Chapter 7, Problem 49AP

One method of pitching a softball is called the “wind-mill” delivery method, in which the pitcher’s arm rotates through approximately 360° in a vertical plane before the 198-gram ball is released at the lowest point of the circular motion. An experienced pitcher can throw a ball with a speed of 98.0 mi/h. Assume the angular acceleration is uniform throughout the pitching motion and take the distance between the softball and the shoulder joint to be 74.2 cm. (a) Determine the angular speed of the arm in rev/s at the instant of release, (b) Find the value of the angular acceleration in rev/s² and the radial and tangential acceleration of the ball just before it is released, (c) Determine the force exerted on the ball by the pitcher’s hand (both radial and tangential components) just before it is released.

To determine

The angular speed of the arm in rev/s at the instant of release.

Answer to Problem 49AP

The angular speed of the arm at the instant of release is $9.4\text{rev/s}$.

Explanation of Solution

Given Info:

The speed of the ball is $98.0\text{mi/h}$.

The distance between the soft ball and the shoulder is $74.2\text{cm}$.

Explanation:

Formula to calculate the angular speed of the arm is,

$\omega =\frac{v}{r}$

• $v$ is the speed of the soft ball
• $r$ is distance between the soft ball and the shoulder

Substitute $98.0\text{mi/h}$ for $v$ and $74.2\text{cm}$ for r to find the angular speed of the arm at the instant of release,

$\begin{array}{c}\omega =\frac{98.0\text{mi/h}}{74.2\times {10}^{-2}\text{m}}\left(\frac{0.447\text{m/s}}{1\text{mi/h}}\right)\left(\frac{1\text{rev}}{2\pi \text{rad}}\right)\\ =9.4\text{rev/s}\end{array}$

Thus, the angular speed of the arm at the instant of release is $9.4\text{rev/s}$.

Conclusion:

The angular speed of the arm at the instant of release is $9.4\text{rev/s}$.

To determine

The angular acceleration in ${\text{rev/s}}^{\text{2}}$, radial and tangential acceleration of the ball just before it is released.

Answer to Problem 49AP

The tangential acceleration of the ball just before it is released is $206{\text{m/s}}^2$.

The radial acceleration of the ball just before it is released is $2.59\times {10}^3{\text{m/s}}^{\text{2}}$.

The angular acceleration is $44.2{\text{rev/s}}^{\text{2}}$.

Explanation of Solution

Given Info:

The angular speed of the arm at the instant of release is $9.4\text{rev/s}$. The speed of the ball is $98.0\text{mi/h}$. The distance between the soft ball and the shoulder is $74.2\text{cm}$, The speed of the ball is $98.0\text{mi/h}$. The distance between the soft ball and the shoulder is $74.2\text{cm}$, The angular acceleration is $44.2{\text{rev/s}}^{\text{2}}$. The distance between the soft ball and the shoulder is $74.2\text{cm}$,

Explanation:

Formula to calculate the angular acceleration is,

$\alpha =\frac{{\omega }^2-{\omega }_i{}^2}{2\Delta \theta }$

• ${\omega }_i$ is the initial angular speed of the arm

Since, the arm is at rest initially; the initial angular speed is zero.

Substitute $9.4\text{rev/s}$ for $\omega$ and $zero$ for ${\omega }_i$ and $1\text{rev}$ for $\Delta \theta$ to find the angular acceleration.

$\begin{array}{c}\alpha =\frac{{\left(9.4\text{rev/s}\right)}^2-0}{2\left(1\text{rev}\right)}\\ =44.2{\text{rev/s}}^{\text{2}}\end{array}$

Thus, the angular acceleration is $44.2{\text{rev/s}}^{\text{2}}$.

Formula to calculate the angular speed of the arm is,

$a_c=\frac{v^2}{r}$

• $v$ is the speed of the soft ball
• $r$ is distance between the soft ball and the shoulder

Substitute $98.0\text{mi/h}$ for $v$ and $74.2\text{cm}$ for r to find the radial acceleration,

$\begin{array}{c}{a}_c=\frac{{\left(\left(98.0\text{mi/h}\right)\left(\frac{0.447\text{m/s}}{1\text{mi/h}}\right)\right)}^2}{74.2\times {10}^{-2}\text{m}}\\ =2.59\times {10}^3{\text{m/s}}^{\text{2}}\end{array}$

Thus, the radial acceleration of the ball just before it is released is $2.59\times {10}^3{\text{m/s}}^{\text{2}}$.

Formula to calculate the tangential acceleration is,

$a_c=r\alpha$

• $\alpha$ is the angular acceleration of the ball
• $r$ is distance between the soft ball and the shoulder

Substitute $44.2{\text{rev/s}}^{\text{2}}$ for $\alpha$ and $74.2\text{cm}$ for rto find the tangential acceleration,

$\begin{array}{c}{a}_c=\left(72.4\times {10}^{-2}\text{m}\right)\left(\left(44.2{\text{rev/s}}^{\text{2}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\right)\\ =206{\text{m/s}}^2\end{array}$

Thus, the tangential acceleration of the ball just before it is released is $206{\text{m/s}}^2$.

Conclusion:

The tangential acceleration of the ball just before it is released is $206{\text{m/s}}^2$.

The radial acceleration of the ball just before it is released is $2.59\times {10}^3{\text{m/s}}^{\text{2}}$.

The angular acceleration is $44.2{\text{rev/s}}^{\text{2}}$.

To determine

Both radial and tangential components of the force exerted on the ball by the pitcher’s hand.

Answer to Problem 49AP

The radial component of the force exerted on the ball is $515\text{N}$.

The tangential component of the force exerted on the ball is $40.8\text{N}$.

Explanation of Solution

Given Info:

The radial acceleration of the ball just before it is released is $2.59\times {10}^3{\text{m/s}}^{\text{2}}$. The mass of the ball is $0.198\text{kg}$, The tangential acceleration of the ball just before it is released is $206{\text{m/s}}^2$, The mass of the ball is $0.198\text{kg}$.

Explanation:

According to the release point, the weight of the ball and the centripetal acceleration is acting in the radial direction.

Thus,

Formula to calculate the radial component of the force is,

$F_r=mg+m{a}_c$

Substitute $0.198\text{kg}$ for m, $9.8{\text{m/s}}^{\text{2}}$ for g and $2.59\times {10}^3{\text{m/s}}^{\text{2}}$ for $a_c$ to find the radial component o the force,

$\begin{array}{c}{F}_r=\left(0.198\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)+\left(0.198\text{kg}\right)\left(2.59\times {10}^3{\text{m/s}}^{\text{2}}\right)\\ =515\text{N}\end{array}$

Thus, the radial component of the force exerted on the ball is $515\text{N}$.

According to the release point, only tangential acceleration will be there in the tangential direction,

Thus,

Formula to calculate the tangential component of the force is,

$F_t=m{a}_t$

Substitute $0.198\text{kg}$ for m, $206{\text{m/s}}^2$ for $a_t$ to find the tangential component of the force.

$\begin{array}{c}{F}_t=\left(0.198\text{kg}\right)\left(206{\text{m/s}}^2\right)\\ =40.8\text{N}\end{array}$

Thus, the tangential component of the force exerted on the ball is $40.8\text{N}$.

Conclusion:

The radial component of the force exerted on the ball is $515\text{N}$.

The tangential component of the force exerted on the ball is $40.8\text{N}$.