What is the velocity on perimeter of a wheel when it is hit by hand tangentially to increase speed?

Question

This is a homework problem in Newtonian mechanics. The problem statement is

A circular wheel (whose radius is $R$ and moment of inertia is $I$) is hit by a hand of mass $m$ repeatedly with a velocity of $v$ (in rest reference frame) tangentially on its circumference so that it speeds up in the direction of rotation. It is hit in such a way that the hand loses all of it's momentum after impact wrt. the reference frame of the point of impact.

The hand is moving along with the elbow, neglecting rotation of forearm. Considering the hand to be a point mass, what is the velocity of the perimeter of the wheel after n hits? [Hint: Solve the recurrence relation]

The whole situation sounds so complicated that I don't know even where to start or how to approach such a problem. Can anyone say how to approach this? Also, does anyone here know the original source (the book where it originated first) of this seemingly hard problem?

Answer 1

From what I can tell, each time the hand hits the wheel its speed in the lab frame is reduced from $v$ to $v_{i}$, where I'll call $v_{i}$ the speed of the wheel in the lab frame after $i$ hits.

So on the $(i+1)^{th}$ hit, the change in momentum of the hand is $\Delta p = m(v_{i}-v)$. The average force applied to the wheel assuming the collision takes place over an interval of time $\Delta t$ then equals $F=\frac{m(v-v_{i})}{\Delta t}$. The average torque acting on the wheel during this interval is consequently $\tau=\frac{mR(v-v_{i})}{\Delta t}$.

Hence the angular impulse equals $mr(v-v_{i})$, which can be set equal to the change in angular momentum $\Delta L$,

$mR(v-v_{i}) = I\Delta{\omega} = \frac{I}{R}\Delta{v}$. This yields the linear recurrence relation

$v_{i+1} = v_{i} + \frac{mR^{2}}{I}(v-v_{i})$

which may then be solved by trialling $v_{i} = A\lambda^{i}$ and finding the general solution. If I have made any errors in the above please let me know!