Gummed up hula-hoop

Hey y'all, here is my solution which works for all values of $m$ , $M$ :

There are no external torques on the system, so the angular momentum of the hoop and bead is conserved.

It is not too hard to calculate the angular momentum of the hoop and bead in the center of mass frame, before the collision. The center of mass of the system moves at a constant velocity along $\hat{x}$ with velocity $\displaystyle v_{com} = v_o\frac{m}{m+M}$ . The vertical position of the center of mass is similarly given by $\displaystyle y_{com} = R\frac{m}{m+M}$ .

In this frame, the velocity of the bead relative to the center of mass is $\displaystyle v_o-v_{com}$ and the velocity of the hoop is $\displaystyle v_{com}$ . The perpendicular distance of the bead to the center of mass is given by $\displaystyle R-y_{com}$ and the perpendicular distance of the center of the hoop to the center of mass of the system is given by $\displaystyle y_{com}$ . The angular momentum of the system is then

$\begin{aligned} L&=m\left(R-y_{com}\right)\left(v_o-v_{com}\right) + My_{com}v_{com} \\ &=m\frac{M}{m+M}\frac{M}{m+M}v_oR+M\frac{m}{m+M}\frac{m}{m+M}v_oR \\ &= v_oR\frac{mM}{m+M} \end{aligned}$

Now we know $L$ which is equal to $I\omega$ . All that remains is to calculate the moment of the inertia of the hoop and bead about the center of mass.

The moment of inertia is an integral quantity, and we can add up the moments of the bead and hoop. The bead is easy, it is located $\displaystyle R-y_{com}$ from the center of mass so its moment is $\displaystyle m\left(R-y_{com}\right)^2$ .

We can work out the moment of inertia for the hoop with the parallel axis theorem. Any object of mass $M$ with moment of inertia $I$ about its own center of mass has moment of inertia $\displaystyle I+Md^2$ about a parallel axis located $d$ from its center of mass. Therefore, the moment of inertia of the hoop about the center of mass is given by $\displaystyle MR^2+My_{com}^2$ .

Altogether:

$\begin{aligned} I &= I_{hoop} + I_{bead} \\ &=MR^2+MR^2\left(\frac{m}{m+M}\right)^2+mR^2\left(\frac{M}{m+M}\right)^2 \\ &=\frac{MR^2}{\left(m+M\right)^2}\left( \left(m+M\right)^2+m^2+mM \right) \\ &=\frac{MR^2}{\left(m+M\right)^2}\left(m^2+2Mm+M^2+m^2+mM\right) \\ &=\frac{MR^2}{\left(m+M\right)^2}\left(2m^2+3Mm+M^2\right) \end{aligned}$

Now, $\displaystyle\omega = \frac{L}{I}$

$\displaystyle\omega = v_oR \frac{mM}{m+M}\frac{\left(m+M\right)^2}{MR^2}\frac{1}{2m^2+3Mm+M^2}$

And just like that $\boxed{\displaystyle\omega = \frac{v_o}{R}\frac{m\left(M+m\right)}{2m^2+3mM+M^2}}$

I used conservation of angular momentum to do this problem, and I ended up with

but I ended up with $w = \frac {1000mv_0}{m+M} = 73.89$ which rounds up to 74.

This solution works because the gumball mass is small compared to the hula hoop, but it does not work in general.

Can you think of how things change when the mass of the gumball is on the same order of magnitude as the mass of the hula hoop?

it is the right answer however. If the gumball is assumed to be a point mass, then answer is 74. But, conserving angular momentum about center of mass of the system is justified.