The component $M_{rr}$ represents the effective generalized mass or radial inertia associated with the generalized velocity $\dot{r}$ (the radial speed of mass $m_1$ and the vertical speed of mass $m_2$). The formula for the component is: $M_{rr} = \frac{\partial^2 T}{\partial \dot{r}^2} = m_1 + m_2$
Here is the explanation for why this result is simply the sum of the two masses:
The total kinetic energy of the system is the sum of the kinetic energies of the two masses, grouped by their associated generalized velocities ($\dot{r}$ and $\dot{\varphi}$):
$$ T = \frac{1}{2} (m_1 + m_2) \dot{r}^2 + \frac{1}{2} (m_1 r^2) \dot{\varphi}^2 $$
To find $M_{rr}$, we first take the partial derivative of $T$ with respect to $\dot{r}$:
$$ \frac{\partial T}{\partial \dot{r}} = \frac{\partial}{\partial \dot{r}} \left[ \frac{1}{2} (m_1 + m_2) \dot{r}^2 + \frac{1}{2} (m_1 r^2) \dot{\varphi}^2 \right] $$
The first term, $\frac{1}{2} (m_1 + m_2) \dot{r}^2$, differentiates to $(m_1 + m_2)\dot{r}$.
The second term, $\frac{1}{2} (m_1 r^2) \dot{\varphi}^2$, is treated as a constant with respect to $\dot{r}$, so its derivative is 0.
$$ \frac{\partial T}{\partial \dot{r}} = (m_1 + m_2) \dot{r} $$
Next, we take the partial derivative of the result with respect to $\dot{r}$ again:
$$ M_{rr} = \frac{\partial^2 T}{\partial \dot{r}^2} = \frac{\partial}{\partial \dot{r}} \left[ (m_1 + m_2) \dot{r} \right] $$
Since ($m_1 + m_2$) is a constant, the derivative is simply:
$$ M_{rr} = m_1 + m_2 $$
The result $M_{rr} = m_1 + m_2$ has a direct physical meaning:
Since the velocity of both masses is directly dependent on the $\dot{r}$ generalized velocity, and the velocity is linearly proportional to $\dot{r}$, their inertias are fully additive in the radial direction. The generalized inertia for motion in the $r$ direction is simply the total mass of the system.