Collision with extended bodies

Question

In this situation, after the collision, how would I determine the angular velocity of both bodies, and how would I determine the velocity of C1, and C2?

Answer 1

At the moment of impact, there is an argument to be made that the net impulse would act through the exact center of the contacting surfaces. This is due to the symmetry in this problem between the two objects and where they contact.

Consider the point of impact as point K with coordinates $(x_k,y_k)$.

The impact speed (relative velocity along the contact direction) is calculated from

$$ \require{cancel} v_{{\rm imp}}=v_{1}-\left(y_{k}-y_{1}\right)\cancelto{0}{\omega_{1}}+v_{2}-\left(y_{k}-y_{2}\right)\cancelto{0}{\omega_{2}} \tag{1}$$

where $v_1$ and $\omega_1$ is the velocity and rotational velocity of object 1, and $v_2$ and $\omega_2$ is the velocity and rotational velocity of object 2. All rotational quantities are positive when counter-clockwise. Initially, no rotation exists.

Similarly, after the impact, the bounce speed is

$$ v_{{\rm bounce}}=v_{1}^{\star}-\left(y_{k}-y_{1}\right)\omega_{1}^{\star}+v_{2}^{\star}-\left(y_{k}-y_{2}\right)\omega_{2}^{\star} \tag{2} $$

where the superscript * denotes after impact quantity.

The law of the contacts states that $v_{\rm bounce} = -v_{\rm imp}$ for an elastic impact, and $v_{\rm bounce} = 0$ for a plastic contact. Since it is not specified, I will assume an elastic impact.

$$ v_{1}^{\star}-\left(y_{k}-y_{1}\right)\omega_{1}^{\star}+v_{2}^{\star}-\left(y_{k}-y_{2}\right)\omega_{2}^{\star}=-\left(v_{1}+v_{2}\right) \tag{3}$$

If an equal and opposite impulse $J$ is applied on the two objects, then the combined momentum and angular momentum (about the origin) before and after impact will not change.

• Before Impact $$\small \begin{array}{r|ll|c} & \text{Object 1} & \text{Object 2} & \text{Total}\\ \hline \text{Momentum} & p_{1}=m\,v_{1} & p_{2}=-m\,v_{2} & p=m\,\left(v_{1}-v_{2}\right)\\ \text{Angular Momentum} & L_{1}=-y_{1}mv_{1} & L_{2}=y_{2}mv_{2} & L=-y_{1}mv_{1}+y_{2}mv_{2} \end{array}$$
• After Impact $$\small \begin{array}{r|ll|c} & \text{Object 1} & \text{Object 2} & \text{Total}\\ \hline \text{Momentum} & p_{1}=m\,v_{1}-J & p_{2}=-m\,v_{2}+J & p=m\,\left(v_{1}-v_{2}\right)\\ \text{Angular Momentum} & L_{1}=-y_{1}mv_{1}+y_{k}J & L_{2}=y_{2}mv_{2}-y_{k}J & L=-y_{1}mv_{1}+y_{2}mv_{2} \end{array}$$

Given the momentum quantities after the impact, the motion quantities can be extracted from the following

$$\small \begin{array}{l|l} \text{Object 1} & \text{Object 2}\\ \hline m\,v_{1}^{\star}=m\,v_{1}-J & m\,\left(-v_{2}^{\star}\right)=-m\,v_{2}+J\\ \;v_{1}^{\star}=v_{1}-\tfrac{1}{m}J & \;v_{2}^{\star}=v_{2}-\tfrac{1}{m}J\\ \hline I\,\omega_{1}^{\star}-y_{1}m\,v_{1}^{\star}=-y_{1}mv_{1}+y_{k}J & I\,\omega_{2}^{\star}+y_{2}m\,v_{2}^{\star}=y_{2}mv_{2}-y_{k}J\\ \;I\,\omega_{1}^{\star}=-y_{1}mv_{1}+y_{1}m\,\left(v_{1}-\tfrac{1}{m}J\right)+y_{k}J & \;I\,\omega_{2}^{\star}=y_{2}mv_{2}-y_{2}m\,\left(v_{2}-\tfrac{1}{m}J\right)-y_{k}J\\ \;I\,\omega_{1}^{\star}=\left(y_{k}-y_{1}\right)J & \;I\,\omega_{2}^{\star}=-\left(y_{k}-y_{2}\right)J\\ \;\omega_{1}^{\star}=\tfrac{1}{I}\left(y_{k}-y_{1}\right)J & \;\omega_{2}^{\star}=-\tfrac{1}{I}\left(y_{k}-y_{2}\right)J \end{array}\tag{4}$$

where $m$ is the mass of the objects, and $I$ the mass moment of inertia of each object. For this example $I = m ( 2^2+4^2)/12 = \tfrac{5}{3}\,m$.

Now use these motion quantities from (4) into the law of contact (3) to find the impulse magnitude $J$.

$$\begin{gathered}v_{1}-\tfrac{1}{m}J-\tfrac{\left(y_{k}-y_{1}\right)^{2}}{I}J+v_{2}-\tfrac{1}{m}J+\tfrac{\left(y_{k}-y_{2}\right)^{2}}{I}J=-\left(v_{1}+v_{2}\right)\\ \tfrac{1}{m}J+\tfrac{\left(y_{k}-y_{1}\right)^{2}}{I}J+\tfrac{1}{m}J-\tfrac{\left(y_{k}-y_{2}\right)^{2}}{I}J=2\left(v_{1}+v_{2}\right)\\ \left(\tfrac{1}{m}+\tfrac{1}{m}+\tfrac{\left(y_{k}-y_{1}\right)^{2}}{I}-\tfrac{\left(y_{k}-y_{2}\right)^{2}}{I}\right)J=2\left(v_{1}+v_{2}\right)\\ J=\frac{2\left(v_{1}+v_{2}\right)}{\tfrac{1}{m}+\tfrac{1}{m}+\tfrac{\left(y_{k}-y_{1}\right)^{2}}{I}-\tfrac{\left(y_{k}-y_{2}\right)^{2}}{I}} \end{gathered} \tag{5}$$

Once $J$ is known, plug into (4) to get $v_1^\star$, $\omega_1^\star$, $v_2^\star$, and $\omega_2^\star$. This will give you the motion of the objects after the impact.