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There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$(which is how changes affect $v$), with $I_i+\tfrac{I_i^2}{m_ir_i^2}$(how changes affect $\omega$). For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass means more energy is going into accelerating the object than before, with less available for rotation?

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$(which is how changes affect $v$), with $I_i+\tfrac{I_i^2}{m_ir_i^2}$(how changes affect $\omega$). For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass mean more energy is going into accelerating the object than before, with less available for rotation?

Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A < v_B$. And a similar ratio, except based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$, how changes affect $v$, with $I_i+\tfrac{I_i^2}{m_ir_i^2}$, how changes affect $\omega$. For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass means more energy is going into accelerating the object than before, with less available for rotation?

In some sense $m$ appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). Similarly for $I$. This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence $r$ appearing once in each ratio, as $r^2$.

Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A < v_B$. And we have a similar radical and ratio, except this time based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$(which is how changes affect $v$), with $I_i+\tfrac{I_i^2}{m_ir_i^2}$(how changes affect $\omega$). For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass means more energy is going into accelerating the object than before, with less available for rotation?

In some sense $m$ appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). Similarly for $I$. This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence we have $r$ appearing only once in each ratio, as $r^2$.

For the same $ \omega$’s, kinetic energy would be higher for case A, so $I_A >I_B,E_A=E_B \implies \omega_A < \omega_B$. This is not surprising. It says the ball with lower moment of inertia ends up spinning faster. Specifically $$ \frac{\omega_A}{\omega_B} = \sqrt{ \frac{ \tfrac{I_B^2}{m_Br_B^2} + I_B}{\tfrac{I_A^2}{m_Ar_A^2} + I_A}} = \sqrt{ \frac{I_B^2 + I_B m r^2}{I_A^2 + I_A m r^2}} $$ When $m_A=m_B,r_A=r_B, I_A>I_B$.

Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A > v_B$. And a similar ratio, except based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$, how changes affect $v$, with $I_i+\tfrac{I_i^2}{m_ir_i^2}$, how changes affect $\omega$. For example, why would increasing mass increase $\omega$? Shouldn’t larger mass means more energy is going into accelerating the object than before, with less available for rotation?

In some sense $m$ (or $I$) appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence $r$ appearing once in each ratio, as $r^2$.

For the same $ \omega$’s, kinetic energy would be higher for case B, so $I_A <I_B,E_A=E_B \implies \omega_A > \omega_B$. This is not surprising. It says the ball with lower moment of inertia ends up spinning faster. Specifically $$ \frac{\omega_A}{\omega_B} = \sqrt{ \frac{ \tfrac{I_B^2}{m_Br_B^2} + I_B}{\tfrac{I_A^2}{m_Ar_A^2} + I_A}} = \sqrt{ \frac{I_B^2 + I_B m r^2}{I_A^2 + I_A m r^2}} $$ When $m_A=m_B,r_A=r_B, I_A<I_B$.

Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A < v_B$. And a similar ratio, except based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.

There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$, how changes affect $v$, with $I_i+\tfrac{I_i^2}{m_ir_i^2}$, how changes affect $\omega$. For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass means more energy is going into accelerating the object than before, with less available for rotation?

In some sense $m$ appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). Similarly for $I$. This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence $r$ appearing once in each ratio, as $r^2$.