# 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 $M$.

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.

• Thank you very much for the clarification. I was also not so sure about what I claimed, that is why it started with a "can". Anyways. Supposing that there is no compression involved and we have a time recording of the collision with some sensors or camera, which happens to be delta t, how can this force be computed? The rotating rigid body is large enough that the point mass m, faces a flat surface.
– man
Mar 21, 2015 at 21:27
• You can find the change in momentum, $\Delta \vec{p}$ of $m$ and divide by $\Delta t$. That would give you the average force. Mar 23, 2015 at 1:00