Question

Why is the following situation impossible? A softball pitcher has a strange technique: she begins with her hand at rest at the highest point she can reach and then quickly rotates her arm backward so that the ball moves through a half-circle path. She releases the ball when her hand reaches the bottom of the path. The pitcher maintains a component of force on the 0.180-kg ball of constant magnitude 12.0 N in the direction of motion around the complete path. As the ball arrives at the bottom of the path, it leaves her hand with a speed of 25.0 m/s

Short answer

We find that her arms would need to be 1.36 m long to perform this task. This is significantly longer than the human arm.

Step-by-step solution

Law of Energy conservation

The Law of conservation of energy states that, for an isolated system working in ideal conditions, the decrease in potential energy of the body will be equal to the increase in kinetic energy of the body

$\mathbf{∆}{\mathit{E}}_{\mathbf{s}\mathbf{y}\mathbf{s}\mathbf{t}\mathbf{e}\mathbf{m}}\mathbf{=}\mathbf{0}$

which can be written as:

$\mathbf{∆}\mathit{K}\mathbf{+}\mathbf{∆}\mathit{U}\mathbf{=}\mathbf{0}$

The potential energy of the system is:

$\mathit{U}\mathbf{=}\mathit{m}\mathit{g}\mathit{R}$

The kinetic energy of the system is:

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

where

U=Potential energy

K= Kinetic energy

v = Speed of body

g = Gravitational acceleration

R = Displacement in meter

m = Mass of body.

Given Data

Mass of the ball: m=0.180Kg

Force on the ball: F=12.0N

Initial speed of the ball: v=25m/s

Calculation

The distance traveled by the ball from the top of the arc to the bottom is πR. The change in internal energy of the system due to the nonconservative force, the force exerted by the pitcher, is

$\Delta E=Fd\cos 0^o=F\left(\pi R\right)$

We shall assign the gravitational energy of the ball to be zero at the bottom of the arc.

Then

$\Delta E_{mech}=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+mg{y}_f-mg{y}_i$

Becomes,

$\begin{array}{c}\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2+mg{y}_i+F\left(\pi R\right)=\frac{1}{2}m{v}_i^2+mg2R+F\left(\pi R\right)\\ \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+\left(2mg+\pi F\right)R\end{array}$

Solve for R, which is the length of her arms.

$\begin{array}{l}R=\frac{\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2}{2mg+\pi F}\\ R=m\frac{v_f^2-v_i^2}{4mg+2\pi F}\end{array}$

For the given values, the above equation becomes

$\begin{array}{c}R=\left(0.180kg\right)\frac{{\left(25.0m/s\right)}^2-0}{4\left(0.180kg\right)\left(9.8m/s^2\right)+2\pi \left(12.0N\right)}\\ =1.36m\end{array}$

We find that her arms would need to be 1.36 m long to perform this task. This is significantly longer than the human arm. Therefore, it is not possible.