Question

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tension of Tmax = 7.5 N before it will break. What isthe maximum possible speed of the ball at the top of the loop, in meters per second?

Answer 1

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,

$\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}$

For the maximum velocity of the ball at the top of the vertical circular motion,

$v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}$

where g is the acceleration due to gravity,

Substituting the known values,

$\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}$

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.