Let the potential energy be zero at the lowest point of the tube.

First, determine the speed of ball B ($v_B$) at the highest point using conservation of mechanical energy. Ball B moves from the bottom (speed $v_0$) to the top (height $2R$).

$$ \frac{1}{2}m_2v_0^2 = \frac{1}{2}m_2v_B^2 + m_2g(2R) $$ $$ v_B^2 = v_0^2 - 4gR $$

The net force on the tube is the vector sum of the forces exerted by ball A and ball B. By Newton's third law, the force exerted by a ball on the tube is equal and opposite to the force exerted by the tube on the ball. For the net force on the tube to be zero, the downward force from ball A must balance the upward force from ball B.

Force from ball A (at the bottom): The tube exerts an upward normal force $N_A$ on ball A. For circular motion at the bottom:

$$ N_A - m_1g = \frac{m_1v_0^2}{R} $$

The force ball A exerts on the tube is $F_A = N_A$ (downward).

$$ F_{A, \text{down}} = m_1g + \frac{m_1v_0^2}{R} $$

Force from ball B (at the top): The tube exerts a normal force $N_B$ on ball B. Assuming the force from the tube is upward, the net force providing centripetal acceleration (downward) is:

$$ m_2g - N_B = \frac{m_2v_B^2}{R} $$

The force ball B exerts on the tube is $F_B = N_B$ (upward).

$$ F_{B, \text{up}} = m_2g - \frac{m_2v_B^2}{R} $$

Substitute $v_B^2$:

$$ F_{B, \text{up}} = m_2g - \frac{m_2(v_0^2 - 4gR)}{R} = m_2g - \frac{m_2v_0^2}{R} + 4m_2g = 5m_2g - \frac{m_2v_0^2}{R} $$

For the net force on the tube to be zero, the magnitudes of the opposing forces must be equal:

$$ F_{A, \text{down}} = F_{B, \text{up}} $$ $$ m_1g + \frac{m_1v_0^2}{R} = 5m_2g - \frac{m_2v_0^2}{R} $$

Rearranging the terms to find the required relationship:

$$ \frac{m_1v_0^2}{R} + \frac{m_2v_0^2}{R} = 5m_2g - m_1g $$ $$ \frac{(m_1+m_2)v_0^2}{R} = (5m_2 - m_1)g $$

This is a valid relationship. An alternative form by grouping $m_1$ and $m_2$ is:

$$ (m_2 - m_1)v_0^2 = (m_1 + 5m_2)gR $$

Wait, let's recheck the derivation. My first derivation was: $-(m_1g + m_1v_0^2/R) + (m_2v_0^2/R - 5m_2g) = 0 \implies (m_2 - m_1)v_0^2/R = (m_1 + 5m_2)g$. My second derivation: $m_1g + m_1v_0^2/R = 5m_2g - m_2v_0^2/R \implies (m_1+m_2)v_0^2/R = (5m_2-m_1)g$. There is a sign error in the second derivation. Let's re-examine the force from ball B. Force on ball B at top: gravity $m_2g$ (down), normal force from tube $N_B$. Net force is centripetal, $m_2v_B^2/R$ (down). Let downward be positive.

$$ N_B + m_2g = \frac{m_2v_B^2}{R} $$

Here $N_B$ is the force from the tube, assumed positive if downward. If $N_B$ is negative, the force from the tube is upward. The force from the ball on the tube, $F_B$, is equal and opposite to $N_B$. So $F_B$ is upward if $N_B$ is negative.

$$ N_B = \frac{m_2v_B^2}{R} - m_2g $$

The upward force on the tube is $F_{B, \text{up}} = -N_B = m_2g - \frac{m_2v_B^2}{R}$. This matches the second derivation, which must be correct. Let's re-verify the final algebra.

$$ m_1g + \frac{m_1v_0^2}{R} = 5m_2g - \frac{m_2v_0^2}{R} $$ $$ \frac{m_1v_0^2}{R} + \frac{m_2v_0^2}{R} = 5m_2g - m_1g $$ $$ (m_1+m_2)v_0^2 = (5m_2 - m_1)gR $$

This is the correct relationship. My first derivation's vector setup was flawed. Let's rewrite the solution to be correct and clear.

To find the speed of ball B ($v_B$) at the highest point, we use conservation of mechanical energy, setting the potential energy at the bottom to zero.

$$ \frac{1}{2}m_2v_0^2 = \frac{1}{2}m_2v_B^2 + m_2g(2R) \implies v_B^2 = v_0^2 - 4gR $$

The net force on the tube is zero. This means the downward force from ball A must balance the upward force from ball B. By Newton's third law, the force a ball exerts on the tube is equal and opposite to the force the tube exerts on the ball.

Force from ball A (at the bottom): The tube must provide an upward normal force $N_A$ to support the weight $m_1g$ and provide the upward centripetal force.

$$ N_A - m_1g = \frac{m_1v_0^2}{R} $$

The force ball A exerts on the tube is $F_A = N_A$ directed downward.

$$ F_{A, \text{down}} = m_1g + \frac{m_1v_0^2}{R} $$

Force from ball B (at the top): Gravity $m_2g$ acts downward. The tube must exert an upward force $N_B$ if the required downward centripetal force is less than the weight. The net downward force is the centripetal force.

$$ m_2g - N_B = \frac{m_2v_B^2}{R} $$

The force ball B exerts on the tube is $F_B = N_B$ directed upward.

$$ F_{B, \text{up}} = m_2g - \frac{m_2v_B^2}{R} $$

Substitute $v_B^2 = v_0^2 - 4gR$:

$$ F_{B, \text{up}} = m_2g - \frac{m_2(v_0^2 - 4gR)}{R} = m_2g - \frac{m_2v_0^2}{R} + 4m_2g = 5m_2g - \frac{m_2v_0^2}{R} $$

For the net force on the tube to be zero, the magnitudes must be equal:

$$ F_{A, \text{down}} = F_{B, \text{up}} $$ $$ m_1g + \frac{m_1v_0^2}{R} = 5m_2g - \frac{m_2v_0^2}{R} $$ $$ \frac{m_1v_0^2}{R} + \frac{m_2v_0^2}{R} = 5m_2g - m_1g $$ $$ (m_1+m_2)v_0^2 = (5m_2 - m_1)gR $$