Can we say that the force of this rotating rigid body is given by $F=MV^2/R$, where R is the radius and M is the mass of the rotating rigid body?
No, you can't. I believe you have a fundamental misunderstanding of what $F=MV^2/R$ represents. It is the magnitude of force required to maintain a particle of mass $M$ on a curved path with instantaneous radius of curvature $R$ at instantaneous speed $V$. If particle $M$ happens to be moving at constant speed in a circle, that magnitude is constant. That force must be directed on particle $M$ toward the center of curvature. That says nothing about any contact force between some other arbitrary particle and $M$.
In this collision, angular momentum will be conserved. Beyond that kinetic energy will be conserved if the collision is elastic (no destruction occurs).
Now let's talk about finding the "force". Are you asking for the average force, $<F>$, or the maximum force, $F_{max}$ during the time of contact? To find force function $F(t)$ you must know the interaction time which depends on the specific materials of the point mass $m$ and your rigid body $M$. Also, is the collision perfectly elastic or partially elastic? Is $m$ colliding with the top of $M$ or some vertical flat part of $M$? The shape of $M$ most definitely will affect the force, too., but the time interval is the key to getting the force.
One can generally easily calculate velocities, momentum changes, and angular momentum changes of each object after collisions if given enough information, forces are more difficult because interaction times (or distances) are needed. Sometimes, those interaction times are estimated and other times they can actually be measured using high speed cameras (look at a golf tournament on TV and they will sometimes show the club impacting the ball...impressive!) Again, the interaction time is highly material dependent.