathrm{platform}}}{m}}.
\]
To solve for the initial radius:
\[
I_i=\frac{I_f\omega_f}{\omega_i}.
\]
\[
r_i=\sqrt{\frac{I_i-I_{\mathrm{platform}}}{m}}.
\]
To solve for the person mass:
\[
(I_p+mr_i^2)\omega_i=(I_p+mr_f^2)\omega_f.
\]
\[
m=
\frac{I_p(\omega_f-\omega_i)}
{\omega_ir_i^2-\omega_fr_f^2}.
\]
8. Worked example
Suppose a platform has
\[
M=100\ \mathrm{kg},\qquad R=2.0\ \mathrm{m}.
\]
A person has mass
\[
m=60\ \mathrm{kg}.
\]
The person walks from
\[
r_i=2.0\ \mathrm{m}
\]
to
\[
r_f=0.5\ \mathrm{m}.
\]
The initial angular speed is
\[
\omega_i=0.8\ \mathrm{rad/s}.
\]
First compute the platform inertia:
\[
I_p=\frac12(100)(2.0)^2=200\ \mathrm{kg\,m^2}.
\]
Initial inertia:
\[
I_i=200+(60)(2.0)^2=440\ \mathrm{kg\,m^2}.
\]
Final inertia:
\[
I_f=200+(60)(0.5)^2=215\ \mathrm{kg\,m^2}.
\]
Apply angular momentum conservation:
\[
\omega_f=\frac{I_i}{I_f}\omega_i.
\]
\[
\omega_f=\frac{440}{215}(0.8).
\]
\[
\omega_f\approx1.64\ \mathrm{rad/s}.
\]
Since the person moved inward, the system's moment of inertia decreased and the angular speed increased.
Key idea: with no external torque, moving mass inward decreases \(I\), so \(\omega\) increases to keep \(I\omega\)
constant.
