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