In Figure 11.96a spherical non-spinning asteroid of mass\(M\)and radius\(R\)moving with speed\({v_1}\)to the right collides with a similar non-spinning asteroid moving with speed\({v_2}\)to the left, and they stick together. The impact parameter is\(d\).Note that\({I_{sphere}} = \frac{2}{5}M{R^2}.\)

After the collision, what is the velocity \({v_{CM}}\) of the center of mass and the angular velocity \(\omega \) about the center of mass? (Note that each asteroid rotates about its own center with this same \(\omega \)).

Angular velocity is a vector quantity and is described as the rate of change of angular displacement which specifies the angular speed or rotational speed of an object and the axis about which the object is rotating. The amount of change of angular displacement of the particle at a given period of time is called angular velocity. The track of the angular velocity vector is vertical to the plane of rotation, in a direction which is usually indicated by the right-hand rule.

The velocity of the center of mass is,

\({\overrightarrow v _{cm}} = \frac{{{m_1}{{\overrightarrow v }_1} + {m_2}{{\overrightarrow v }_2}}}{{{m_1} + {m_2}}}\)

Thanks to\({m_1} = {m_2}\),the above equation simplifies as,

\({\overrightarrow v _{cm}} = \frac{{{{\overrightarrow v }_1} + {{\overrightarrow v }_2}}}{2}\)

There are no external forces acting on the system so the velocity of the center of mass remains a constant that is,

\({\overrightarrow v _{cm}} = \frac{{{{\overrightarrow v }_1} + {{\overrightarrow v }_2}}}{2}\)

Now, we must calculate the angular momentum of the system. The balls are initially not spinning, so calculate the angular momentum each ball has about the system’s center of mass just as they touch.

At this moment the asteroid’s centers are a distance\(R\)away from the system’s center of mass. Using geometry and trig, you can solve for the angle between the lever arms and the center of mass to be,

\(\theta = {\sin ^{ - 1}}\frac{d}{{2R}}\)

The angular momentum of first asteroid is,

\(\begin{aligned}{}{\overrightarrow L _1}& = \overrightarrow r \times \left( {M{{\overrightarrow v }_1}} \right)\\ &= M\left| {\overrightarrow r } \right|\left| {{{\overrightarrow v }_1}} \right|\sin \theta \cdot {\widehat L_1}\end{aligned}\)

The right hand rule tells us that it points in the z-direction (out of the page), so

\({\overrightarrow L _1} = \left\langle {0,0,MR\left| {{{\overrightarrow v }_1}} \right|\sin \theta } \right\rangle \)

The same applies to\({\overrightarrow L _2}\).

The second angular momentum is,

\({\overrightarrow L _2} = \left\langle {0,0,MR\left| {{{\overrightarrow v }_2}} \right|\sin \theta } \right\rangle \)

And these both together to get the total angular momentum,

\(\overrightarrow L = \left\langle {0,0,MR\left( {\left| {{{\overrightarrow v }_1}} \right| + \left| {{{\overrightarrow v }_2}} \right|} \right)\sin \theta } \right\rangle \)

Angular momentum is conserved, so the two rotating rocks together will have a total angular momentum equal to the initial state.

For a spinning object, angular momentum is equal to\(\overrightarrow L = I\omega \).

As each rock has an angular speed of\(\omega \)and a momentum of inertia is,

\(I = \frac{2}{5}M{R^2}\)

Each sphere will have one-half the total angular momentum of the system.

\(\frac{{MR\left( {\left| {{{\overrightarrow v }_1}} \right| + \left| {{{\overrightarrow v }_2}} \right|} \right)\sin \theta }}{2} = \frac{2}{5}M{R^2}\omega \) ….. (1)

Solve equation (1) for\(\omega \).

\(\omega = \frac{5}{4}\frac{{\left( {\left| {{{\overrightarrow v }_1}} \right| + \left| {{{\overrightarrow v }_2}} \right|} \right)\sin \theta }}{{{R^2}}}\)

Here,\(\theta = {\sin ^{ - 1}}\frac{d}{{2R}}\)

Therefore, the angular velocity of center mass after the collision will be,\(\omega = \frac{5}{4}\frac{{\left( {\left| {{{\overrightarrow v }_1}} \right| + \left| {{{\overrightarrow v }_2}} \right|} \right)\sin \theta }}{{{R^2}}}\)