Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
[tex]\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}[/tex]For the maximum velocity of the ball at the top of the vertical circular motion,
[tex]v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}[/tex]Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
