# Force acting on a small point mass dropped over a rotating rigid body

I came across a situation where I must find the force exerted by a rotating rigid body on a point mass $$m$$, dropped over it. The rotating rigid body will smash the point mass away in a direction depending on the the collision point.

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? The above force is for particle rotating in circle where $$V$$ is the tangential velocity. Does this apply to a rigid body as well? Or am I missing something big?

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 $M$.
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.
• You can find the change in momentum, $\Delta \vec{p}$ of $m$ and divide by $\Delta t$. That would give you the average force. – Bill N Mar 23 '15 at 1:00