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

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

`v_(\max )=\sqrt[]{(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*9.8)}{0.31}} \\ v_(\max )=\sqrt[]{(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.