# Treating Angular Momentum as Linear?

Scenario: a ball of mass $$m_B$$ collides elastically collides with a door. The door is a uniform rectangle of mass $$m_D$$, moment of inertia $$I$$, length $$R$$, and height $$h$$, and it is free to rotate about an axis through its left edge. The ball impacts the door perpendicularly with an initial velocity $$v_i$$, and bounces away from the door with a final velocity $$v_f$$ that is also perpendicular to the door. The point of impact is a distance $$d$$ away from the left edge of the door. The collision imparts the door with an angular velocity $$\omega$$. In terms of the initial conditions, what are $$v_f$$ and $$\omega$$?

My approach: I wrote a conservation of energy equation and a conservation of momentum equation, which together would let me solve for the two unknowns. Since I was presented this problem before I knew about angular momentum, I tried to linearize the angular momentum of the door instead of trying to angularize the linear momentum of the ball. To do this, I reasoned that, at the moment of collision, I each vertical strip of the door would gain an instantaneous tangential velocity $$v_T$$, which has the same direction as $$v_i$$. If $$r$$ is the distance from the strip to the left edge, then we would have $$v_T=\omega r$$. To get a momentum from this, I would just have to multiply $$v_T$$ times the mass of the strip. Since the door is uniform, this mass would be the mass density of the door, $$\frac{m_D}{Rh}$$, times the the area of the strip, $$A$$. Integrating over all strips, I found that the total linear momentum of the door was $$\int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh}\cdot h \; \mathrm{d}r\right)$$. So in all I got that

\begin{align*} \text{Conservation of Energy: } \frac{1}{2}m_Bv_i^2 &= \frac{1}{2}m_Bv_f^2 + \frac{1}{2}\omega^2I \\ \text{Conservation of Momentum: } m_Bv_i &= m_Bv_f + \int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh} \; \mathrm{d}A\right) = m_Bv_f + \int\limits_0^R\left(\omega r\right)\left(\frac{m_D}{Rh}\cdot h \; \mathrm{d}r\right)\end{align*}

but this is clearly wrong since it implies that $$v_f$$ and $$\omega$$ don't depend on $$d$$, which they obviously do.

After learning a bit more about angular momentum I rewrote the second equation as $$m_bv_id=m_bv_fd+\omega I$$, which after solving gave me a completely reasonable answer, but I'm still not sure why the my first approach isn't valid. Is there a way to linearize the door's angular momentum in a way that depends on $$d$$, or is it necessarily a hopeless endeavor?

Let me know if the question is unclear, and thank you for your responses.

• Can you explain what you mean by the word "linearize" here? You appear to be using it in a non-standard way.
– d_b
Nov 15, 2021 at 20:27
• @d_b I just mean that I am trying to transform the angular momentum into a linear momentum. Nov 16, 2021 at 16:25

When analyzing the problem using angular momentum, we cleverly choose the coordinate system so that the origin is located at the position of the hinge. With this choice of origin the torque of the hinge, $$r\times F$$, is zero, since $$r$$ is zero. This is what allows us to use conservation of angular momentum.