A rope has a maximum tension of $220 \mathrm{~N}$ before it will break. An $11 \mathrm{~kg}$ ball attached to this rope sits on a horizontal, frictionless surface and is swung around in a horizontal circle with radius $0.80 \mathrm{~m}$ at constant speed using the maximum tension. If the ball is now replaced with a heavier $33 \mathrm{~kg}$ ball and again swung around at the same constant speed and the same maximum tension, what is the radius of the $33 \mathrm{~kg}$ ball's path? $0.80 \mathrm{~m}$ $0.27 \mathrm{~m}$ $0.89 \mathrm{~m}$ $2.4 \mathrm{~m}$

Step 1 ：Given that the tension in the rope provides the centripetal force necessary to keep the ball moving in a circle, the centripetal force is given by the equation \(F = m \cdot v^2 / r\), where \(m\) is the mass of the ball, \(v\) is the velocity of the ball, and \(r\) is the radius of the circle.

Step 2 ：In this case, the tension in the rope is equal to the centripetal force, so we have \(T = m \cdot v^2 / r\). We can rearrange this equation to solve for the radius: \(r = m \cdot v^2 / T\).

Step 3 ：Since the tension and the speed of the ball are the same in both cases, we can set up a proportion to find the radius of the path of the heavier ball: \(r_1 / r_2 = m_1 / m_2\), where \(r_1\) and \(m_1\) are the radius and mass of the path of the lighter ball, and \(r_2\) and \(m_2\) are the radius and mass of the path of the heavier ball.

Step 4 ：Given that the mass of the lighter ball \(m_1\) is 11 kg, the mass of the heavier ball \(m_2\) is 33 kg, and the radius of the path of the lighter ball \(r_1\) is 0.8 m, we can substitute these values into the proportion to find \(r_2\).

Step 5 ：\(r_2 = r_1 \cdot m_2 / m_1 = 0.8 \cdot 33 / 11 = 0.26666666666666666\) m

Step 6 ：\(\boxed{0.27 \, m}\) is the radius of the path of the heavier ball.