There is a bicycle on a bicycle stand. We know that its rear wheel’s moment of inertia around its axis is I = 0.5 ${\mathrm{kgm}}^2$ and its radius is r = 30 cm. We speed up the wheel by repeatedly hitting it with a hand tangentially on its circumference in the direction of its rotation. The speed of each hit in the rest reference frame is v = 5 ${\mathrm{ms}}^{-1}$, the mass of the hand is m = 1.5 kg, the hand loses all its velocity in the reference frame of the point of impact after the impact (the hand is moving around an elbow, but we neglect rotation of the forearm). If velocity of a point on the perimeter of the wheel after 10 hits be ${\mathrm{v}}_{\mathrm{n}}=5\left(1-{\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)}^{10}\right)$ then value of $\frac{9\mathrm{\alpha }}{\mathrm{\beta }}$ to the nearest integer

The hand is a point mass which has a moment of inertia with respect to the impact. The angular momentum transferred from the hand to the when in the n-th hit (after the hit, the hand is at rest with respect to the point of impact) in $\mathrm{mr}\left(\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right)$ where v_n is the velocity of th point of impact after the n-th hit. The angular momentum of the wheel after the n-th hit is therefore.

 ${\mathrm{L}}_{\mathrm{n}}=\mathrm{mr}\left(\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}\right)+{\mathrm{L}}_{\mathrm{n}-1}$         

That give a recurrent formula for the velocities of the point of impact based on the formula ${\mathrm{L}}_{\mathrm{n}}={\mathrm{I\omega }}_{\mathrm{n}}$.

${\mathrm{Av}}_{\mathrm{n}}=\mathrm{w}-{\mathrm{v}}_{\mathrm{n}}+{\mathrm{Av}}_{\mathrm{n}-1}$

Where we used the substitution $\mathrm{A}=\mathrm{I}/\left({\mathrm{mr}}^2\right)$

Now, we will try to compute the first few terms of the progression. We get

${\mathrm{v}}_1=\frac{\mathrm{W}}{1+\mathrm{A}},{\mathrm{v}}_2=\frac{\mathrm{W}}{1+\mathrm{A}}\left(1+\frac{\mathrm{A}}{1+\mathrm{A}}\right),{\mathrm{v}}_3=\frac{\mathrm{W}}{1+\mathrm{A}}\left(1+\frac{\mathrm{A}}{1+\mathrm{A}}+{\left(\frac{\mathrm{A}}{1+\mathrm{A}}\right)}^2\right),\dots \dots$

The formula for v_n will clearly be

${\mathrm{v}}_{\mathrm{n}}=\frac{\mathrm{W}}{1+\mathrm{A}}\left(1+\frac{\mathrm{A}}{1+\mathrm{A}}+\dots ..+{\left(\frac{\mathrm{A}}{1+\mathrm{A}}\right)}^{\mathrm{n}-1}\right)$

Summing up a geometric series and simplifying, we obtain

${\mathrm{v}}_{\mathrm{n}}=\mathrm{W}\left(1-{\left(\frac{\mathrm{A}}{1+\mathrm{A}}\right)}^{\mathrm{n}}\right)$

The velocity is ${\mathrm{v}}_{10}=4.54 {\mathrm{ms}}^{-1}$