Question

In fast-pitch softball, a pitcher swings her arm from straight overhead in a circle, releasing the 0.196 kg ball at the bottom of the swing. If a pitcher's arm has a length of 0.984 m and mass of 11.3 kg, and the ball goes from initially at rest to a tangential speed of 29.8 m/s before release, what average torque must the pitcher apply to the ball? Model the pitcher's arm as a uniform thin rod swung about one end, and make sure you don't forget the mass of the softball (but you can ignore its shape; i.e., treat it like a point mass). Hint: you will need to make use of the equations of circular motion for constant angular acceleration that we derived back in week 2.

Answer 1

Answer:

560.06714 Nm

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = $\pi$ (Half rotation)

v = Velocity of bat = 29.8 m/s

M = Mass of bat = 11.3 kg

m = Mass of ball = 0.196 kg

R = Radius of swing = 0.984 m

$\omega_f=\dfrac{v}{r}\\\Rightarrow \omega_f=\dfrac{29.8}{0.984}\\\Rightarrow \omega_f=30.28455\ rad/s$

From equation of rotatational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{30.28455^2-0^2}{2\times \pi}\\\Rightarrow \alpha=145.96958\ rad/s^2$

Moment of inertia is given by

$I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm$

The torque the pitcher applies is 560.06714 Nm

Answer 2

The average torque this pitcher must apply to the softball is 560.64 Newton.

Given the following data:

• Mass of ball = 0.196 kg.

• Mass of pitcher's arm = 11.3 kg.

• Length of pitcher's arm (radius) = 0.984 m.

• Initial speed = 0 m/s (since it starts from rest).

• Tangential speed = 29.8 m/s.

To determine the average torque this pitcher must apply to the ball:

How to calculate the average torque.

First of all, we would determine the final angular speed of the ball by using this formula:

$\omega_f =\frac{v}{r} \\\\\omega_f =\frac{29.8}{0.984} \\\\\omega_f = 30.29\;rad/s$

Next, we would determine the constant angular acceleration by using the equation of circular motion:

$\alpha =\frac{\omega^2_f - \omega^2_i}{2\theta} \\\\\alpha =\frac{30.29^2 - 0^2}{2\times \pi} \\\\\alpha =\frac{917.48}{2\times 3.142} \\\\\alpha =\frac{917.48}{6.284}\\\\\alpha = 146\;rad/s^2$

For the moment of inertia:

For a point mass, moment of inertia is given by this formula:

$I =\frac{1}{3} Mr^2 + mr^2\\\\I =\frac{1}{3} \times 11.3 \times 0.984^2 + 0.196 \times 0.984^2\\\\I = 3.84 \;Kgm^2$

Now, we can determine the average torque:

$Torque = I\alpha \\\\Torque =3.84 \times 146$

Torque = 560.64 Newton.

Read more on torque here: https://brainly.com/question/14839816