Question

In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder.

Answer 1

Answer:

K.E = 1490.73 J

Explanation:

Given:

Speed of the ball (V)= 139 km/h = 38.61 m/s

Moment of inertia, I = 0.720 kg-m²

Distance, r = 0.600 m

Angular velocity (ω) = $\frac{V}{r}$

(ω) = $\frac{38.61}{0.600}$ =64.35 rad/sec

The kinetic energy (K.E) is given as

K.E = $\frac{1}{2}\times I\omega^{2}$

substituting the values in the above equation we get

K.E = $\frac{1}{2}\times 0.720\times 64.35^{2}$

K.E = 1490.73 J