A rotational collision happens when two rotating bodies interact about the same axis. Examples include clutching
disks, rotating flywheels that couple together, and inelastic rotational impacts.
\[
L_i=L_f
\]
if the net external torque about the axis is negligible.
1. Angular momentum before and after collision
For two rotating bodies,
\[
L_i=I_1\omega_{1i}+I_2\omega_{2i}.
\]
After collision,
\[
L_f=I_1\omega_{1f}+I_2\omega_{2f}.
\]
Conservation of angular momentum gives
\[
I_1\omega_{1i}+I_2\omega_{2i}
=
I_1\omega_{1f}+I_2\omega_{2f}.
\]
2. Perfectly inelastic rotational collision
If the two disks stick together, they rotate with one common final angular velocity:
\[
\omega_{1f}=\omega_{2f}=\omega_f.
\]
Then
\[
I_1\omega_{1i}+I_2\omega_{2i}
=
(I_1+I_2)\omega_f.
\]
Solving for the common final angular velocity:
\[
\omega_f=
\frac{I_1\omega_{1i}+I_2\omega_{2i}}
{I_1+I_2}.
\]
3. Partially inelastic and elastic rotational collisions
A coefficient of restitution can describe how much relative angular motion remains after collision:
\[
e=\frac{\omega_{2f}-\omega_{1f}}{\omega_{1i}-\omega_{2i}}.
\]
The usual range is
\[
0\le e\le 1.
\]
The important cases are:
- \(e=0\): perfectly inelastic, disks stick together.
- \(0
- \(e=1\): ideal elastic collision.
Combining angular momentum conservation with the restitution equation gives
\[
\omega_{1f}
=
\frac{I_1\omega_{1i}+I_2\omega_{2i}-I_2e(\omega_{1i}-\omega_{2i})}
{I_1+I_2},
\]
\[
\omega_{2f}
=
\frac{I_1\omega_{1i}+I_2\omega_{2i}+I_1e(\omega_{1i}-\omega_{2i})}
{I_1+I_2}.
\]
4. Rotational kinetic energy
Rotational kinetic energy is
\[
K=\frac12I\omega^2.
\]
Before collision:
\[
K_i=\frac12I_1\omega_{1i}^2+\frac12I_2\omega_{2i}^2.
\]
After collision:
\[
K_f=\frac12I_1\omega_{1f}^2+\frac12I_2\omega_{2f}^2.
\]
The rotational kinetic energy lost is
\[
E_{\mathrm{lost}}=K_i-K_f.
\]
In a perfectly inelastic rotational collision, angular momentum is conserved, but kinetic energy is usually not
conserved. The lost mechanical energy becomes heat, sound, deformation, or internal energy.
5. Common moments of inertia
| Body |
Axis |
Moment of inertia |
| Solid disk |
Central symmetry axis |
\(I=\frac12MR^2\) |
| Thin hoop |
Central symmetry axis |
\(I=MR^2\) |
| Direct input |
Given axis |
\(I=\text{given}\) |
6. Worked example: two disks stick together
Suppose disk 1 has
\[
I_1=0.8\ \mathrm{kg\,m^2},
\qquad
\omega_{1i}=20\ \mathrm{rad/s}.
\]
Disk 2 has
\[
I_2=0.4\ \mathrm{kg\,m^2},
\qquad
\omega_{2i}=0.
\]
They collide and stick together, so
\[
\omega_{1f}=\omega_{2f}=\omega_f.
\]
Initial angular momentum:
\[
L_i=(0.8)(20)+(0.4)(0).
\]
\[
L_i=16\ \mathrm{kg\,m^2/s}.
\]
Final inertia:
\[
I_f=I_1+I_2=0.8+0.4=1.2\ \mathrm{kg\,m^2}.
\]
Final common angular velocity:
\[
\omega_f=\frac{16}{1.2}.
\]
\[
\omega_f\approx13.3\ \mathrm{rad/s}.
\]
Initial rotational kinetic energy:
\[
K_i=\frac12(0.8)(20)^2+\frac12(0.4)(0)^2=160\ \mathrm{J}.
\]
Final rotational kinetic energy:
\[
K_f=\frac12(1.2)(13.3)^2\approx106.7\ \mathrm{J}.
\]
Energy lost:
\[
E_{\mathrm{lost}}=160-106.7=53.3\ \mathrm{J}.
\]
Key idea: angular momentum is conserved when external torque is negligible, but rotational kinetic energy depends on
the collision type.
