A good baseball pitcher can throw a baseball toward home plate at ${}^{\mathbf{85}\mathbf{}\mathbf{mi}\mathbf{/}\mathbf{h}}$with a spin${}^{\mathbf{1800}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{min}}$. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the${}^{\mathbf{60}\mathbf{}\mathbf{ft}}$ path is a straight line.

The velocity of the baseball,${}^{\mathrm{v}=85\mathrm{mi}/\mathrm{h}}$

The straight path covered by the baseball,${}^{∆\mathrm{x}=60\mathrm{ft}}$

The angular velocity of the baseball,${}^{\mathrm{\omega }=1800\mathrm{rev}/\min }$

Angular velocity represents the rate at which an object rotates about an axis. Angular velocity is a vector quantity, with direction given by right hand rule.

The expression for linear velocity is given as:

$v=\frac{∆x}{∆t}$ … (i)

Here,${}^{∆x}$is the linear displacement and${}^{∆t}$is the time interval.

The expression for angular velocity is given as:

localid="1654330564056" $\omega =\frac{∆\theta }{∆t}$ … (ii)

Here,${}^{∆\theta }$is the angular displacement.

Convert the velocity from mi/h to ft/min.

$$\begin{gathered} \mathrm{v}=85\mathrm{mi}/\mathrm{h}\left(\frac{5280\mathrm{ft}}{1{\mathrm{m}}_{\mathrm{i}}}\right)\left(\frac{1\mathrm{h}}{60\min }\right) \\ =7480\mathrm{ft}/\min \end{gathered}$$

Using equation (i), the time interval is calculated as:

$$\begin{gathered} ∆\mathrm{t}=\frac{∆\mathrm{x}}{\mathrm{v}} \\ =\frac{60\mathrm{ft}}{7480\mathrm{ft}/\min } \\ =0.008\min \end{gathered}$$

Thus, the time taken to reach home plate is$0.008\min$.

Using equation (ii), the number of revolutions can be calculated as:

$$\begin{gathered} ∆\mathrm{\theta }=\mathrm{\omega }\times ∆\mathrm{t} \\ =1800\mathrm{rev}/\min \times 0.008\min \\ =14.4\mathrm{rev} \\ \approx 14\mathrm{rev} \end{gathered}$$

Thus, the baseball makes 14 revolutions on its way to home plate.